Class 12 Commerce Applied Mathematics - Probability Distribution And Its Mean And Variance

The probability mass function (p.m.f) of $X$ is given below:
$X=x$ $1$ $2$ $3$
$P(X=x)$ $\cfrac{1}{5}$ $\cfrac{2}{5}$ $\cfrac{2}{5}$

We know,

$E({ X }^{ 2 })=\sum { { X }^{ 2 } } P(X)$
Therefore,

$E({ X }^{ 2 })={ 1 }^{ 2 }\times \cfrac { 1 }{ 5 } +{ 2 }^{ 2 }\times \cfrac { 2 }{ 5 } +{ 3 }^{ 2 }\times \cfrac { 2 }{ 5 }$

$=\cfrac { 1 }{ 5 } +\cfrac { 8 }{ 5 } +\cfrac { 18 }{ 5 } =\cfrac { 27 }{ 5 }$

The mean and variance of a Binomial distribution is $4$ and $3$ respectively. Find its parameters.

Fact: In binomial distribution, mean

$=np$ and variance

$=npq$

Here given

$np=4$ and

$npq=3$

$q=\dfrac{3}{4}\Rightarrow p=1-p=\dfrac{1}{4}$

$np = \cfrac14$

$\Rightarrow n\left(\cfrac14\right) = 4$

$\Rightarrow n = 16$

The random variable x has a probability distribution $P(x)$ of the following form where $k$ is some number:
$P(x)=\begin{cases}k & if\, x=0\\2k&if\, x=1 \\3k&if\, x=2\\0&otherwise \end{cases}$
Determine the value of $k$ and $P(X \le 2).$

$\sum _{ x=0 }^{ \infty }{ P(x)=1 }$

$P(1)+P(2)+......=1$

$k+2k+3k=1$

$k=\dfrac { 1 }{ 6 }$

$P(x\le 2)=P(0)+P(1)+P(2)=k+2k+3k=6k=1$

Find the variance and standard deviation of the random variable $X$ whose probability distribution is given below:
$x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\dfrac {1}{8}$ $\dfrac {3}{8}$ $\dfrac {3}{8}$ $\dfrac {1}{8}$

Let mean of

$X$ be

$\mu$

$\therefore \mu=\sum_{x=0}^3 X\times P(X)=\dfrac{12}{8}=\dfrac{3}{2}$

$\therefore \text{Var}(X)=\sigma^2=\sum_{x=0}^3 X^2\times P(X)-\mu^2 =\dfrac{24}{8}-\dfrac{9}{4}=\dfrac{3}{4}$

$\therefore \text{Standard Deviation}=\sigma=\sqrt{\text{Var}(X)}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}$

These are the required answers.

The probability that a doctor successfully performs an operation is $80\%$ . What is the value of $625$ times the probability that at least $3$ operations out of $4$ conducted by him will be successful?

The probability that a doctor successfully perform an operation is 80% =

$\cfrac { 4 }{ 5 }$

Case(1): All 4 operations are successful

Probability =

${ \left( \cfrac { 4 }{ 5 } \right) }^{ 4 }$

Case(2): One of the 4 operations is unsuccessful

Probability

$=\quad _{ 1 }^{ 4 }{ C }.{ \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }\left( \cfrac { 1 }{ 5 } \right)$

$\therefore$ Total Probability

$=\quad { \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }\left( \cfrac { 8 }{ 5 } \right)$

$=\quad { \left( \cfrac { 4 }{ 5 } \right) }^{ 3 }\left( \cfrac { 8 }{ 5 } \right) \\ \cong \quad 81.92%$

Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Find the Probability that both are king.

No .of favourable outcomes

$4$

No.of possible out comes

$52$

Probabilty that it is a king is

$\dfrac{4}{52} * \dfrac{3}{52}=\dfrac{3}{676}$

Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Find the probability distribution of number of red balls drawn.

$5$ white

$4$ red

$P(red)=\cfrac { red }{ red+white } =\cfrac { 4 }{ 9 }$

Find $k$ such that the function
$P(x)=\quad \begin{cases} k\begin{pmatrix} 4 \\ x \end{pmatrix};\quad x=0,1,2,3,4,k>0 \\ 0;\quad otherwise \end{cases}$ is a probability mass function.

As sum of all probabilities

$=1$

$\Rightarrow R({^{4}C_0}+{^{4}C_1}+{^{4}C_2}+{^{4}C_3}+{^{4}C_4})=1$

$=R(1+4+6+4+1)=1$

$\Rightarrow R(16)=1$

$\Rightarrow R=\dfrac{1}{16}$ .

1114475_1195282_ans_70b8f54aa26943aaa796a84af79d034f.jpg

Given $X \sim B(n, p)$ , If $E(X)=6, \text{ Var. }(X)=4.2$ , find $n$ and $p$ .

We have,

$E(x)=6$

$np=6$

$Var.(X)=4.2$

$npq=4.2$

Therefore,

$\dfrac{npq}{np}=\dfrac{4.2}{6}$

$q=0.7$

$p=1-q=0.3$

Therefore,

$n=\dfrac{6}{p}=\dfrac{6}{0.3}=20$

Hence, this is the answer.

In a dice game , a player pays a stake of Re $1$ for each throw of a die. She receives Rs. $5$ if the die shows a $3$ , Rs $2$ if the die shows a $1$ or $6$ , and nothing otherwise . What is the player’s expected profit per throw over a long series of throws ?

X	$-1$	$5$	$2$
$P(X)$	$\cfrac{1}{2}\\(either 4,5)$	$\cfrac{1}{6}\\(only \quad 3)$	$\cfrac{1}{3}\\ (either\quad 1 \quad or\quad 6)$

From the above table

$\sum P(X)=-\cfrac{1}{2}+\cfrac{5}{6}+\cfrac{2}{3}\\ \quad\quad =Rs1$

A random variable X has the following probability distribution
X 0 1 2 3 4
P(X) 0 k 2k 3k 1/2
Determine P(0 < X < 3)

Since,

$P(X)=1$

$k+2k+3k+\dfrac{1}{2}=1$

$6k=\dfrac{1}{2}$

$k=\dfrac{1}{12}$

Now,

$P(0<X<3)=P(1)+P(2)$

$=k+2k=3k=3\times \dfrac{1}{12}=\dfrac{1}{4}$

A random variable has the following probability distribution :
Find P

1306788_1466792_ans_49250299a43c4febb801684109392119.jpeg

Given below is the probability distribution of $X$ :
$x$ $0$ $1$ $2$ $3$ $4$
$P[X=x]$ $k$ $2k$ $4k$ $2k$ $k$
Find:
$P(X\ge 2), P(X < 3), P(X\le 1)$ .

1303399_1367564_ans_82aa2e6ebe404d169c9253131dbf4a54.jpeg

A purse contains four coins each of which is either a rupee or two rupees coin. Find the expected value of a coin in that purse.

Various possibilities of coins in the purse can be

	$p_{i}$	$M_{i}$	$p_{i}M_{i}$
(i) 4 1 rupee coins	$\displaystyle \frac{1}{16}$	4	$\displaystyle \frac{4}{16}$
(ii) 3 one Rs.+1 two Rs.	$\displaystyle \frac{4}{16}$	5	$\displaystyle \frac{20}{16}$
(iii) 2 one Rs.+2 two Rs.	$\displaystyle \frac{6}{16}$	6	$\displaystyle \frac{36}{16}$
(iv) 1 one Rs.+3 two Rs.	$\displaystyle \frac{4}{16}$	7	$\displaystyle \frac{28}{16}$
(iv) 4 two Rs.	$\displaystyle \frac{1}{16}$	8	$\displaystyle \frac{8}{16}$
			6 / -

Hence expected value is Rs,

$6 / -$

A pair of dice is thrown $5$ times. If getting a doublet is considered as a success, then find the mean and variance of successes.

In a single throw of a pair of dice, probability of getting a doublet

$\displaystyle =\frac{1}{6}$
considering it to be a success,

$\displaystyle p=\frac{1}{6}$

$\therefore$

$\displaystyle q=1-\frac{1}{6}=\frac{5}{6}$
mean

$\displaystyle =5\times \frac{1}{6}=\frac{5}{6},$ variance

$\displaystyle =5\times \frac{1}{6}.\frac{5}{6}=\frac{25}{36}$

There are $100$ tickets in a raffle (Lottery). There is $1$ prize each of Rs. $1000/-,$ Rs. $500/-$ and Rs. $200/-.$ Remaining tickets are blank. Find the expected price of one such ticket.

Expectation

$=\sum p_{i}M_{i}$ outcome of a ticket can be

$\displaystyle \begin{matrix} & p_{i} & M_{i} & p_{i}M_{i}\\ (i)\: I\: prize & \cfrac{1}{100} & 1000 & 10\\ (ii)\: II\: prize & \cfrac{1}{100} & 500 & 5\\ \ (iii)\: III\: prize & \cfrac{1}{100} & 200 & 2\\ (iv)\: Blank & \cfrac{97}{100} & 0 & 0\\ & & & \sum p_{i}M_{i}=17 \end{matrix}$
Hence expected price of one such ticket Rs.

$17$

A pair of dice is thrown $4$ times. If getting a total of $9$ in a single throw is considered as a success then find the mean and variance of successes.

$P=$ probability of getting a total of

$\displaystyle 9=\frac{4}{36}=\frac{1}{9}$

$\therefore$

$\displaystyle 1-\frac{1}{9}=\frac{8}{9}$

$\therefore$ mean

$\displaystyle =np=4\times \frac{1}{9}=\frac{4}{9}$
variance

$\displaystyle =npq=4\times \frac{1}{9}\times \frac{8}{9}=\frac{32}{81}$

A fair die is tossed. If $2, 3$ or $5$ occurs, the player wins that number of rupees, but if $1, 4$ or $6$ occurs, the player loses that number of rupees. Then find the possible payoffs for the player.

$A_{i}$	$2$	$3$	$5$	$1$	$4$	$6$
$M_{i}$	$2$	$3$	$6$	$-1$	$-4$	$-6$
$P_{i}$	$1/6$	$1/6$	$1/6$	$1/6$	$1/6$	$1/6$

Then expected value

$E$ of the game payoffs for the player

$\displaystyle =2\left ( \frac{1}{6} \right )+3\left ( \frac{1}{6} \right )+5\left ( \frac{1}{6} \right )-1\left ( \frac{1}{6} \right )-4\left ( \frac{1}{6} \right )-6\left ( \frac{1}{6} \right )=-\left ( \frac{1}{6} \right )$
Since

$E$ is negative therefore game is unfavorable to the player.

Find the mean number of heads in three tosses of a fair coin.

Let

$X$ denote the success of getting heads.
Therefore, the sample space is

$S = \left\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\right\}$
It can be seen that

$X$ can take the value of 0, 1, 2 or 3.

$\therefore P\left(X = 0\right) = P\left(TTT\right)$

$= P\left(T\right) \cdot P\left(T\right) \cdot P\left(T\right)$

$= \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2}$

$= \displaystyle\frac{1}{8}$

$\therefore P\left(X = 1\right) = P\left(HHT\right) + P\left(HTH\right) + P\left(THH\right)$

$= \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} + \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} + \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2}$

$= \displaystyle\frac{3}{8}$

$\therefore P\left(X = 2\right) = P\left(HHT\right) + P\left(HTH\right) + P\left(THH\right)$

$= \displaystyle\frac{3}{8}$

$\therefore P\left(X =3\right) = P\left(HHH\right)$

$= \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2} \times \displaystyle\frac{1}{2}$

$= \displaystyle\frac{1}{8}$
Therefore, the required probability distribution is as follows.
Mean of

$X E\left(X\right), \mu = \sum { { X }_{ i } } P\left( { X }_{ i } \right)$

$=0\times \displaystyle\frac { 1 }{ 8 } +1\times \displaystyle\frac { 3 }{ 8 } +2\times \displaystyle\frac { 3 }{ 8 } +3\times \displaystyle\frac { 1 }{ 8 }$

$=\displaystyle\frac { 3 }{ 8 } +\displaystyle\frac { 3 }{ 4 } +\displaystyle\frac { 3 }{ 8 }$

$=\displaystyle\frac { 3 }{ 2 }$

$= 1.5$

Find the mean and variance for the first $10$ multiples of $3$

The first 10 multiples of 3 are

$3,6,9,12,15,18,21,24,27,30$
Here number of observation are

$n=10$

$\therefore$ Mean,

$\displaystyle \bar{X}=\frac{\sum_{i=1}^{10}x_{i}}{10}=\frac{165}{10}=16.5$
Now,

$\displaystyle x_{i}$	$\displaystyle \left ( X_{i}-\bar{X} \right )$	$\displaystyle \left ( X_{i}-\bar{X} \right )^{2}$
3	-13.5	182.25
6	-10.5	110.25
9	-7.5	56.25
12	-4.5	20.25
15	-1.5	2.25
18	1.5	2.25
21	4.5	20.25
24	7.5	56.25
27	10.5	110.25
30	13.5	182.25
		742.5

$\therefore$

Variance

$\displaystyle \left (\sigma ^{2} \right )=\frac{1}{n}\sum_{i=1}^{10}\left ( X_{i}-\bar{X} \right )^{2}=\frac{1}{10}\times 742.5=74.25$

Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X$ .

When two fair dice are rolled,

$6 \times 6 = 36$ observations are obtained.

$P\left(X = 2\right) = P\left(1, 1\right) = \displaystyle\frac{1}{36}$

$P\left( X=3 \right) =P\left( 1,2 \right) +P\left( 2,1 \right) =\displaystyle\frac { 2 }{ 36 } =\displaystyle\frac { 1 }{ 18 }$

$P\left( X=4 \right) =P\left( 1,3 \right) +P\left( 2,2 \right) +P\left( 3,1 \right) =\displaystyle\frac { 3 }{ 36 } =\displaystyle\frac { 1 }{ 12 }$

$P\left( X=5 \right) =P\left( 1,4 \right) +P\left( 2,3 \right) +P\left( 3,2 \right) +P\left( 4,1 \right) =\displaystyle\frac { 4 }{ 36 } =\displaystyle\frac { 1 }{ 9 }$

$P\left( X=6 \right) =P\left( 1,5 \right) +P\left( 2,4 \right) +P\left( 3,3 \right) +P\left( 4,2 \right) +P\left( 5,1 \right) =\displaystyle\frac { 5 }{ 36 }$

$P\left( X=7 \right) =P\left( 1,6 \right) +P\left( 2,5 \right) +P\left( 3,4 \right) +P\left( 4,3 \right) +P\left( 5,2 \right) +P\left( 6,1 \right) =\displaystyle\frac { 6 }{ 36 } =\displaystyle\frac { 1 }{ 6 }$

$P\left( X=8 \right) =P\left( 2,6 \right) +P\left( 3,5 \right) +P\left( 4,4 \right) +P\left( 5,3 \right) +P\left( 6,2 \right) =\displaystyle\frac { 5 }{ 36 }$

$P\left( X=9 \right) =P\left( 3,6 \right) +P\left( 4,5 \right) +P\left( 5,4 \right) +P\left( 6,3 \right) =\displaystyle\frac { 4 }{ 36 } =\displaystyle\frac { 1 }{ 9 }$

$P\left( X=10 \right) =P\left( 4,6 \right) +P\left( 5,5 \right) +P\left( 6,4 \right) =\displaystyle\frac { 3 }{ 36 } =\displaystyle\frac { 1 }{ 12 }$

$P\left( X=11 \right) =P\left( 5,6 \right) +P\left( 6,5 \right) =\displaystyle\frac { 2 }{ 36 } =\displaystyle\frac { 1 }{ 18 }$

$P\left( X=12 \right) =P\left( 6,6 \right) =\displaystyle\frac { 1 }{ 36 }$
Therefore, the required probability distribution is as follows.
Then,

$E\left( X \right) =\sum { { X }_{ i }\cdot P\left( { X }_{ i } \right) }$

$=2\times \displaystyle\frac { 1 }{ 36 } +3\times \displaystyle\frac { 1 }{ 18 } +4\times \displaystyle\frac { 1 }{ 12 } +5\times \displaystyle\frac { 1 }{ 9 } +6\times \displaystyle\frac { 5 }{ 36 } +7\times \displaystyle\frac { 1 }{ 6 } +8\times \displaystyle\frac { 5 }{ 36 } +9\times \displaystyle\frac { 1 }{ 9 } +10\times \displaystyle\frac { 1 }{ 12 } +11\times \displaystyle\frac { 1 }{ 18 } +12\times \displaystyle\frac { 1 }{ 36 }$

$=\displaystyle\frac { 1 }{ 18 } +\displaystyle\frac { 1 }{ 6 } +\displaystyle\frac { 1 }{ 3 } +\displaystyle\frac { 5 }{ 9 } +\displaystyle\frac { 5 }{ 6 } +\displaystyle\frac { 7 }{ 6 } +\displaystyle\frac { 10 }{ 9 } +1+\displaystyle\frac { 5 }{ 6 } +\displaystyle\frac { 11 }{ 18 } +\displaystyle\frac { 1 }{ 3 }$

$= 7$

$E\left( { X }^{ 2 } \right) =\sum { { X }_{ i }^{ 2 }\cdot P\left( { X }_{ i } \right) }$

$=4\times \displaystyle\frac { 1 }{ 36 } +9\times \displaystyle\frac { 1 }{ 18 } +16\times \displaystyle\frac { 1 }{ 12 } +25\times \displaystyle\frac { 1 }{ 9 } +36\times \displaystyle\frac { 5 }{ 36 } +49\times \displaystyle\frac { 1 }{ 6 } +64\times \displaystyle\frac { 5 }{ 36 } +81\times \displaystyle\frac { 1 }{ 9 } +100\times \displaystyle\frac { 1 }{ 12 } +121\times \displaystyle\frac { 1 }{ 18 } +144\times \displaystyle\frac { 1 }{ 36 }$

$=\displaystyle\frac { 1 }{ 9 } +\displaystyle\frac { 1 }{ 2 } +\displaystyle\frac { 4 }{ 3 } +\displaystyle\frac { 25 }{ 9 } +5+\displaystyle\frac { 49 }{ 6 } +\displaystyle\frac { 80 }{ 9 } +9+\displaystyle\frac { 25 }{ 3 } +\displaystyle\frac { 121 }{ 18 } +4$

$=\displaystyle\frac { 987 }{ 18 } =\displaystyle\frac { 329 }{ 6 } =54.833$
Then,

$Var\left( X \right) =E\left( { X }^{ 2 } \right) -{ \left[ E\left( X \right) \right] }^{ 2 }$

$=54.833-{ \left( 7 \right) }^{ 2 }$

$= 54.833 - 49$

$= 5.833$

$\therefore$ Standard deviation

$=\sqrt { Var\left( X \right) }$

$=\sqrt { 5.833 }$

$=2.415$

If $F\left( x \right) =\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ x } \right)$ , $-\infty < x < \infty$ is a distribution function of a continuous variable $X$ , then find $P\left( 0\le x\le 1 \right)$

Given,

$F\left( x \right) =\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ \left( x \right) } \right)$
Therefore,

$P\left( 0\le x\le 1 \right) =F\left( 1 \right) \sim F\left( 0 \right)$

$=\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ \left( 1 \right) } \right) -\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ \left( 0 \right) } \right)$

$=\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\dfrac { \pi }{ 4 } \right) -\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +0 \right)$

$=\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\dfrac { \pi }{ 4 } -\dfrac { \pi }{ 2 } \right) =\dfrac { 1 }{ 4 }$

Given the p.d.f. (probability density function) of a continuous random variable $x$ as
$f(x) =\begin{cases} \dfrac{x^2}{3}\,\,\, -1 < x < 2\\\\ 0 \,\,\, \text{otherwise}\end{cases}$
Determine the c.d.f. (cumulative distribution function) of $x$ and hence find $P(x < 1), P(x \leq -2), P(x > 0), P(1 < x < 2)$ .

c.d.f. of the continuous variable is given by,

$f(x) = \displaystyle \int_{-1}^{x} \cfrac{y^2}{3} dy$

$= \left [ \cfrac{y^3}{9} \right ]^x_{-1}$

$= \cfrac{(x^3 + 1)}{9} , x \in R$

Consider

$P(x < 1) = f(1) = \cfrac{(1)^3 + 1}{9} = \cfrac{2}{9}$

$P(x \leq -2) = 0$

$P(x > 0) = 1 - P(x \leq 0)$

$= 1 - f(0)$

$= 1 - \left( \cfrac{1}{9} \right ) = \cfrac{8}{9}$

$P(1 < x < 2) = f(2) - f(1)$

$= 1 - \cfrac{2}{9}$

$= \cfrac{7}{9}$

Find the mean and variance of distribution
$f(x) = \left\{\begin{matrix}3e^{-x} & 0<x<\infty \\ 0 & \text{elsewhere} \end{matrix}\right.$

$f(x)=\begin{cases} 3{ e }^{ -x },0<x<\infty \quad \\ 0,-\infty <x\le 0 \end{cases}$

$Mean = \int _{ -\infty }^{ \infty }{xf(x)\quad dx }$

$\mu =$

$\int _{ -\infty }^{ 0 }{ x.0\quad dx\quad +\quad \int _{ 0 }^{ \infty }{ 3x{ e }^{ -x }\quad dx } }$

$\mu =$

$0\quad - \quad$

$3{ \left| x{ e }^{ -x } \right| }_{ 0 }^{ \infty }\quad -\quad 3\int _{ 0 }^{ \infty }{ -{ e }^{ -x }\quad dx }$

$\mu = 0\quad +\quad (-3){ \left| { e }^{ -x } \right| }_{ 0 }^{ \infty }$

$\boxed{ \mu = 3 }$

$Variance = \int _{ -\infty }^{ \infty }{ { (x-\mu ) }^{ 2 }f(x)\quad dx }$

${ \sigma }^{ 2 } =$

$0\quad +\quad \int _{ 0 }^{ \infty }{ { (x-3) }^{ 2 }.3{ e }^{ -x }\quad dx }$

${ \sigma }^{2} =$

$\int _{ 0 }^{ \infty }{ 3{ x }^{ 2 }{ e }^{ -x }dx\quad +\quad \int _{ 0 }^{ \infty }{ 27{ e }^{ -x }dx\quad -\quad \int _{ 0 }^{ \infty }{ 6.3x{ e }^{ -x }dx } } }$

${ \sigma }^{2} =$

$6-27 \left| {e}^{-x} \right | _{0}^{\infty} - 6*3$

$\boxed{ { \sigma }^{ 2 } = 15 }$

The following is the p.d.f. (Probability Density Function) of a continuous random variable $X$ :
$F(x) = \dfrac {x}{32}, 0 < x < 8; = 0$ , otherwise
Also find its value at $x = 0.5$ and $9$ .

Values of

$F(x)$ at different values of

$x$ .
At

$x = 0.5$

$F(x) = F(0.5) = \dfrac {(0.5)^{2}}{64}$ .............(integration of F(x))

$= \dfrac {0.25}{64}$

$= \dfrac {1}{256}$
For any value of

$x$ greater than or equal to

$8, F(x) = 1$

$\therefore F(9) = 1$

A random variable $X$ has the following probability distribution:
$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$
(a) Find $k$
(b) Find $P(0< X< 4)$
(c) Obtain cumulative distribution function (c.d.f) of $X$ .

(a)

$K+3K+5K+7K+9K+11K+13K=1$

$\therefore$

$49K=1$

$\therefore$

$K=\cfrac{1}{49}$
(b)

$P(0< x< 4)=P(1)+P(2)+P(3)$

$\cfrac { 3 }{ 49 } +\cfrac { 10 }{ 49 } +\cfrac { 21 }{ 49 } =\cfrac { 34 }{ 49 }$
(c) C.D.F

$=(2x+1)K,0\le x$

The table gives the probability distribution of a random variable $X$ .
$x$ 1 2 3 4 5
$P(X=x)$ $0.2$ $0.1$ $0.3$ $0.3$ $p$
(i) Find $P$ .
(ii) Find the mean of $X$
(iii) Find the variance of $X$ .

$\textbf{Using the concept of probability distribution}$

$\text{The given data is,}$

$x$

$1$

$2$

$3$

$4$

$5$

$P(X=x)$

$0.2$

$0.1$

$0.3$

$p$

$\text{Now, we know}$

$\sum_{i=1}^{5}P(X=i)=1$

$\Rightarrow (0.2+0.1+0.3+0.3+p)=1$

$\Rightarrow0.9+p=1$

$\Rightarrow p=0.1$

$\text{Mean of X, M=}$

$\dfrac { 1+2+3+4+5 }{ 5 } = 3$

$\text{Variance of}$

$X=\dfrac { { (1-3) }^{ 2 }+{ (2-3) }^{ 2 }+{ (3-3) }^{ 2 }+{ (4-3) }^{ 2 }+{ (5-3) }^{ 2 } }{ 4 } = \dfrac { 4+1+1+4 }{ 4 } =\dfrac { 5 }{ 2 }$

$\textbf{Hence the respective values of p, M and variance are}$

$\boldsymbol{0.1, 3}$

$\textbf{and}$

$\boldsymbol{\dfrac{5}{2}}$

The range of a random variable $X$ is $\{0, 1, 2\}$ .
Given that $P(X = 0) = 3C^3, P(X = 1) = 4C - 10C^2, P(X = 2) = 5C - 1$ . Find the value of:
(i) $C$
(ii) $P(X < 1)$
(iii) $P(1 < X \leq 2)$ and $P(0< X \leq 3)$ .

(i) We must have

$\sum P(X)=1$

$\Rightarrow P(X=0)+P(X=1)+P(X=2)=1$

$\Rightarrow (3C^3)+(4C-10C^2)+(5C-1)=1$

$\Rightarrow 3C^3-10C^2+9C-2=0$

$\Rightarrow (C-1)(3C^2-7C+2)=0$

$\Rightarrow (C-1)(C-2)(3C-1)=0$

$\Rightarrow C=1,2,\dfrac{1}{3}$

Also

$0\le P(X=0), P(X=1), P(X=2)\le 1$

So only acceptable value of

$C$ is

$\dfrac{1}{3}$

(ii)

$P(X<1)=P(X=0)=3\left(\dfrac{1}{3}\right)^3=\dfrac{1}{9}$

(iii)

$P(1<X\le 2)=P(X=2)=4C-10C^2=4\cdot \dfrac{1}{3}-10\cdot \dfrac{1}{9}=\dfrac{2}{9}$

and

$P(0<X\le 3)=1-P(X=0)=1-\dfrac{1}{9}=\dfrac{8}{9}$

X = 1 2 3 4 5
P(X =) $\dfrac{1}{2}$
$\dfrac{1}{4}$ $\dfrac{1}{8}$ $\dfrac{1}{16}$ P
The probability distribution of a random variable X taking values 1, 2, 3, 4, 5 is given
(a) Find the value of P.
(b) Find the mean of X.
(c) Find the variance of X.

$\sum _{ i=1 }^{ 5 }{ P({ X }=i) } =1$

$\Rightarrow P({ X }=1)+P({ X }=2)+P({ X }=3)+P({ X }=4)+P({ X }=5) =1$

$\displaystyle \Rightarrow \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +P=1$

$\Rightarrow P=1-\cfrac { 1 }{ 16 }$

$\Rightarrow P= \cfrac { 15 }{ 16 }$

Mean of

$X=\cfrac { 1+2+3+4+5 }{ 5 } = 3$

Variance of

$\displaystyle X=\frac { { (1-3) }^{ 2 }+{ (2-3) }^{ 2 }+{ (3-3) }^{ 2 }+{ (4-3) }^{ 2 }+{ (5-3) }^{ 2 } }{ 4 } = \frac { 4+1+1+4 }{ 4 } =\frac { 5 }{ 2 }$

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Fix the distribution and find $P(X \geq 1)$ .

Fact: In binomial distribution , mean

$=np$ and variance

$=npq$

Here given

$np=4$ and

$npq=3$

$q=\dfrac{3}{4}\Rightarrow p=1-p=\dfrac{1}{4}$

Thus

$n=16$

Hence

$P(X\ge 1)=1-P(X=0)=1-^{16}C_0p^0q^{16}=1-\left(\dfrac{3}{4} \right)^{16}$

Verify $f(x)=\begin{cases} 30{ x }^{ 4 }{ e }^{ -6{ x }^{ 5 } };\quad \quad x>0 \\ 0;\quad otherwise \end{cases}$ for p.d.f if $f(x)$ is a p.d.f then find $P(1)$ .

$\displaystyle p.d.f \text{ of } f(x)=\int_0^\infty30x^4e^{-6x^5}dx$

Let

$6x^5=t\Rightarrow 30x^4=dt$ , we get

$\Rightarrow \displaystyle \int_0^\infty e^{-t} dt=[-e^{-t}]_0^\infty=1$

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is draw at random with replacement.Find the probability distribution of the number defective bulbs.

We have a lot of

$30$ bulbs, out of which

$6$ are defective, so,

$24$ are non defective.

probability of drawing a defective bulb

$=\dfrac{6}{30}=\dfrac{1}{5}$ .

$\Rightarrow$ probability of drawing a non-defective bulb =

$\dfrac{4}{5}$ .

Let

$X$ be the random variable denoting the drawing of

$4$ defective bulbs from the lot with replacement.

Then the probability distribution is

$X\ :$

$0$

$1$

$2$

$3$

$4$

$P(X)\ :$

$\left(\dfrac{1}{5}\right)^4$

$\left(\dfrac{1}{5}\right)^3\times$

$\left(\dfrac{4}{5}\right)$

$\left(\dfrac{1}{5}\right)^2\times$

$\left(\dfrac{4}{5}\right)^2$

$\left(\dfrac{1}{5}\right)^1\times$

$\left(\dfrac{4}{5}\right)^3$

$\left(\dfrac{4}{5}\right)^4$

or,

$X\ :$

$0$

$1$

$2$

$3$

$4$

$P(X)\ :$

$\dfrac{1}{625}$

$\dfrac{4}{625}$

$\dfrac{16}{625}$

$\dfrac{64}{625}$

$\dfrac{256}{625}$

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

We have a lot of

$6$ items, 2 are defective

$\rightarrow 6-2=4$ are non-defective.

P (drawing a defective item) =

$\dfrac{2}{6}=\dfrac{1}{3}$

P (drawing a non-defective item) = 1 - P (drawing a defective item)

$=1-\dfrac{1}{3}=\dfrac{2}{3}$

Note

$: 4$ items can be drawn from a lot of 6 items in

$^{6}C_4$ ways.

Let

$X$ be the random variable of drawing 2 defective items from the lot.

$P (X = 0) =$ P (no defective item in the sample) =

$\dfrac{^{4}C_4}{^{6}C_4}=\dfrac{1}{15}$

$P (X = 1) =$ P (one defective item from the lot) = P(one defective, 3 non-defective items)

$=(\large\frac{^{2}C_1\;^{4}C_3}{^{6}C_4}) = \dfrac{8}{15}$

$P (X = 2) =$ P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =

$\large\frac{^{2}C_2\;^{4}C_2}{^{6}C_4} = \dfrac{6}{15}$

Therefore the required probability distribution is as follows:

$X: \space\space\space\space\space\space\space\space\space 0\space\space\space\space 1\space\space\space\space 2$

$P(X) :\dfrac{1}{15}\space\dfrac{8}{15}\space\dfrac{6}{15}$

So Variance of

$X$ will be

$:E(X^2)-(E(X))^2$

$E(X)=\sum_{i=0}^{2}x_iP(X=x_i)$

$\rightarrow 0\times \dfrac{1}{15}+1\times\dfrac{8}{15}+2\times\dfrac{6}{15}=\dfrac{20}{15}$

$E(X^2)=\sum_{i=0}^{2}x_i^2P(X=x_i)$

$\rightarrow 0^2\times\dfrac{1}{15}+1^2\times\dfrac{8}{15}+2^2\times\dfrac{6}{15}=\dfrac{32}{15}$

Variance of

$X=\dfrac{32}{15}-(\dfrac{20}{15})^2=\dfrac{16}{45}$

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Mean of X.

We have a lot of

$6$ items, 2 are defective

$\rightarrow 6-2=4$ are non-defective.

P (drawing a defective item) =

$\dfrac{2}{6}=\dfrac{1}{3}$

P (drawing a non-defective item) = 1 - P (drawing a defective item)

$=1-\dfrac{1}{3}=\dfrac{2}{3}$

Note

$: 4$ items can be drawn from a lot of 6 items in

$^{6}C_4$ ways.

Let

$X$ be the random variable of drawing 2 defective items from the lot.

$P (X = 0) =$ P (no defective item in the sample) =

$\dfrac{^{4}C_4}{^{6}C_4}=\dfrac{1}{15}$

$P (X = 1) =$ P (one defective item from the lot) = P(one defective, 3 non-defective items)

$=(\large\frac{^{2}C_1\;^{4}C_3}{^{6}C_4}) = \dfrac{8}{15}$

$P (X = 2) =$ P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =

$\large\frac{^{2}C_2\;^{4}C_2}{^{6}C_4} = \dfrac{6}{15}$

Therefore the required probability distribution is as follows:

$X: \space\space\space\space\space\space\space\space\space 0\space\space\space\space 1\space\space\space\space 2$

$P(X) :\dfrac{1}{15}\space\dfrac{8}{15}\space\dfrac{6}{15}$

So Mean of

$X$ will be

$:\sum_{i=0}^{2}x_iP(X=x_i)$

$\rightarrow 0\times \dfrac{1}{15}+1\times\dfrac{8}{15}+2\times\dfrac{6}{15}=\dfrac{4}{3}$

A random variable $x$ has the following probability distribution:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$
Find $'k'$ .

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$3k$	$5k$	$7k$	$9k$	$11k$	$13k$

We know,

$\displaystyle\sum _{ r=0 }^{ 6 }{ P(X=x) } =1$

$\therefore k+3k+5k+7k+9k+11k+13k=1$

$49k=1$

$k=\cfrac { 1 }{ 49 }$

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find.
$E\left(\left(\dfrac{x-10}{5}\right)^2\right)$ .

$E(X)=10$

We know that

$E\left(\dfrac{x-10}{5}\right)=E\left(\dfrac{x}{5}-2\right)$

$=\dfrac{E(X)}{5}-2$

$=\dfrac{10}{5}-2$

$=0$

Standard deviation is

$\sigma=5$

$E\left(\left(\dfrac{x-10}{5}\right)^2\right)-\left[E\left(\dfrac{x-10}{5}\right)\right]^2=\sigma^2$

$\implies E\left(\left(\dfrac{x-10}{5}\right)^2\right)=25$

The P.M.F of a r.v. $X$ is $P\left( x \right) = \frac{1}{5}$ for $x = 1,2,3, \ldots 14,15;$ $P(x)=0$ otherwise.
Find (i) $E(X)$ (ii) $Var(X)$

i)E(X)=

$\sum { xP(X=x) } =1*\frac { 1 }{ 5 } +2*\frac { 1 }{ 5 } +3*\frac { 1 }{ 5 } ...+15*\frac { 1 }{ 5 } =\frac { 15*16 }{ 2*5 } =24\\$

ii)Var(X)=

$=E({ X }^{ 2 })-{ E(X) }^{ 2 }$

$\Rightarrow E({ X }^{ 2 })=\sum { { x }^{ 2 }P(X=x) } ={ 1 }^{ 2 }*\frac { 1 }{ 5 } +2^{ 2 }*\frac { 1 }{ 5 } +{ 3 }^{ 2 }*\frac { 1 }{ 5 } ...+{ 15 }^{ 2 }*\frac { 1 }{ 5 } =\frac { 15*16*31 }{ 6*5 } =31*8=248$

Var(X)=248-576=-328

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find.
$E\left(\left(\dfrac{x-10}{5}\right)\right)$ .

$E(X)=10$

We know that

$E\left(\dfrac{x-10}{5}\right)=E\left(\dfrac{x}{5}-2\right)$

$=\dfrac{E(X)}{5}-2$

$=\dfrac{10}{5}-2$

$=0$

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find $E(X^2)$

$Mean=E(X)=10$

standard deviation

$=\sigma=5$

We know that

$\sigma^2=E(X^2)-(E(X))^2$

$\implies 5^2=E(X^2)-10^2$

$\implies E(X^2)=125$

Find the mean of the probability distribution.

Given

$P(X=x)=\begin{cases} K\quad ,\quad x=0\quad or\quad 1 \\ 2Kx\quad ,\quad x=2 \\ K(5-x)\quad ,\quad x=3\quad or\quad 4 \end{cases}$

Mean

$=0.P(X=0)+1.P(X=1)+2.P(X=2)+3.P(X=3)+4.P(X=4)\\ \Rightarrow 1.K.1+2.2K.2+3.K(5-3)+4.K(5-4)\\ \Rightarrow 0+K+8K+6K+4K\\ \Rightarrow 19K\\ \mu =19K$

Find k, the probability distribution of X is as follows?
X 1 2 3 4
P(X) 2k 3k 3k 2k

$\because\sum{X.P(X)}=1$

$1(2k)+2(3k)+3(3k)+4(2k)=1$

$2k+6k+9k+8k=1$

$25k=1$

$\therefore k=\dfrac{1}{25}$

Calculate the expected value and variance of $x$ , if $x$ denotes the number obtained on the uppermost face when a fair die is thrown.

Let

$x$ denote the number on the uppermost face of the dice.

And the probability of any number on the uppermost face is

$1/6$

when

$x=1\quad P(1)=1/6$

$x=2\quad P(2)=1/6$

$x=3\quad P(3)=1/6$

$x=4\quad P(4)=1/6$

$x=5\quad P(5)=1/6$

$x=6\quad P(6)=1/6$

Hence adding all we get mean as

$6/6$

Events A and B are such that $P\left( A \right) = \frac{1}{2},P\left( B \right) = \frac{1}{{12}}$ and $P\left( {notA\, \text{OR} \, not\,B} \right) = \frac{1}{4}$ state whether A and B are independent?

P(A'UB')=1/4=1-P(A

$\cap$ B)

P(A

$\cap$ B)=3/4

P(A)*P(B)=1/24 not equals toP(A

$\cap$ B)

A and B are not independent

A r.v. $X$ has the following probability distribution:
$X=x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(x)$ $0.1$ $k$ $0.2$ $2k$ $0.3$ $k$
Find the value of $k$ and calculate mean and variance of $X$ .

$x=x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(x)$	$0.1$	$k$	$0.2$	$2k$	$0.3$	$k$

Since sum of probabilities

$=1$

$0.1+k+0.2+2k+0.3+k=1$

$4k=1-0.6$

$k=0.1$

Mean

$=(-2)0.1+(-1)k+(0.2)(o)+1(2k)+2(0.3)+3(k)$

$=4k+0.4$

$=4(0.1)+0.4$

$(\mu)=0.8$

Variance

$\Rightarrow (6^{2})=\displaystyle\sum_{i=1}^{n}x^{2}(p(x=xi)-\mu i$

$6^{2}=4(0.1)+1(k)+0(0.2)+1(2k)+4(0.3)+9k-\mu^{2}$

$=12(0.1)+1.6-(0.8)^{2}$

$=1.2+1.6-0.64$

$6^{2}\Rightarrow 2.16$

X 1 2 3 4 5 6
P(X=x) k 2k 3k 4k 5k 6k
The probability distribution of discrete r.v. $X$ is
Then $P(X\le 4)$ =

Sum of the probabilities is equal to 1

$\therefore k+2k+3k+4k+5k+6k=1$

$21k=1$

$k=\dfrac{1}{21}$

$P(X\leq 4)$

$=P(1)+P(2)+P(3)+P(4)$

$=k+2k+3k+4k$

$=\dfrac{1}{21}+\dfrac{2}{21}+\dfrac{3}{21}+\dfrac{4}{21}$

$=\dfrac{1+2+3+4}{21}$

$=\dfrac{10}{21}$

$\therefore P(X\leq 4)=0.476$

Two number are selected at random (without replacement ) from the first six positive integers. Let X denote the larger of the numbers obtained. Find E (X).

The first 6 positive integers are

$1,2,3,4,5,6$ .

We can select the two positive numbers in

$6\times 5 =30$ different ways.

Out of this, 2 numbers are selected at random and let X denote the larger of the two numbers.

Since X is the larger of the two numbers, X can assume the value of

$2,3,4,5$ or

$6$ .

P(X=2) = P(larger number is 2) = {(1,2), (2,1)} =

$\dfrac{2}{30}$

P(X=3) = P(larger number is 3) = {(1,3), (3,1), (2,3), (3,2)} =

$\dfrac{4}{30}$

P(X=4) = P(larger number is 4) = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)} =

$\dfrac{6}{30}$

P(X=5) = P(larger number is 5) = {(1,5), (5,1), (2,5), (5,2), (3,5), (5,3), (4,5), (5,4)} =

$\dfrac{8}{30}$

P(X=6) = P(larger number is 6) = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)} =

$\dfrac{10}{30}$

The expected value of mean can be calculated as :

Mean =

$\Sigma(X_{i}\times P(X_{i})) =$

$=2\times \dfrac{2}{30} + 3\times \dfrac{4}{30}+4 \times \dfrac{6}{30} +5\times \dfrac{8}{30}+6\times \dfrac{10}{30} = \dfrac{14}{3}$

The random variable $X$ has probability distribution $P(X)$ of the following form.
$P(X)=\begin{cases} k\quad if\quad X=0 \\ 2k\quad if\quad X=1 \\ 3k\quad if\quad X=2 \\ 0\quad otherwise \end{cases}$
(a) determine value of $K$
(b) Find $P(X< 2), P(X\le 2), P(X\ge 2)$

$(a)$ As Sum of all probabilities should be

$\Rightarrow k+2k+3k=1 \Rightarrow k=1/6$

$(b)$

$P(x<2)=p(x=0)+p(x=1)=k+2k=3\left(\dfrac{1}{6}\right)=\dfrac{1}{2}$

$p(x \le 2)=\dfrac{1}{2}+3k=\dfrac{1}{2}+\dfrac{3}{6}=\dfrac{1}{2}+\dfrac{1}{2}=1$

$P(x\ge 2)=3k+0=3\left(\dfrac{1}{6}\right)+0=\dfrac{1}{2}$

The p. m. f. X- number of major defects randomly selected appliance of a certain type is :
X-x 0 1 2 3 4
P(a) 0.08 0.15 0.45 0.27 0.05
Find the expected value and standard deviation of X.

x	0	1	2	3	4
$P(\mu )$	0.08	0.15	0.45	0.27	0.05

expected value .

$E(x) = 0\times 0.008+1\times 0.15+2\times 0.45+3\times 0.27+4\times 0.05$

$= 0.15+0.90+0.81+0.20$

$E(x) = 2.06 = \mu$

$\rightarrow$ variance of x.

$V(x)= \dfrac{E(x-\mu )2}{\sim }$

$= (0-\mu )^{2}0.08+(1-\mu )^{2}\times 0.15+(2-\mu )^{2}\times 0.45$

$+(3-\mu )^{2}\times 0.27+(4-\mu )^{2}\times 0.05$

$v(x) = 0.9364$

$\rightarrow$ standard deviation of x

$SD (x) = \sqrt{V(x)}$

$= \sqrt{0.9364}$

$= 0.9677$

Let a pair of dice be thrown and the random variable $X$ be the sum of the numbers that appear on the two dice. Find the mean of $X$ .

$P(X=2)=P(1,1)=\cfrac{1}{36}$

$P(X=3)=P[(1,2),(2,1)]=\cfrac{2}{36}$

$P(X=4)=P[)1,3),(2,2),(3,1)]=\cfrac{3}{36}$

$P(X=5)=P[(1,4),(2,3),(3,2),(4,1)]=\cfrac{4}{36}$

$P(X=6)=P[(1,5),(2,4),(3,3),(4,2),(5,1)]=\cfrac{5}{36}$

$P(X=7)=P[(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)]=\cfrac{6}{36}$

$P(X=8)=P[(2,6,(3,5),(4,4),(5,3),(6,2)]=\cfrac{5}{36}$

$P(X=9)=P[(3,6),(4,5),(5,4),(6,3)]=\cfrac{4}{36}$

$P(X=10)=P[(4,6),(5,5),(6,4)]=\cfrac{3}{36}$

$P(X=11)=P[(5,6),(6,5)]=\cfrac{2}{36}$

$P(X=12)=P[(6,6)]=\cfrac{1}{36}$

$X$	$P=P(X)$	${p}_{i}{x}_{i}$
2	$\cfrac { 1 }{ 36 }$	$\cfrac { 2 }{ 36 }$
3	$\cfrac { 2 }{ 36 }$	$\cfrac { 6 }{ 36 }$
4	$\cfrac { 3 }{ 36 }$	$\cfrac { 12 }{ 36 }$
5	$\cfrac { 4 }{ 36 }$	$\cfrac { 20 }{ 36 }$
6	$\cfrac { 5 }{ 36 }$	$\cfrac { 30 }{ 36 }$
7	$\cfrac { 6 }{ 36 }$	$\cfrac { 42 }{ 36 }$
8	$\cfrac { 5 }{ 36 }$	$\cfrac { 40 }{ 36 }$
9	$\cfrac { 4 }{ 36 }$	$\cfrac { 36 }{ 36 }$
10	$\cfrac { 3 }{ 36 }$	$\cfrac { 30 }{ 36 }$
11	$\cfrac { 2 }{ 36 }$	$\cfrac { 22 }{ 36 }$
12	$\cfrac { 1 }{ 36 }$	$\cfrac { 12 }{ 36 }$

Total

$1$

$\cfrac { 252 }{ 36 }$
Mean of

$X==\cfrac { \sum { { p }_{ i }{ x }_{ i } } }{ \sum { { p }_{ i } } } =\cfrac { 252 }{ 36\times 1 } =7$
Mean

$=7$

12 different objecs are to be distributed among 9 persons. Find the number of ways in which objects can be distributed . (All objects must be distributed and each person should get atleast 1 object)

No. of ways of distributing

$12$ objects among

$9$ persons to that each gets atleast one

$={ C }_{ 9-1 }^{ 12+9-1 }={ C }_{ 8 }^{ 20 }\\ =\cfrac { 20! }{ 8!\times 12! } =\cfrac { 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12! }{ 8\times 7\times 6\times 5\times 4\times 3\times 2\times 12! } \\ =19\times 17\times 15\times 13\times 2$

$=125970$ ways

Four red balls are numbered from $1$ to $4$ and three blue balls are numbered from $3$ to $5$ . If these balls are placed in a bag and a ball is drawn from a bag, what is the probability of choosing a red ball and a blue ball numbered $5$ ?

Given,

4 red balls, 3 blue balls.

Therefore the probability of choosing a red ball

$=\dfrac{1}{4+3}=\dfrac{1}{7}$

Total of blue balls 3, numbered from 3,4,5.

Therefore the probability of choosing a 5 numbered blue ball

$=\dfrac{1}{1+1+1}=\dfrac{1}{3}$

The probability distribution of a random variable X is given below
$X=x_i$ $P(X=x_i)$
$1$ k
$2$ $2k$
$3$ $3k$
$4$ $4k$
$5$ $5k$
Find the value of k and the mean and variance of X.

As we know that

$\sum{{P}_{i}} = 1$

$\therefore k + 2k + 3k + 4k + 5k = 1$

$\Rightarrow k = \cfrac{1}{15}$

Thus the value of

$k$ is

$\cfrac{1}{15}$

$X = {x}_{i}$	$P{\left( X = {x}_{i} \right)}$	$XP$	${X}^{2} P$
$1$	$k$	$k$	$k$
$2$	$2k$	$4k$	$8k$
$3$	$3k$	$9k$	$27k$
$4$	$4k$	$16k$	$64k$
$5$	$5k$	$25k$	$125k$
		$\sum{XP} = 55k = 55 \times \cfrac{1}{15} = 3.66$	$\sum{{X}^{2} P} = 225 k 225 \times \cfrac{1}{15} = 15$

Mean of the given data

$\left( \bar{x} \right) = \sum{XP} = 3.66$

Variance of the given data

$= \sum{{X}^{2}P} - {\bar{x}}^{2} = 15 - {\left( 3.66 \right)}^{2} = 15 - 13.4 = 1.6$

Hence the mean and variance of the given data is

$3.66$ and

$1.6$ respectively.

Two unbiased dice are thrown together at random. Find the expected value of the total number of points shown up.

Let $"x"$ indicate the sum of the points on the two dice

[Since you are required to find the expected value the sum of the scores on the two dice, the variable would represent the sum of the points on the dice]

The sum of the points on the two dice would be

$2$ if the dice show up $\{(1,1)\}$
$3$ if the dice show up $\{(1,2), (2,1)\}$
$4$ if the dice show up $\{(1,3), (2,2), (3,1)\}$
$5$ if the dice show up $\{(1,4), (2,3), (3,2), (4,1)\}$
$6$ if the dice show up $\{(1,5), (2,4), (3,3), (4,2), (5,1)\}$
$7$ if the dice show up $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$
$8$ if the dice show up $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$
$9$ if the dice show up $\{(3,6), (4,5), (5,4), (6,3)\}$
$10$ if the dice show up $\{(4,6), (5,5), (6,4)\}$
$11$ if the dice show up $\{(5,6), (6,5)\}$
$12$ if the dice show up $\{(6,6)\}$

⇒ The values carried by the variable ("x") would be either $1, 2, 3, ... , 12$

⇒ "X" is a discrete random variable with range $= \{1, 2, 3, ... , 12\}$

$"X"$ represents the random variable and $P(X = x)$ represents the probability that the value within the range of the random variable is a specified value of $"x"$

In the experiment of tossing $2$ dice

Total no. of possible choices	=	$6 × 6$
	=	$36$

$x$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$p(X = x)$	$\dfrac{1}{36}$	$\dfrac{2}{36}$	$\dfrac{3}{36}$	$\dfrac{4}{36}$	$\dfrac{5}{36}$	$\dfrac{6}{36}$	$\dfrac{5}{36}$	$\dfrac{4}{36}$	$\dfrac{3}{36}$	$\dfrac{2}{36}$	$\dfrac{1}{36}$

$E(x) = \Sigma px=\dfrac{252}{36}$ (or)

$7$

Detailed calculation for mean and variance table is shown above:

Variance of the sum of the points on the two dice

⇒ var (x)	=	E (x²) − (E(x))²
⇒ var (x)	=	Σ px² − (Σ px)²
	=	54.83 − (7)²
	=	54.83 − 49
	=	5.83

Standard Deviation of the sum of the points on the dice

$⇒ SD (x)=+ \sqrt {Var (x)}$

$=+\sqrt \{5.83\}$

$=+2.412$

1198706_1514408_ans_3add4c3aeb3c446cb7a443a1be67c793.PNG

Two cards are drawn (without replacement) from a well shuffled deck of $52$ cards. Find probability distribution and mean of number of cards numbered $4$

Total no. of cards in the pack

$= 52$

The number of cards numbered

$4$ in the pack of cards

$= 4$

Let

$X$ be the discrete random variable denoting the number of cards numbered

$4$ when two cards are drawn without replacement.

$X$ takes the values

$0, 1, 2$ .

Now,

Case I:- Probability of no

$4$ being drawn

$P{\left( X = 0 \right)} = \cfrac{{^{48}{C}_{2}}}{{^{52}{C}_{2}}} = \cfrac{188}{221}$

Case II:- Probability of one

$4$ being drawn

$P{\left( X = 1 \right)} = \cfrac{{^{48}{C}_{1}} \times {^{4}{C}_{1}}}{{^{52}{C}_{2}}} = \cfrac{32}{221}$

Case III:- Probability of two

$4$ being drawn

$P{\left( X = 2 \right)} = \cfrac{{^{4}{C}_{2}}}{{^{52}{C}_{2}}} = \cfrac{1}{221}$

Therefore, the probability distribution is-

$X$	$0$	$1$	$2$
$P{\left( x \right)}$	$\cfrac{188}{221}$	$\cfrac{32}{221}$	$\cfrac{1}{221}$

Now, mean of probability distribution-

Mean

$= \sum{{x}_{i} {P}_{i}} \\ = 0 \times \cfrac{188}{221} + 1 \times \cfrac{32}{221} + 2 \times \cfrac{1}{221} \\ = \cfrac{0 + 32 + 2}{221} = \cfrac{34}{221}$

Find the probability distribution of the number of doublets in three throws of a pair of dice

Let

$X$ denote the number of doublets. Possible doublets are

$\left(1, 1\right), \left(2, 2\right), \left(3, 3\right), \left(4, 4\right), \left(5, 5\right), \left(6, 6\right)$ .

Since, the dice thrown trice,

$\therefore\, X$ can take the value

$0, 1, 2$ , or

$3$ .

Probability of getting a doublet

$= p = \dfrac{1}{6}$ and

$q = 1 – \dfrac{1}{6}=\dfrac{5}{6}$

$P\left(0\right)=P\left(no\,doublet\right)=qqq=\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}=\dfrac{125}{216}$

$P\left(1\right)=P\left(one\,doublet\right)=pqq+qpq+qqp$

$=\dfrac{1}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{1}{6}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}$

$=\dfrac{25}{216}+\dfrac{25}{216}+\dfrac{25}{216}=\dfrac{75}{216}$

$P\left(2\right)=P\left(two\,doublet\right)=pqq+ppp+qpp$

$=\dfrac{1}{6}\times\dfrac{1}{6}\times\dfrac{5}{6}+\dfrac{1}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}+\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}$

$=\dfrac{5}{216}+\dfrac{5}{216}+\dfrac{5}{216}=\dfrac{15}{216}$

$P\left(3\right)=P\left(three\,doublet\right)=ppp=\dfrac{1}{6}\times\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{216}$

Mean

$\mu=E\left(X\right)=\displaystyle\sum_{i=0}^{3}{P\left({x}_{i}\right){x}_{i}}$

$=0\times\dfrac{125}{216}+1\times\dfrac{75}{216}+2\times\dfrac{15}{216}+3\times\dfrac{1}{216}$

$=\dfrac{0+75+30+3}{216}=\dfrac{108}{216}=\dfrac{1}{2}$

Find expected value of the random variable $X$ whose probability mass function is :
$X=x$ 1 2 3
$P(X=x)$ $\dfrac{1}{5}$ $\dfrac{2}{5}$ $\dfrac{2}{5}$

$\displaystyle E(x)=\sum x_iP(x_i)$

$=1(\dfrac{1}{5})+2(\dfrac{2}{5})+3(\dfrac{2}{5})$

$=\dfrac{1+4+6}{5}$

$=\dfrac{11}{5}$

$\boxed{\therefore E(x)=\dfrac{11}{5}}$

If $f(x)=kx,0 < x < 2$
$=0$ , otherwise,

is a probability density function of a random variable $X$ , then find:
(i) Value of $k$ ,
(ii) $P(1 < X< 2)$

$\because$ the value of a probability density function over the whole range is equal to

$'1'$

$\displaystyle \therefore \int^2_0 f(x)dx=\int^2_0 kx dx=1$

$\Rightarrow K\left[\dfrac{x^2}{2}\right]^2_0=1$

$\Rightarrow k\left[\dfrac{4}{2}-0\right]=1$

$\Rightarrow k\times 2=1$

$\Rightarrow k=\dfrac{1}{2}$

ii)

$P(1< x<2)$

$\displaystyle \int^2_1 f(x) dx=\int^2_1\dfrac{1}{2}x dx$ Putting value of

$k=\dfrac{1}{2}$

$=\dfrac{1}{2}\left[\dfrac{x^2}{2}\right]^2_1$

$=\dfrac{1}{2}\left[\dfrac{4}{2}-\dfrac{1}{2}\right]$

$=\dfrac{1}{2}\times \dfrac{3}{2}=\dfrac{3}{4}$

$\therefore \boxed{P(1< x< 2)=\dfrac{3}{4}}$

There are 5 cards , numbered 1 to 5, one number of each card . two cards are drawn at random without replacement . Let X denote the sum of the numbers on the two cards drawn.find the mean and variance of X.

1569607_1708260_ans_9344c1864efe43ee9d28a6f089f601b8.png

The probability distribution of $X$ is :
$X$ $0$ $1$ $2$ $3$
$P(X)$ $0.2$ $K$ $K$ $2K$

Write the value of $K$ .

We have ,

$SP(X) = 1$

$\Rightarrow \,\, 0.2 + 4K = 1$ where

$SP$ is the sum of probabilities

$\Rightarrow \,\, 4K = 0.8 \,\,\, \Rightarrow$

$K = 0.2$

Consider the probability distribution of a random variable $x$ :
$\begin{array}{|l|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\\hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\\hline\end{array}$
Calculate
(ii) Variance of $\mathrm{X}$

we have,

$\begin{array}{|l|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\\hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\\mathrm{XP(X)}&0&0.25&0.6&0.6&0.60 \\\hline \mathrm{X^2P(X)}&0&0.25&1.2&1.8&2.40 \\\hline\end{array}$

$\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}$

Where,

$E(X)=\mu=\sum_{i=1}^{n} x_{i} P_{i}(x i)$

And

$E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)$

$\therefore E(X)=0+0.25+0.6+0.6+0.60=2.05$

$E\left(X^{2}\right)=0+0.25+1.2+1.8+2.40=5.65$

(ii)

$\mathrm{V}(\mathrm{X})=1.44475$

A biased die is such that $\mathrm{P}(4)=\dfrac{1}{10}$ and other scores being equally likely. The die is tossed twice. If $X$ is the 'number of four seen', find the variance of the random variable $\mathbf{X}$

since,

$\mathrm{X}=$ number of four seenOn tossing two die,

$\mathrm{X}=0,1,2$

Also,

$P_{(4)}=\dfrac{1}{10}$ and

$P_{(not 4)}=\dfrac{9}{10}$

$\mathrm{So}$

$P(X=0)=P_{(not 4)} \cdot P_{(not 4)}=\dfrac{9}{10} \cdot \dfrac{9}{10}=\dfrac{81}{100}$

$P(X=1)=P_{(not 4)}, P_{(4)}+P_{(4)} \cdot P_{(not4)}=\dfrac{9}{10} \cdot \dfrac{1}{10}+\dfrac{1}{10} \cdot \dfrac{9}{10}=\dfrac{18}{100}$

$P(X=2)=P_{(4)} \cdot P_{(4)}=\dfrac{1}{10} \cdot \dfrac{1}{10}=\dfrac{1}{100}$
Thus, we get following table

$\begin{array}{|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 \\\hline \mathrm{P}(\mathrm{X}) & \dfrac{81}{100} & \dfrac{18}{100} & \dfrac{1}{100} \\\hline \mathrm{XP}(\mathrm{X}) & 0 & 18 / 100 & 2 / 100 \\\hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & 18 / 100 & 4 / 100 \\\hline\end{array}$

$\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\sum X^{2} P(X)-\left[\sum X P(X)\right]^{2}$

$=\left[0+\dfrac{18}{100}+\dfrac{4}{100}\right]-\left[0+\dfrac{18}{100}+\dfrac{2}{100}\right]^{2}$

$=\dfrac{22}{100}-\left(\dfrac{20}{100}\right)^{2}=\dfrac{11}{50}-\dfrac{1}{25}$

$=\dfrac{11-2}{50}=\dfrac{9}{50}=\dfrac{18}{100}=0.18$

The probability distribution of a random variable X is given below:
$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$
(i) Determine the value of $\mathbf{k}$

We have,

$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$
(i) since,

$\displaystyle \sum_{i=1}^{n} P_{i}=1, i=1,2, \ldots, n$ and

$p_{i} \geq 0$

$\therefore k+\dfrac{k}{2}+\dfrac{k}{4}+\dfrac{k}{8}=1$

$\Rightarrow 8 k+4 k+2 k+k=8$

$k=\dfrac{8}{15}$

If $\mathrm{X}$ is the number of tails in three tosses of a coin, determine the standard deviation of $x$

Given that, random variable

$X$ is the number of tails in three tosses of a coin.

$\mathrm{So}, X=0,1,2,3$

$\Rightarrow P(X-x)=^{n} C_{x}(p)^{x} q^{n-x} \Rightarrow P(X=r)=^{n} C_{r}(p)^{r}(q)^{n-r}$

Where

$n=3, p=1 / 2, q=1 / 2$ and

$x=0,1,2,3 n=3, p=\dfrac{1}{2}, q=\dfrac{1}{2}, r=0,1,2,3$

$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} &0&1&2&3 \\\hline \mathrm{P(X)}&\dfrac{1}{8}&\dfrac{3}{8}&\dfrac{3}{8}&\dfrac{1}{8} \\\hline \mathrm{XP(X)}&0&\dfrac{3}{8}&\dfrac{3}{4}&\dfrac{3}{8}\\\hline \mathrm{X^2P(X)}&0&\dfrac{3}{8}&\dfrac{3}{2}&\dfrac{9}{8} \\\hline\end{array}$

We know that,

$\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}$ Eq....

$(\mathrm{i})$

Where,

$E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)$ and

$E(X)=\sum_{i=1}^{n} x_{i} P\left(x_{i}\right)$

$\therefore E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} P\left(X_{i}\right)=0+\dfrac{3}{8}+\dfrac{3}{2}+\dfrac{9}{8}=\dfrac{24}{8}=3$

And

$[E(X)]^{2}=\left[\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)\right]^{2}=\left[0+\dfrac{3}{8}+\dfrac{3}{4}+\dfrac{3}{8}\right]^{2}=\left[\dfrac{12}{8}\right]^{2}=\dfrac{9}{4}$

$\therefore \operatorname{Var}(X)=3-\dfrac{9}{4}=\dfrac{3}{4}[\text { using Eq. }(\mathrm{i})]$
And standard deviation of

$X=\sqrt{\operatorname{Var}(X)}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}$

Consider the probability distribution of a random variable $x$ :
$\begin{array}{|l|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\\hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\\hline\end{array}$
Calculate:
(i) $\mathrm{V}\left(\dfrac{\mathrm{X}}{2}\right)$

we have,

$\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}$

Where,

$\displaystyle E(X)=\mu=\sum_{i=1}^{n} x_{i} P_{i}(x i)$

And

$E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)$

$\therefore E(X)=0+0.25+0.6+0.6+0.60=2.05$

$E\left(X^{2}\right)=0+0.25+1.2+1.8+2.40=5.65$

(i)

$V\left(\dfrac{X}{2}\right)=\dfrac{1}{4} V(X)=\dfrac{1}{4}\left[5.65-(2.05)^{2}\right]$

$\dfrac{1}{4}[5.65-4.2025]=\dfrac{1}{4} \times 1.4475=0.361875$

A discrete random variable X has the probability distribution given as below:
$\begin{array}{|l|ll|l|l|}\hline \mathrm{X} &0.5 & 1 & 1.5 & 2 \\\hline \mathrm{P(X)}& k & k^2 & 2k^2 & k \\\hline\end{array}$
(ii) Determine the mean of distribution.

We have,

$\begin{array}{|l|ll|l|l|}\hline \mathrm{X} &0.5 & 1 & 1.5 & 2 \\\hline \mathrm{P(X)}& k & k^2 & 2k^2 & k \\\hline\end{array}$
We know that,

$\displaystyle \sum_{i=1}^{n} P_{i}=1$ , where

$P_{i} \geq 0$

$\Rightarrow P_{1}+P_{2}+P_{3}+P_{4}=1$

$\Rightarrow k+k^{2}+2 k^{2}+k=1$

$\Rightarrow 3 k^{2}+2 k-1=0$

$\Rightarrow 3 k^{2}+3 k-k-1=0$

$\Rightarrow 3 k(k+1)-1(k+1)=0$

$\Rightarrow(3 k-1)(k+1)=0$

$\Rightarrow k=1 / 3 \Rightarrow k=-1$

$\mathrm{k}=-1$ not possible because

$k$ is

$\geq 0 \Rightarrow \mathrm{k}=1 / 3$

Mean of the distribution

$(\mu)=E(X)=\sum_{i=1}^{n} x_{i} p_{i}$

$=0.5(k)+1\left(k^{2}\right)+1.5\left(2 k^{2}\right)+2 k=4 k^{2}+2.5 k$

$=4 \cdot \dfrac{1}{9}+2.5 \cdot \dfrac{1}{3}\left[\because k=\dfrac{1}{3}\right]$

$=\dfrac{4+7.5}{9}=\dfrac{23}{18}$

A die is thrown three times. Let $x$ be 'the number of two seen'. Find the expectation of $x$

We have,

$X=$ number of two seen

So, on throwing a die three times, we will have

$X=0,1,2,3$

$\therefore P(X=0)=P_{(\operatorname{not} 2)} \cdot P_{(\operatorname{not} 2)} \cdot P_{(\operatorname{not} 2)}=\dfrac{5}{6} \cdot \dfrac{5}{6} \cdot \dfrac{5}{6}=\dfrac{125}{216}$

$P(X=1)=P_{(\operatorname{not} 2)} \cdot P_{(\operatorname{not} 2)} \cdot P_{(2)}+P_{(\operatorname{not} 2)} \cdot P_{(2)} \cdot P_{(\operatorname{not} 2)}+P_{(2)} \cdot P_{(\operatorname{not} 2)} \cdot P_{(\operatorname{not} 2)}$

$=\dfrac{5}{6} \cdot \dfrac{5}{6} \cdot \dfrac{1}{6}+\dfrac{5}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6}+\dfrac{1}{6} \cdot \dfrac{5}{6} \cdot \dfrac{5}{6}=\dfrac{25}{36} \cdot \dfrac{3}{6}=\dfrac{25}{72}$

$P(X=2)=P_{(\operatorname{not} 2)} \cdot P_{(2)} \cdot P_{(2)}+P_{(2)} \cdot P_{(2)} \cdot P_{(\operatorname{not} 2)}+P_{(2)} \cdot P_{(\operatorname{not} 2)} \cdot P_{(2)}$

$=\dfrac{5}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6}+\dfrac{1}{6} \cdot \dfrac{5}{6} \cdot \dfrac{1}{6}$

$=\dfrac{1}{36} \cdot\left[\dfrac{15}{6}\right]=\dfrac{15}{216}$

and

$P(X=3)=P_{(2)} \cdot P_{(2)} \cdot P_{(2)}=\dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6}=\dfrac{1}{216}$

We know that,

$E(X)=\sum X P(X)=0 \cdot \dfrac{125}{216}+1 \cdot \dfrac{25}{216}+3 \cdot \dfrac{1}{216}$

$=\dfrac{75+30+3}{216}=\dfrac{108}{216}=\dfrac{1}{2}$

Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

Let $E_1 =$ Event that first ball being red

And $E_2 =$ Event that exactly two of three balls being red

$\therefore P(E_1) = P_R \cdot P_{R} \cdot P_{R} + P_{R} \cdot P_{R} \cdot P_{\bar R} + P_{R} \cdot P_{\bar R} \cdot P_R + P_{R} \cdot P_{\bar R} \cdot P_{\bar R}$

$= \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} + \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} + \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{4}{6} + \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{2}{6}$

$\dfrac{60 + 60 + 60 + 30}{336} = \dfrac{210}{336}$

$P (E_1 \cap E_2) = P_R \cdot P_{\bar R} \cdot P_R + P_R \cdot P_R \cdot P_{\bar R}$

$= \dfrac{5}{8} \cdot \dfrac{3}{7} \cdot \dfrac{4}{6} + \dfrac{5}{8} \cdot \dfrac{4}{7} \cdot \dfrac{3}{6} = \dfrac{120}{336}$

$\therefore P(E_2 / E_1) = \dfrac{P(E_1 \cap E_2)}{P(E_1)} = \dfrac{120/336}{210/336} = \dfrac{4}{7}$

1781158_1802795_ans_010246d7428f49489b3c65a32c68f1c6.PNG

Find the variance of the distribution:Vedantu
$\begin{array}{|l|l|l|l|l|l|l|}\hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \mathrm{P}(\mathrm{x}) & \dfrac{1}{6} & \dfrac{5}{18} & \dfrac{2}{9} & \dfrac{1}{6} & \dfrac{1}{9} & \dfrac{1}{18} \\\hline\end{array}$

We have,

$\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \mathrm{P}(\mathrm{X}) & \dfrac{1}{6} & \dfrac{5}{18} & \dfrac{2}{9} & \dfrac{1}{6} & \dfrac{1}{9} & \dfrac{1}{18} \\\hline \mathrm{XP}(\mathrm{X}) & 0 & \dfrac{5}{18} & \dfrac{4}{9} & \dfrac{1}{2} & \dfrac{4}{9} & \dfrac{5}{18} \\\hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & \dfrac{5}{18} & \dfrac{8}{9} & \dfrac{3}{2} & \dfrac{16}{9} & \dfrac{25}{18} \\\hline \end{array}$

$\therefore$ Variance

$=E\left(X^{2}\right)-[E(X)]^{2}=\sum X^{2} P(X)-\left[\sum X P(X)\right]^{2}$

$=\left[0+\dfrac{5}{18}+\dfrac{8}{9}+\dfrac{3}{2}+\dfrac{16}{9}+\dfrac{25}{18}\right]-\left[0+\dfrac{5}{18}+\dfrac{4}{9}+\dfrac{1}{2}+\dfrac{4}{9}+\dfrac{5}{18}\right]^{2}$

$=\left[\dfrac{5+16+27+32+25}{18}\right]-\left[\dfrac{5+8+9+8+5}{18}\right]^{-2}$

$=\dfrac{105}{18}-\dfrac{35 \cdot 35}{18 \cdot 18}=\dfrac{18 \cdot 105-35 \cdot 35}{18 \cdot 18}$

$=\dfrac{35}{18 \cdot 18}[54-35]=\dfrac{19 \cdot 35}{324}=\dfrac{665}{324}$

Two probability distributions of the discrete random variable $X$ and $Y$ are given below.
$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{1}{5} & \dfrac{1}{5} \\\hline\end{array}$
$\begin{array}{|l|l|l|l|l|}\hline \mathrm{Y} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{Y}) & \dfrac{1}{5} & \dfrac{3}{10} & \dfrac{2}{5} & \dfrac{1}{5} \\\hline\end{array}$
Prove that $E\left(Y^{2}\right)=2 E(X)$

$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{1}{5} & \dfrac{1}{5} \\\hline\end{array}$

$\begin{array}{|l|l|l|l|l|}\hline \mathrm{Y} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{Y}) & \dfrac{1}{5} & \dfrac{3}{10} & \dfrac{2}{5} & \dfrac{1}{5} \\\hline\end{array}$
since, we have to prove that,

$E\left(Y^{2}\right)=2 E(X)$

$\therefore E(X)=\sum X P(X)$

$=0 \cdot \dfrac{1}{5}+1 \cdot \dfrac{2}{5}+2 \cdot \dfrac{1}{5}+3 \cdot \dfrac{1}{5}=\dfrac{7}{5}$

$\Rightarrow 2 E(X)=\dfrac{14}{5} \ldots(i)$

$E(Y)^{2}=\sum Y^{2} P(Y)$

$=0 \cdot \dfrac{1}{5}+1 \cdot \dfrac{3}{10}+4 \cdot \dfrac{2}{5}+9 \cdot \dfrac{1}{10}$

$=\dfrac{3}{10}+\dfrac{8}{5}+\dfrac{9}{10}=\dfrac{28}{10}=\dfrac{14}{5}$

$\Rightarrow E(Y)^{2}=\dfrac{14}{5}$ From Eqs. (i) and (ii),

$E(Y)^{2}=2 E(X)$ Hence proved.

The probability distribution of a discrete random variable X is given as under:
Calculate :
(i) The value of A if E(X) = 2.94
(ii) Variance of X.

(i) We have,

$\sum XP(X) = \dfrac{1} {2} + \dfrac{2} {5} + \dfrac{12} {25} + \dfrac{2A} {10} + \dfrac{3A} {25} + \dfrac{5A} {25}$

$= \dfrac{25 + 20 + 24 + 10A + 6A + 10A} {50} = \dfrac{69 + 26A} {50}$
Since,

$E(X) = \sum XP(X)$

$\Rightarrow 2.94 = \dfrac{69 + 26A} {50}$

$\Rightarrow 26 A = 50 \times 2.94 - 69$

$\Rightarrow A = \dfrac{147 - 69} {26} = \dfrac{78} {26} = 3$

(ii) We know that,

$Var(X) = E(X)^{2} - [E(X)]^{2}$

$\sum X^{2}P(X) - [\sum XP(X)]^{2}$

$= \dfrac{1} {2} + \dfrac{4} {5} + \dfrac{48} {25} + \dfrac{4A^{2}} {10} + \dfrac{9A^{2}} {25} + \dfrac{25A^{2}} {25} - [E(X)]^{2}$

$= \dfrac{25 + 40 + 96 + 20A^{2} + 18A^{2} + 50A^{2}} {50} - [E(X)]^{2}$

$= \dfrac{161 + 88 A^{2}} {50} - [E(X)]^{2} = \dfrac{161 + 88 (3)^{2}} {50} - [E(X)]^{2} [\therefore A = 3]$

$\dfrac{953} {50} - [2.94]^{2} [\therefore E(X) = 2.94]$

$= 19.0600 - 8.6436 = 10.4164$

Find the probability distribution of the maximum of the two scores obtained when die is thrown twice. Determine also the mean of the distribution

Let X is the random variable score obtained when a die is thrown twice.

$\begin{array}{l}\therefore \mathrm{x}=1,2,3,4,5,6 \\\text { Here, } S=\{(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3), \ldots,(6,6)\}\end{array}$

$P(X=1)=\dfrac{1}{6} \cdot \dfrac{1}{6}=\dfrac{1}{36}$

$P(X=2)=\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}=\dfrac{3}{36}$

$P(X=3)=\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}+\dfrac{1}{6} \cdot \dfrac{1}{6}=\dfrac{5}{36}$
similarly

$P(X=4)=\dfrac{7}{36}$

$P(X=5)=\dfrac{9}{36}$

$P(X=6)=\dfrac{11}{36}$
So, the required distribution is

$\begin{array}{|l|l|l|l|l|l|l|}\hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \mathrm{P}(\mathrm{X}) & 1 / 36 & 3 / 36 & 5 / 36 & 7 / 36 & 9 / 36 & 11 / 36 \\\hline\end{array}$
Also, we know that, Mean

$\{E(X)\}=\sum X P(X)$

$=\dfrac{1}{36}+\dfrac{6}{36}+\dfrac{15}{36}+\dfrac{28}{36}+\dfrac{45}{36}+\dfrac{66}{36}=\dfrac{161}{36}$

Let X be a discrete random variable whose probability distribution is defined as follows:
$P(X = x)$ = $\begin{matrix} k(x + 1) & for \space x = 1, 2, 3, 4 \\ 2kx & for \space x = 5, 6, 7 \\ 0 & otherwise \end{matrix}$
where k is a constant. Calculate
(i) the value of k
(ii) E(x)
(iii) Standard deviation of X.

$\begin{matrix} k(x + 1) & for \space x = 1, 2, 3, 4 \\ 2kx & for \space x = 5, 6, 7 \\ 0 & otherwise \end{matrix}$
Thus, we have following table

(i) Since,

$\sum P_i = 1$

$\Rightarrow K (2 + 3 + 4 + 5 + 10 + 12 + 14) = 1 \Rightarrow K = \frac{1} {50}$
(ii)

$\therefore E(X) = \sum XP (X)$

$\therefore E(X) = 2k + 6k + 12k + 20k + 50k + 72k + 98k + 0 = 260k$

$= 260 \times \dfrac{1} {50} = \dfrac{26} {5} = 5.2 \left [ \therefore k = \dfrac{1} {50} \right ]$ .....(i)

(iii) We know that,

$Var(X) = [E(X^{2})] - [E(X)]^{2} = \sum X^{2} P(X) - [ \sum \left \{XP(X) \right \} ]^{2}$

$\therefore E(X) = [2k + 6k + 12k + 20k + 50k + 72k + 98k + 0] - [5.2]^{2}$ [using Eq. (i)]

$[1498k] - 27.04 = \left[ 1498 \times \dfrac{1} {50}\right ] - 27.04 \left[ \therefore k = \dfrac{1} {50}\right ]$

$= 29.96 - 27.04 = 2.92$

We know that, standard deviation of

$X = \sqrt{Var(X)} = \sqrt{2.92} = 1.7088 = 1.7$ (approx)

1781134_1802139_ans_268dbd30a767494b9b486e07c84fdb53.PNG

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$X$ $0$ $1$ $2$ $3$ $4$
$P(X)$ $0.1$ $0.5$ $0.2$ $-0.1$ $0.3$

1960761_1860091_ans_7dde8e50c00a4db6b4d09b5d50be3188.gif

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$Y$ $-1$ $0$ $1$
$P(Y)$ $0.6$ $0.1$ $0.2$

1960763_1860096_ans_5e438c0e1af946199d3669aaa3249066.gif

Let X be a random variable taking values $x_1, x_2,....... x_n$ with probabilities
$p_1, p_2,......p_n$ respectively. Then var(X) = __________.

$Var(X) = (X^{2}) - [E(X)]^{2}$

$= \sum_{i=1}^{n} X^{2}P(X) - \left [\sum_{i=1}^{n} XP(X) \right ]^{2}$

$= \sum P_i x_i - \left (\sum P_i x_i \right )^{2}$

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$Z$ $3$ $3$ $1$ $0$ $-1$
$P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

1960765_1860104_ans_269eb51a350344cf85c1d505f4612124.gif

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$X$ $0$ $1$ $2$
$P(X)$ $0.4$ $0.4$ $0.2$

1959850_1860079_ans_8c7abd86c7eb4f3886b8b106a6e83bf5.gif

Find the probability distribution of $X$ , the number of heads in a simultaneous toss of two coins.

Possible values of

$X$ are

$0,1$ or

$2$
Now

$P(X=0)=$ P(No head)

$=$ P(Tail first time and tail second time)

$=$ P(Tail first time).P(tail in second time)

$=\cfrac{1}{2}\times {1}{2}=\cfrac{1}{4}$

$=$ P(TH or HT)=P(TH)+P(HT)

$=\cfrac{1}{4}+\cfrac{1}{4}=\cfrac{1}{2}$

$P(x=2)=$ P(2 Head)=P(HH)

$=\cfrac{1}{4}$
Required is distribution

$X$	0	1	2
$P(X)$	$\cfrac { 1 }{ 4 }$	$\cfrac { 1 }{ 2 }$	$\cfrac { 1 }{ 4 }$

Complete the following table:

Completing the table:
Possibility of throwing 2 dice with

$\operatorname{six}$ faces,

$n(S)=6 \times 6=36$
1. Sum on 2 dice will be 3 means (1,2),(2,1)

$\therefore \mathrm{n}(\mathrm{A})=2$

$\therefore P(A) \quad=\dfrac{n(A)}{n(S)}=\dfrac{2}{36}$
2. Sum on 2 dice, getting 4 means(1,3),(2,2),(3,1)

$\therefore n(B)=3$

$\therefore P(B) \quad=\dfrac{n(B)}{n(S)}=\dfrac{3}{36}$
3. Sum of 2 dice, getting 5 means (1,4),(2,3),(3,2),(4,1)

$\therefore \mathrm{n}(\mathrm{C})=4$

$\therefore P(C) \quad=\dfrac{n(C)}{n(S)}=\dfrac{4}{36}$
4. Sum of 2 dice, getting 6 means ( 1, 5), ( 2, 4), (3, 3), (4, 2), (5, 1)

$\therefore \mathrm{n}(\mathrm{D})=5$

$\therefore P(D) \quad=\dfrac{n(D)}{n(S)}=\dfrac{5}{36}$
5. Sum of 2 dice, getting 7 means ( 1,6 ), (2,5),(3,4),(4,3),(5,2),(6,1)

$\therefore \mathrm{n}(\mathrm{E})=6$

$\therefore P(E) \quad=\dfrac{n(E)}{n(S)}=\dfrac{6}{36}$
6. Sum of 2 dice, getting 9 means (3,6),(4,5),(5,4),(6,3)

$\therefore \mathrm{n}(\mathrm{F})=4$

$\therefore P(F) \quad=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}$
7. Sum of 2 dice, getting 10 means (4,6),(5,5),(6,4)

$\therefore \mathrm{n}(\mathrm{G})=3$

$\therefore P(G) \quad=\dfrac{n(G)}{n(S)}=\dfrac{3}{36}$
8. Sum of 2 dice, getting sum 11 means (5,6),(6,5)

$\therefore \mathrm{n}(\mathrm{H})=2$

$\therefore P(H) \quad=\dfrac{n(H)}{n(S)}=\dfrac{2}{36}$

$\therefore$ The Table is completed as follows:
1957640_1869364_ans_0d17aecaa6a24a869186b7560f86f27c.png

The random variable X has a probability distribution $P(X)$ of the following form, where $k$ is some number.
$P(X = x) = \begin{cases}k,& \text{if} \ x = 0 \\ 2k, & \text{if} \ x=1 \\ 3k, & \text{if} \ x=2 \\ 0, & \text{otherwise} \end{cases}$
Determine the value of $k$ .

As Sum of all probabilities should be

$1$

$⇒k+2k+3k=1⇒k=\dfrac 16$

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(X>6)$

1965290_1860171_ans_4a7d44b825584b52b8d4971df38e46c0.gif

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(X<3)$

1965046_1860168_ans_a187032c28be41c7a9cc3d31e4144700.png

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$K$

1964807_1860164_ans_a9c741db28294f299c26f8c1b22550bd.png

The random variable $X$ has a probability distribution $P(X)$ of the following from, where $k$ is some number:
$P(X)=\left\{ \begin{matrix} k,\quad if\quad x=0 \\ 2k,\quad if\quad x=1 \\ 3k,\quad if\quad x=2 \\ 0,\quad otherwise \end{matrix} \right .$
Find $P(X<2),P(X\le 2),P(X\ge 2)$

1965314_1860200_ans_b0f3bf83040940d5b4f51b80b9e21aae.gif

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(0<X<3)$

1965293_1860175_ans_d0c4c9f30d0c41e99334012a14c2bb28.gif

A fair coin is tossed until a head or five tails occur. If $X$ denotes the number of tosses of the coin, find the mean of $X$ .

Given

$X$ denotes the number of coins. Coins is tossed until a head or five tails accur therefere, it is clear if on

$X=1$ , head comes then the process will be stopped and if tail comes then coin will be tossed second time. Clearly it will be repeated again and again till 5 tails come maximum
Then the value of

$X$ will be

$1,2,3,4,5$

$S=\left\{ H,TH,TTH,TTTH\quad or\quad TTTTH,TTTTT \right\}$
Probability that heat comes in first throw

$P(X=1)=\cfrac{1}{2}$
Similarly

$P(X=2)=P(TH)=P(T)P(H)=\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { 1 }{ 4 }$

$P(X=3)=P(TTH)=P(T)P(T)P(H)$

$=\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { 1 }{ 8 }$

$P(X=4)=P(TTTH)=P(T)P(T)P(T)P(H)$

$=\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { 1 }{ 16 }$

$P(X=5)=P(TTTTH\cup TTTTT)$

$=P(T)P(T)P(T)P(T)P(H)+P(T)P(T)P(T)P(T)P(T)$

$=\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } =\cfrac { 1 }{ 32 } +\cfrac { 1 }{ 32 } =\cfrac { 1 }{ 16 }$
Hence probability distribution will be like this

$X$	1	2	3	4	5
$P(X)$	$\cfrac { 1 }{ 2 }$	$\cfrac { 1 }{ 4 }$	$\cfrac { 1 }{ 8 }$	$\cfrac { 1 }{ 16 }$	$\cfrac { 1 }{ 16 }$

Mean

$=\sum _{ i=1 }^{ 5 }{ { p }_{ i }{ x }_{ i } } =1\times \cfrac { 1 }{ 2 } +2\times \cfrac { 1 }{ 4 } +3\times \cfrac { 1 }{ 8 } +4\times \cfrac { 1 }{ 16 } +5\times \cfrac { 1 }{ 16 } =\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } +\cfrac { 3 }{ 8 } +\cfrac { 1 }{ 4 } +\cfrac { 5 }{ 16 }$

$=\cfrac { 16+16+12+8+5 }{ 32 } =\cfrac { 31 }{ 16 } =1.9$

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age $X$ of the selected student is recorded. What is the probability distribution of the random variable $X$ ? Find mean, variance and standard deviation of $X$ .

There are 15 students in the class. Each student has the same chance to be chosen.
Therefore, the probability of each student to be selected is

$\displaystyle\frac{1}{15}$ .
The given information can be compiled in the frequency table as follows.

$P\left( X=14 \right) =\displaystyle\frac { 2 }{ 15 } , P\left( X=15 \right) =\displaystyle\frac { 1 }{ 15 } , P\left( X=16 \right) =\displaystyle\frac { 2 }{ 15 } , P\left( X=16 \right) =\displaystyle\frac { 3 }{ 15 } , P\left( X=18 \right) =\displaystyle\frac { 1 }{ 15 } , P\left( X=19 \right) = \displaystyle\frac { 2 }{ 15 } , P\left( X=20 \right) =\displaystyle\frac { 3 }{ 15 } , P\left( X=21 \right) =\displaystyle\frac { 1 }{ 15 }$
Therefore, the probability distribution of random variable

$X$ is as follows.
Then, mean of

$X = E\left( X \right)$

$=\sum { { X }_{ i }P\left( { X }_{ i } \right) }$

$=14\times \displaystyle\frac { 2 }{ 15 } +15\times \displaystyle\frac { 1 }{ 15 } +16\times \displaystyle\frac { 2 }{ 15 } +17\times \displaystyle\frac { 3 }{ 15 } +18\times \displaystyle\frac { 1 }{ 15 } +19\times \displaystyle\frac { 2 }{ 15 } +20\times \displaystyle\frac { 3 }{ 15 } +21\times \displaystyle\frac { 1 }{ 15 }$

$=\displaystyle\frac { 1 }{ 15 } \left( 28+15+32+51+18+38+60+21 \right)$

$=\displaystyle\frac { 263 }{ 15 }$

$=17.53$

$E\left( { X }^{ 2 } \right) =\sum { { X }_{ i }^{ 2 }P\left( { X }_{ i } \right) }$

$={ \left( 14 \right) }^{ 2 }\cdot \displaystyle\frac { 2 }{ 15 } +{ \left( 15 \right) }^{ 2 }\cdot \displaystyle\frac { 1 }{ 15 } +{ \left( 16 \right) }^{ 2 }\cdot \displaystyle\frac { 2 }{ 15 } +{ \left( 17 \right) }^{ 2 }\cdot \displaystyle\frac { 3 }{ 15 } +{ \left( 18 \right) }^{ 2 }\cdot \displaystyle\frac { 1 }{ 15 } +{ \left( 19 \right) }^{ 2 }\cdot \displaystyle\frac { 2 }{ 15 } +{ \left( 20 \right) }^{ 2 }\cdot \displaystyle\frac { 3 }{ 15 } +{ \left( 21 \right) }^{ 2 }\cdot \displaystyle\frac { 1 }{ 15 }$

$=\displaystyle\frac { 1 }{ 15 } \cdot \left( 392+225+512+867+324+722+1200+441 \right)$

$=\displaystyle\frac { 4683 }{ 15 }$

$=312.2$

$\therefore$ Variance

$\left( X \right) =E\left( { X }^{ 2 } \right) { \left[ E\left( X \right) \right] }^{ 2 }$

$=312.2-{ \left( \displaystyle\frac { 263 }{ 15 } \right) }^{ 2 }$

$=312.2-307.4177$

$=4.7823$

$\approx 4.78$
Standard deviation

$=\sqrt { Variance\left( X \right) }$

$=\sqrt { 4.78 }$

$=2.186$

$\approx 2.19$

In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take $X = 0$ if he opposed, and $X = 1$ if he is in favour. Find $E\left(X\right)$ and $Var\left(X\right)$ .

It is given that

$P\left(X = 0\right) = 30$ % = \displaystyle\frac{30}{100} = 0.3$$

$P\left(X = 1\right) = 70$ % = \displaystyle\frac{70}{100} = 0.7$$
Therefore, the probability distribution is as follows.
Then,

$E\left( X \right) =\sum { { X }_{ i }P\left( { X }_{ i } \right) }$

$=0\times 0.3+1\times 0.7$

$=0.7$

$E\left( { X }^{ 2 } \right) =\sum { { X }_{ i }^{ 2 }P\left( { X }_{ i } \right) }$

$={ 0 }^{ 2 }\times 0.3+{ \left( 1 \right) }^{ 2 }\times 0.7$

$=0.7$
It is known that,

$Var\left( X \right) =E\left( { X }^{ 2 } \right) -{ \left[ E\left( X \right) \right] }^{ 2 }$

$=0.7-{ \left( 0.7 \right) }^{ 2 }$

$=0.7-0.49$

$=0.21$

A die is thrown, a man $C$ gets a prize of Rs. $5$ if the die turns up $1$ and gets a prize of Rs. $3$ if the die turns up $2,$ else he gets nothing, A man $A$ whose probability of speaking the truth is $\displaystyle \frac{1}{2}$ tells $C$ that the die has turned up $1$ and another man $B$ whose probability of speaking the truth is $\displaystyle \frac{2}{3}$ tells $C$ that the die has turned up $2.$ Find the expectation value of $C.$

Let

$A$ and

$B$ be the events '

$A$ speaks the truth' and '

$B$ speaks the truth' respectively, Then

$\displaystyle P\left ( A \right )=\frac{1}{2}$ and

$P\left ( B \right )=\frac{2}{3}.$
The experiment consists of three hypothesis
(i) the die turns up

$1$
(ii) the die turns up

$2$
(iii) the die turns up

$3, 4, 5$ or

$6$
Let these be the events

$E_{1}, E_{2}$ and

$E_{3}$ respectively then

$\displaystyle P\left ( E_{1} \right )=P\left ( E_{2} \right )=\frac{1}{6}$ and

$\displaystyle P\left ( E_{3} \right )=\frac{4}{6}.$
Let

$E$ be the event that the statements made by

$A$ and

$B$ agree to the same conclusion.

$\therefore$

$\displaystyle P\left ( E/E_{1} \right )=P\left ( A \right ).P\left ( \bar{B} \right )=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

$\displaystyle P\left ( E/E_{2} \right )=P\left ( \bar{A} \right ).P\left ( B \right )=\frac{1}{2}\times \frac{2}{3}=\frac{2}{6}$

$\displaystyle P\left ( E/E_{3} \right )=P\left ( \bar{A} \right ).P\left ( \bar{B} \right )=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$

$P\left ( E \right )=P\left ( E_{1} \right )P\left ( E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{2} \right )+P\left ( E_{3} \right )P\left ( E/E_{3} \right )$

$\displaystyle =\frac{1}{6}.\frac{1}{6}+\frac{1}{6}.\frac{2}{6}+\frac{4}{6}.\frac{1}{6}=\frac{7}{36}$
Thus

$P\left ( E_{1}/E \right )=\frac{P\left ( E_{1} \right )P\left ( E/E_{1} \right )}{P\left ( E \right )}=\frac{1}{7}$

$\displaystyle P\left ( E_{2}/E \right )=\frac{P\left ( E_{2} \right )P\left ( E/E_{2} \right )}{P\left ( E \right )}=\frac{2}{7}$

$\displaystyle P\left ( E_{3}/E \right )=\frac{P\left ( E_{3} \right )P\left ( E/E_{3} \right )}{P\left ( E \right )}=\frac{4}{7}$

$\therefore$ expectation of

$\displaystyle C=\frac{1}{7}\times 5+\frac{2}{7}\times 3+0=$ Rs.

$\displaystyle \frac{11}{7}$

The numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X and hence find the means of the distribution.

Variable (X)	$2$	$3$	$4$	$5$	$6$
Probability P(X)	$\dfrac{1}{15}$	$\dfrac{2}{15}$	$\dfrac{3}{15}$	$\dfrac{4}{15}$	$\dfrac{5}{15}$

First six numbers are

$1, 2, 3, 4, 5, 6.$

X is bigger number among two numbers.

If X

$= 2$ for P(X)

$=$ probability of event that bigger of the two chosen number is

$2.$

So, Cases

$= (1, 2)$

So,

$P(X)=\dfrac{1}{^6C_2}=\dfrac{1}{15}$ ---(1)

If X

$= 3$

So, favourable cases are

$= (1, 3), (2, 3)$

$P(X)=\dfrac{2}{^6C_{2}}=\dfrac{2}{15}$ ---(2)

If X

$= 4$

So, favourable cases are

$= (1, 4), (2, 4), (3, 4)$

$P(X)=\dfrac{3}{^6C_{2}}=\dfrac{3}{15}$ ---(3)

If X

$= 5$

So, favourable cases are

$= (1, 5), (2, 5), (3, 5), (4, 5)$

$P(X)=\dfrac{4}{^6C_{2}}=\dfrac{4}{15}$ ---(4)

If X

$= 6$

So, favourable cases are

$= (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)$

$P(X)=\dfrac{5}{^6C_{2}}=\dfrac{5}{15}$ ---(5)

We can put all value of

$P(X),$

Required mean =

$2.\left ( \dfrac{1}{15} \right )+3.\left ( \dfrac{2}{15} \right )+4.\left ( \dfrac{3}{15} \right )+5.\left ( \dfrac{4}{15} \right )+6.\left ( \dfrac{5}{15} \right )$

$=\dfrac{70}{15}\Rightarrow \dfrac{14}{3}$

Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.

The numbers are 1,2,3,4,5,6

Let X be the bigger number among the two number.

i.e., X = 2,3,4,5,6

$P(x=2)=P(1,2)$

$\implies =\dfrac 1{^6C_2}=\dfrac 1{15}$

$P(x=3)=P(1,3)(2,3)$

$\implies =\dfrac 2{^6C_2}=\dfrac 2{15}$

$P(x=4)=P(1,4)(2,4)(3,4)$

$\implies =\dfrac 3{^6C_2}=\dfrac 3{15}=\dfrac 15$

$P(x=5)=P(1,5)(2,5)(3,5)(4,5)$

$\implies =\dfrac 4{^6C_2}=\dfrac 4{15}$

$P(x=6)=P(1,6)(2,6)(3,6)(4,6)(5,6)$

$\implies =\dfrac 5{^6C_2}=\dfrac 5{15}=\dfrac 13$

Hence the probability distribution is :

X	2	3	4	5	6
p(X=x)	$\dfrac 1{15}$	$\dfrac 2{15}$	$\dfrac 15$	$\dfrac 4{15}$	$\dfrac 13$

Mean =

$2\left(\dfrac 1{15}\right)+3\left(\dfrac 2{15}\right)+4\left(\dfrac 15\right)+5\left(\dfrac 4{15}\right)+6\left(\dfrac 13\right)$

$\implies =\dfrac 2{15}+\dfrac 65+\dfrac {10}3$

$\implies =\dfrac 2{15}+\dfrac {68}{15}\\\implies \dfrac {70}{15}$

Hence, Mean =

$\dfrac {14}3$

Using Variance formula,

$Var(X)=(X-\mu)^2\times P(X=x)\\\implies Var(X)=\left(2-\dfrac {14}3\right)^2\times \dfrac 1{15}+\left(3-\dfrac {14}3\right)^2\times \dfrac 2{15}+\left(4-\dfrac {14}3\right)^2\times \dfrac 1{5}+\left(5-\dfrac {14}3\right)^2\times \dfrac 4{15}+\left(6-\dfrac {14}3\right)^2\times \dfrac 1{3}\\\implies Var(X)=\left(-\dfrac {8}3\right)^2\times \dfrac 1{15}+\left(-\dfrac {5}3\right)^2\times \dfrac 2{15}+\left(-\dfrac {2}3\right)^2\times \dfrac 1{5}+\left(\dfrac {1}3\right)^2\times \dfrac 4{15}+\left(\dfrac {4}3\right)^2\times \dfrac 1{3}\\\implies Var(X)=\dfrac {64}{9\times 15}+\dfrac {50}{9\times 15}+\dfrac {4}{9\times 5}+\dfrac {4}{9\times 15}+\dfrac {16}{9\times 3}\\\implies Var(X)=\dfrac {64+50+12+4+80}{9\times 15}=\dfrac {210}{135}=\dfrac {14}{9}$

Hence, the variance is

$\dfrac {14}9$

From a lot of $15$ bulbs which include $5$ defectives, a sample of $4$ bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.

Let

$X =$ number of defective bulbs out of 4 drawn

$= 0, 1, 2, 3, 4$

Probability of defective bulb =

$\dfrac{5}{15}=\dfrac{1}{3}$

Probability of non-defective bulb =

$1-\dfrac{1}{3}=\dfrac{2}{3}$

$P(X=0)=^4C_0. (\dfrac{2}{3})^4=\dfrac{16}{81}$

$P(X=1)=^4C_1. \dfrac{1}{3}. (\dfrac{2}{3})^3=\dfrac{32}{81}$

Similarly, find the probability for

$x=2,3,4.$

Probability distribution is shown in the table:

Mean

$=$

$\sum xP(x)=\dfrac{108}{81}\Rightarrow \dfrac{4}{3}.$

Three numbers are selected at random (without replacement) from first six positive integers. Let $X$ denote the largest of the three numbers obtained. Find the probability distribution of $X$ . Also, find the mean and variance of the distribution.

$x$	$p(x)$	$xp(x)$	$x^2p(x)$

First six positive integers are

$1, 2, 3, 4, 5, 6.$

Therefore, Three numbers are selected at random without replacement so, total no. of ways

$^6C_3=20$ .

Let,

$x$ denote the larger of three numbers.

So,

$x$ can take values

$3, 4, 5, 6.$

$p(x=3)=\dfrac{1}{20}$

$p(x=4)=\dfrac{3}{20}$

$p(x=5)=\dfrac{6}{20}$

$p(x=6)=\dfrac{10}{20}$

Refer the tabular column:

$mean = \sum xp(x)$

$=\dfrac{3}{20}+\dfrac{12}{20}+\dfrac{30}{20}+\dfrac{60}{20}$

$=\dfrac{150}{20}=5.25$

Variance

$= \sum x^2p(x)-\left ( \sum xp(x) \right )^2$

$=\left ( \dfrac{567}{20} \right )-\left ( \dfrac{105}{20} \right )^2$

$=28.35-27.56=0.79.$

Let the p.m.f. (probability mass function) of random variable $x$ be
$P(x) = \begin{cases} \left( \dfrac{4}{x} \right ) \left( \dfrac{5}{9}\right)^x \left(\dfrac{4}{x} \right )^{4 - x},\ x = 0, 1, 2, 3, 4\\ 0,\,\,\, \text{otherwise} \end{cases}$
Find $E(x)$ and Var $(x)$ .

$P(x) =\left( \dfrac{4}{x} \right ) \left( \dfrac{5}{9} \right )^x \left( \dfrac{4}{x} \right )^{4 - x}$ ,

$x = 0, 1, 2, 3, 4$ .

$= 0$ otherwise
The probability distribution table of the function is

$x$	$0$	$1$	$2$	$3$	$4$
$P(x)$	$\infty$	$\dfrac{1280}{9}$	$\dfrac{200}{81}$	$\dfrac{2000}{6561}$	$\dfrac{625}{6561}$
$x^2$	$0$	$1$	$4$	$9$	$16$
$P(x). x^2$	$0$	$\dfrac{1280}{9}$	$\dfrac{800}{81}$	$\dfrac{18000}{6561}$	$\dfrac{10000}{6561}$

$E (x) =\displaystyle \sum_{x =1}^{4} x . P(x)$

$= 0 \times \infty + 1 \times \dfrac{1280}{9} + 2 \times \dfrac{200}{81} + 3 \times \dfrac{2000}{6561} + 4 \times \dfrac{625}{6561}$

$= 148.4560$

Var

$(x)=\displaystyle \sum x^2P(x) - \left(\sum xP(x)\right)^2= \left( 0 + \dfrac{1280}{9} + \dfrac{800}{81} + \dfrac{18000}{6561} + \dfrac{10000}{6561} \right ) - (148.5660)^2$

$= 156.366 - (148.4560)^2$

The time (in minute) for a lab assistant to prepare the equipment for a certain experiment is a random variable $X$ taking values between $25$ and $35$ minutes with p.d.f.
$f(x) = \begin{cases} \dfrac {1}{10}, 25\leq x \leq 35\\ \\ 0,\,\,\, \text{otherwise} \end{cases}$ .
What is the probability that preparation time exeeds $33$ minutes? Also find the c.d.f. of $X$ .

$f(x) = \cfrac {1}{10}$

Required probability

$= P(X > 33)$

$= \displaystyle \int _{33}^{\infty} f(x) dx$

$= \displaystyle \int _{33}^{35} f(x) dx + \displaystyle \int _{35}^{\infty} f(x) dx$

$\displaystyle \int _{33}^{35} f(x) dx + 0 [\because f(x) = 0$ when

$x > 35]$

$= \displaystyle \int _{33}^{35} \cfrac {1}{10}dx$ .

$= \cfrac {1}{10} \displaystyle \int _{33}^{35} 1dx$ .

$= \cfrac {1}{10} [x]_{33}^{35}$

$= \cfrac {1}{10} [35 - 33] = \cfrac {1}{5}$

Let

$F(x)$ be the c.d.f. of

$X$ .

$\therefore F(x) = P[X\leq x]$

$= \displaystyle \int _{-\infty}^{\infty} f(x) dx$

$= \displaystyle \int _{-\infty}^{25} f(x)dx + \displaystyle \int _{25}^{x}f(x) dx$

$= 0 + \displaystyle \int _{25}^{x} f(x) dx [\because f(x) = 0$ , when

$f(x) < 25]$

$= \displaystyle \int _{25}^{x} \cfrac {1}{10}dx$

$= \cfrac {1}{10}[x]_{25}^{x}$

$= \cfrac {1}{10}[x - 25]$

$\therefore F(x) = \cfrac {x - 25}{10}$ .

The range of a random variable $X$ is $\left \{0, 1, 2\right \}$ . Given that $P(X = 0) = 3C^{3}, P(X = 1) = 4C - 10C^{2}, P(X = 2) = 5C - 1$
i) Find the value of $C$ .
ii) $P(X < 1), P(1 < X \leq 2)$ and $P(0 < X \leq 3)$

(i) We must have

$\sum P(X)=1$

$\Rightarrow P(X=0)+P(X=1)+P(X=2)=1$

$\Rightarrow (3C^3)+(4C-10C^2)+(5C-1)=1$

$\Rightarrow 3C^3-10C^2+9C-2=0$

$\Rightarrow (C-1)(3C^2-7C+2)=0$

$\Rightarrow (C-1)(C-2)(3C-1)=0$

$\Rightarrow C=1,2,\dfrac{1}{3}$

Also

$0\le P(X=0), P(X=1), P(X=2)\le 1$

So only acceptable value of

$C$ is

$\dfrac{1}{3}$ .

(ii)

$P(X<1)=P(X=0)=3\left(\dfrac{1}{3}\right)^3=\dfrac{1}{9}$

(iii)

$P(1<X\le 2)=P(X=2)=4C-10C^2=4\cdot \dfrac{1}{3}-10\cdot \dfrac{1}{9}=\dfrac{2}{9}$

And (iv)

$P(0<X\le 3)=1-P(X=0)=1-\dfrac{1}{9}=\dfrac{8}{9}$

A random variable ' $X$ ' has the following probability distribution:
X=x 0 1 2 3 4 5 6 7
P(X=x) $0$ $k$ $2k$ $2k$ $3k$ ${k}^{2}$ $2{k}^{2}$ $7{k}^{2}+k$
Find:
(i) $k$
(ii) The Mean and
(iii) $P(0< X< 5)$

Sum of all probabilities must be equal to

$1$

So we get

$K+2K+2K+3K+{K}^{2}+2{K}^{2}+7{K}^{2}+K=1$

$\Rightarrow 10{K}^{2}+9K=1$

$\Rightarrow K = -1,\dfrac{1}{10}$

But probability cannot be negative , therefore the value of

$K=\dfrac{1}{10}$

The mean is

$1.K+2.2K+3.2K+4.3K+5{K}^{2}+6.2{K}^{2}+7.7{K}^{2}+7K$

$= 30K+66{K}^{2}=3+0.66=3.66$

$P(0<x<5)=P(x=1)+P(x=2)+P(x=3)+P(x=4)=K+2K+2K+3K=8K=0.8$

Find the mean and variance for the probability density function : $f(x)=\begin{cases} \alpha { e }^{ -\alpha x }\,\,\, \text{if}\, x>0 \\ 0\,\,\, \text{otherwise} \end{cases}$

$f(x)=\begin{cases} \alpha { e }^{ -\alpha x }\,\,\, \text{if}\, x>0 \\ 0\,\,\, \text{otherwise} \end{cases}$

$E(X)=\displaystyle \int _{ -\infty }^{ \infty }{ xf(x) } dx=\int _{ 0 }^{ \infty }{ x(\alpha )\left( { e }^{ -\alpha x } \right) dx }$

$=\alpha \displaystyle \int _{ 0 }^{ \infty }{ x{ e }^{ -\alpha x } } dx=\cfrac { \alpha (1) }{ { \alpha }^{ 2 } } =\cfrac { 1 }{ \alpha }$
[By Bernoulli's formula

$\displaystyle \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -\alpha x } } dx=\cfrac { n }{ { a }^{ n+1 } }$ ]

$E({ X }^{ 2 })=\displaystyle \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \alpha { e }^{ -\alpha x } \right) dx }$

$\left[\displaystyle \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -\alpha x } } dx=\cfrac { n }{ { a }^{ n+1 } } \right] =\alpha \displaystyle \int _{ 0 }^{ \infty }{ { x }^{ 2 }{ e }^{ -\alpha x } } dx=\cfrac { \alpha (2) }{ { \alpha }^{ 3 } } =\cfrac { 2 }{ { \alpha }^{ 2 } }$
[By Bernoulli's formula]
Mean

$=E(X)=\cfrac { 1 }{ \alpha }$
Variance

$=E({ X }^{ 2 })-{ \left[ E(X) \right] }^{ 2 }=\cfrac { 2 }{ { \alpha }^{ 2 } } -\cfrac { 1 }{ { \alpha }^{ 2 } } =\cfrac { 1 }{ { \alpha }^{ 2 } }$

A random variable $X$ has the following probability distribution.
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^{2}$ $2k^{2}$ $7k^{2} + k$
Find (i) $k$ (ii) The mean and (iii) $P(0 < X < 5)$

(i) The total probability should add up to

$1$ .

$\therefore 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$

$\therefore 10k^2 + 9k - 1 = 0$

$\therefore 10k^2 + 10k - k - 1 = 0$

$\therefore (10k - 1)(k + 1) = 0$

Since

$k$ cannot be negative,

$k = \cfrac{1}{10}$

(ii) Mean

$= \sum x_i P_i$

$= 0 \times 0 + 1 \times k + 2 \times 2k + 3 \times 2k + 4 \times 3k + 5 \times k^2 + 6 \times 2k^2 + 7 \times (7k^2 + k)$

$= k + 4k + 6k + 12k + 5k^2 + 12k^2 + 49k^2 + 7k$

$= 30k + 66k^2$

$= 3 + 0.66$

$= 3.66$

(iii)

$P(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$\therefore P(0 < X < 5) = k + 2k + 2k + 3k = 9k = 0.9$

Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.

Let

$X$ denote the number of aces drawn in drawing two cards with replacement.

$\therefore X$ can take the values

$0, 1$ and

$2$ .
Let

$P(S) = P$ (getting aces)

$= \dfrac {4}{52}\Rightarrow P(F)$

$= P$ (not getting aces)

$= \dfrac {48}{52}$

$P(X = 0) = P (no\ ace) = P(FF) = P(F)$

$P(F) = \dfrac {48}{52}\cdot \dfrac {48}{52} =\dfrac {144}{169}$

$P(X = 1) = P$ (one ace)

$= P(SF\ or\ FS) = P(S) .P(F) + P(F) P(S)$

$= 2(PS) . P(F) = 2\times \dfrac {4}{52}\times \dfrac {48}{52} = \dfrac {24}{169}$

$P(X - 2) = P$ (two aces)

$= P(SS) = P(S)$ .

$P(S) = \dfrac {4}{52}\cdot \dfrac {4}{52} = \dfrac {1}{169}$

$X$	$0$	$1$	$2$
$P(X = x)$	= $\dfrac {144}{169}$	$\dfrac {24}{169}$	$\dfrac {1}{169}$

Mean

$= E[X] = \displaystyle \sum_{-\infty}^{\infty} x_{i}p_{i} =(0) \left (\dfrac {144}{169}\right ) + (1)\left (\dfrac {24}{169}\right ) + (2)\left (\dfrac {1}{169}\right )$

$= \dfrac {24 + 2}{169} =\dfrac {26}{169} = \dfrac {2}{13}$

$E(X^{2}) = \displaystyle \sum_{x = 0}^{2} x_{i}^{2} p_{i} = (0^{2}) \left (\dfrac {144}{169}\right ) + (1^{2})\left (\dfrac {24}{169}\right ) + (2^{2})\left (\dfrac {1}{169}\right )$

$= \dfrac {24 + 4}{169} = \dfrac {28}{169}$
Variance

$(X) = E[X^{2}] - (E(X))^{2}$

$= \dfrac {28}{169} - \left (\dfrac {2}{13}\right )^{2} = \dfrac {28}{169} - \dfrac {4}{169} = \dfrac {24}{169}$

If the probability density function of a random variable is given by,
$f(x) = \left\{\begin{matrix} k(1 - x^2),& 0 < x < 1\\ 0, & \text{elsewhere}\end{matrix} \right.$ find $k$ and the distribution function of the random variable.

(i) Since

$f(x)$ is a p.d.f.

$\displaystyle \int_{- \infty}^{\infty} f(x) dx = 1$

$\Rightarrow \displaystyle \int_0^1 k(1 - x^2) dx =1$

$\Rightarrow k \left [ x - \dfrac{x^3}{3} \right ]_0^1 = 1$

$\Rightarrow k \left( 1 - \dfrac{1}{3} \right ) = 1$

$\Rightarrow \dfrac{2k}{3} = 1$

$\Rightarrow k = \dfrac{3}{2}$
(ii) The distribution function

$F(x) = \displaystyle \int_{- \infty}^{x} f(t) dt$
(a) When

$x \in (-\infty, 0]$

$F(x) = \displaystyle \int_{-\infty}^{x} f(t) dt = 0$
(b) When

$x \varepsilon (0, 1]$

$F(x) =\displaystyle \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^0 f(t) dt + \int_0^x f(t) dt$

$= 0 + \dfrac{3}{2}\displaystyle \int_0^x (1 - t^2) dt$

$F(x) = \dfrac{3}{2} (x - \dfrac{x^3}{3})$
(c) When

$x \in [1, \infty)$

$F(x) = \displaystyle \int_{-\infty}^x f(t) dt$

$=\displaystyle \int_{-\infty}^0 f(t) dt + \int_0^1 f(t) dt + \int_1^x f(t) dt$

$= 0 +\displaystyle \int_0^1 \dfrac{3}{2} (1 - t^2) dt + 0$

$= \dfrac{3}{2} \left [ t \dfrac{t^3}{3} \right ]_0^1 = 1$

$\therefore F(x) = \left\{\begin{matrix} 0& -\infty < x \leq 0\\ \dfrac{3}{2} \left( x- \dfrac{x^3}{3} \right ), & 0 < x < 1 \\ 1 & 1 \leq x \infty \end{matrix}\right.$

A random variable $X$ has the following probability mass function.
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$
(a) Find $k$ .
(b) Evaluate $P(X < 4), P (X$ $\geq$ $5)$ and $P(3< X$ $\leq$ $6)$ .
(c) What is the smallest value of $x$ for which $P(X$ $\leq$ $x) >$ $\dfrac{1}{2}$ ?

(a)

$P(X = x)$ is a probability mass function.

$\displaystyle \sum_{x = 0}^{6} P(X = x) = 1$
(i.e.,)

$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 1$ .

$\Rightarrow k + 3k + 5k + 7k + 9k + 11 k + 13k = 1$

$\Rightarrow 49 k = 1 \Rightarrow k = \dfrac{1}{49}$
(b)

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$= \dfrac{1}{49} + \dfrac{3}{49} + \dfrac{5}{49} + \dfrac{7}{49} = \dfrac{16}{49}$

$P(X \geq 5) = P(X = 5) + P(X = 6) = \dfrac{11}{49} + \dfrac{13}{49} = \dfrac{24}{49}$

$P(3 < x \leq 6) = P(X = 4) +P(X = 5) + P(X = 6)$

$= \dfrac{9}{49} + \dfrac{11}{49} + \dfrac{13}{49} = \dfrac{33}{49}$
(c) The minimum value of

$x$ may be determined by trial and error method.

$P(X \leq 0) = \dfrac{1}{49} < \dfrac{1}{2} ; P (X \leq 1) = \dfrac{4}{49} < \dfrac{1}{2}$

$P(X \leq 2) = \dfrac{9}{49} < \dfrac{1}{2} ; P (X \leq 3) = \dfrac{16}{49} < \dfrac{1}{2}$

$P(X \leq 4) = \dfrac{25}{49} > \dfrac{1}{2}$

$\therefore$ the smallest value of

$x$ for which

$P(X \leq x) > \dfrac{1}{2}$ is

$4$ .

Given the probability density function (p.d.f) of a continuous random variable $X$ as,
$f(x) = \dfrac {x^{2}}{3}, -1 < x < 2 = 1$ , otherwise
Determine the cumulative distribution function (c.d.f.) of $X$ and hence find $P(X < 1), P(X > 0), P(1 < X < 2)$

Given the PDF of a continuous random variable

$X.$

$\Rightarrow f\left( x \right) =\dfrac { { x }^{ 3 } }{ 3 } ,-1<x<2=1$

$f\left( x \right) =\int _{ -\infty }^{ \alpha }{ f\left( x \right) dx } =\int _{ \alpha }^{ \beta }{ \dfrac { 1 }{ \beta -\alpha } dx } =\left[ \dfrac { 1 }{ \beta -\alpha } x \right] _{ \infty }^{ x }=\dfrac { x-\alpha }{ \beta -\alpha }$

$f\left( x \right) =\int { \dfrac { 1 }{ 2-(-1) } ,-1<x<2}$

$=\int { \dfrac{1}{3},-1<x<2}$

${0,otherwise }$

$\Rightarrow$ When,

$P(1<x<2)=\int _{ 1 }^{ 2 }{ \dfrac{1}{3} }\;dx=\left[ \dfrac { 1 }{ 3 } x \right] _{ 0 }^{ 1 }=\dfrac { 1 }{ 3 }$

When

$P(x>0)=0$

When

$P(1<x<2)=\int _{ 1 }^{ 2 }{ \dfrac{1}{3} }\;dx=\left[ \dfrac { x }{ 3 } x \right] _{ 1 }^{ 2 }=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 }$

Hence, solved.

A die is tossed twice. A success is getting an odd number on a toss. Find the mean and the variance of the probability distribution of the number of successes.

Let

$x$ be the random variable denoting the number of times an odd number (the number of successes) when a die is tossed twice.

Then

$x$ takes the values

$0,1,2$

Let

$P(X=0)$ be probability of getting no odd number (both times showing even).

$\therefore P(X=0)=\dfrac{3}{6}\times \dfrac{3}{6}=\dfrac{1}{4}$

Let

$P(X=1)$ be probability of getting odd number once.

$\therefore P(X=1)=2C_1\dfrac{3}{6}\times \dfrac{3}{6}=\dfrac{6}{6}\times \dfrac{3}{6}=\dfrac{1}{2}$

Let

$P(X=2)$ be probability of getting odd number twice.

$\therefore P(X=2)=\dfrac{3}{6}\times \dfrac{3}{6}=\dfrac{1}{4}$

Thus the probability distribution of

$X$ is given by

$X=x$	$x=0$	$x=1$	$x=2$
$P(X=x)$	$\dfrac{1}{4}$	$\dfrac{1}{2}$	$\dfrac{1}{4}$

We know that mean

$E(X)=\sum x_iP_i=0\times \dfrac{1}{4}+1\times \dfrac{1}{2}+2\times \dfrac{1}{4}$

$\therefore E(X)=0+\dfrac{1}{2}+\dfrac{1}{2}=1$

Thus mean

$E(X)=1$ .

We know that

$var(X)=E(X^2)-[E(X)]^2$

$E(X^2)=\sum x_i^2P_i=0\times \dfrac{1}{4}+1^2\times \dfrac{1}{2}+2^2\times \dfrac{1}{4}$

$\therefore E(X^2)=0+\dfrac{1}{2}+4\times \dfrac{1}{4}=\dfrac{3}{2}$

Thus

$var(X)=\dfrac{3}{2}-[1]^2=\dfrac{3}{2}-1=\dfrac{1}{2}$

Hence mean is

$1$ and variance is

$\dfrac{1}{2}$ .

A player tosses $2$ fair coins. He wins Rs $5$ if $2$ heads appear, Rs $2$ if head appears and Rs $1$ if no head appears. Find his expected winning amount and variance of winning amount.

No. of coin

$=2$

$\Rightarrow$ Sample space

$=2^2=4=[HH,HT,TH,TT]$

$\Rightarrow$ Probability of getting

$2$ heads, i.e,

$P_1=\dfrac{1}{4}$

$\therefore n_1=Rs. 5$

Probability of getting

$1$ head, i.e,

$P_1=\dfrac{2}{4}$

$\therefore n_2=Rs. 2$

Probability of getting no head, i.e,

$P_3=\dfrac{1}{4}$

$\therefore n_3=Rs. 1$

$\Rightarrow$ Expectation

$=n_1p_1+n_2p_2+n_3p_3$

$=\dfrac{5}{4}+\dfrac{2\times2}{4}+\dfrac{1}{4}$

$=\dfrac{5+4+1}{4}$

$=\dfrac{10}{4}$

$=\dfrac{5}{2}$

Hence, the answer is

$\dfrac{5}{2}.$

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(X\leq 3/X > 0)$ .

$P\left( X\le 3\right)=P\left( X=-3\right) +P\left( X=-1\right) +P\left( X=0\right)+P\left( X=1\right) +P\left( X=3\right)$

$=0.1+0.2+0.2+0.5+0.1$

$=0.75$

$\Rightarrow P\left( X>0\right)=P\left( X=1\right) +P\left( X=3\right) +P\left( X=5\right)+P\left( X=7\right) +P\left( X=9\right)$

$=0.15+0.1+0.1+0.05+0.1$

$=0.5$

$\Rightarrow \dfrac{P\left( X\le 3\right)}{P\left( X>0\right)}=\dfrac{0.75}{0.5}=\dfrac{7.5}{5}=1.5$

Hence, the answer is

$1.5.$

1229829_1347460_ans_88f3d900989944e986e0009c37468389.png

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(-1 \leq X < 2)$ .

$P\left( -1 \le X\le 2\right) =P\left( X=-1\right) +P\left( X=0\right) +P\left( X=1\right)$

$=0.2+0.2+0.15$

$=0.55$

Hence, the answer is

$0.55$

1229235_1347442_ans_0e56af1e276e4b5a8816bf5e38e86f35.png

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(X \geq 5)$ .

$P\left( X\ge 5\right)=1-P\left( X<5\right)$

$=1-0.75$

$=0.25$

Hence, the answer is

$0.25.$

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find the p.m.f. of X.

1379962_1347434_ans_7ca975f3bb364528a1ca0f2a2a8f264f.png

A die is thrown $120$ times and getting $1$ or $5$ is considered a success. Find the mean and variance of the number of sucess.

Given: dice is thrown

$120$ times

$1$ or

$5$ is success

$\therefore\,p=$ Probability of success

$=\dfrac{2}{6}=\dfrac{1}{3}$

$\therefore\,q=$ Probability of failure =

$1-\dfrac{1}{3}=\dfrac{2}{3}$

$\therefore Mean =np$

$120\times \dfrac{1}{3}=40$

$(\because n=120)$

$Variance =npq$

$120\times \dfrac{1}{3}\times \dfrac{2}{3}=\dfrac{80}{3}$

$X=x$	$1$	$2$	$3$
$P(X=x)$	$\cfrac{1}{5}$	$\cfrac{2}{5}$	$\cfrac{2}{5}$

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\dfrac {1}{8}$	$\dfrac {3}{8}$	$\dfrac {3}{8}$	$\dfrac {1}{8}$

X =	1	2	3	4	5
P(X =)	$\dfrac{1}{2}$	$\dfrac{1}{4}$	$\dfrac{1}{8}$	$\dfrac{1}{16}$	P

X=x	0	1	2	3	4	5	6	7
P(X=x)	$0$	$k$	$2k$	$2k$	$3k$	${k}^{2}$	$2{k}^{2}$	$7{k}^{2}+k$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^{2}$	$2k^{2}$	$7k^{2} + k$

X	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

X	1	2	3	4
P(X)	2k	3k	3k	2k

Probability Distribution And Its Mean And Variance - Class 12 Commerce Applied Mathematics - Extra Questions

The probability mass function (p.m.f) of XX is given below:X=xX=x112233P(X=x)P(X=x)15\cfrac{1}{5}25\cfrac{2}{5}25\cfrac{2}{5}

The mean and variance of a Binomial distribution is 44 and 33 respectively. Find its parameters.

The random variable x has a probability distribution P(x)P(x) of the following form where kk is some number:P(x)={kifx=02kifx=13kifx=20otherwiseP(x)=\begin{cases}k & if\, x=0\\2k&if\, x=1 \\3k&if\, x=2\\0&otherwise \end{cases}Determine the value of kk and P(X≤2).P(X \le 2).

Find the variance and standard deviation of the random variable XX whose probability distribution is given below:xx00112233P(X=x)P(X = x)18\dfrac {1}{8}38\dfrac {3}{8}38\dfrac {3}{8}18\dfrac {1}{8}

The probability that a doctor successfully performs an operation is 80%80\%. What is the value of 625625 times the probability that at least 33 operations out of 44 conducted by him will be successful?

Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 5252 cards. Find the Probability that both are king.

Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Find the probability distribution of number of red balls drawn.

Find kk such that the functionP(x)={k(4x);x=0,1,2,3,4,k>00;otherwiseP(x)=\quad \begin{cases} k\begin{pmatrix} 4 \\ x \end{pmatrix};\quad x=0,1,2,3,4,k>0 \\ 0;\quad otherwise \end{cases} is a probability mass function.

Given X∼B(n,p)X \sim B(n, p), If E(X)=6, Var. (X)=4.2E(X)=6, \text{ Var. }(X)=4.2, find nn and pp.

In a dice game , a player pays a stake of Re 11 for each throw of a die. She receives Rs. 55if the die shows a 33, Rs 22 if the die shows a 11 or 66 , and nothing otherwise . What is the player’s expected profit per throw over a long series of throws ?

A random variable X has the following probability distribution X01234P(X) 0k2k3k1/2Determine P(0 < X < 3)

A random variable has the following probability distribution :Find P

Given below is the probability distribution of XX:xx0011223344P[X=x]P[X=x]kk2k2k4k4k2k2kkkFind:P(X≥2),P(X<3),P(X≤1)P(X\ge 2), P(X < 3), P(X\le 1).

A purse contains four coins each of which is either a rupee or two rupees coin. Find the expected value of a coin in that purse.

A pair of dice is thrown 55 times. If getting a doublet is considered as a success, then find the mean and variance of successes.

There are 100100 tickets in a raffle (Lottery). There is 11 prize each of Rs. 1000/−,1000/-, Rs. 500/−500/- and Rs. 200/−.200/-. Remaining tickets are blank. Find the expected price of one such ticket.

A pair of dice is thrown 44 times. If getting a total of 99 in a single throw is considered as a success then find the mean and variance of successes.

A fair die is tossed. If 2,32, 3 or 55 occurs, the player wins that number of rupees, but if 1,41, 4 or 66 occurs, the player loses that number of rupees. Then find the possible payoffs for the player.

Find the mean number of heads in three tosses of a fair coin.

Find the mean and variance for the first 1010 multiples of 33

Let XX denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of XX.

If F(x)=1π(π2+tan−1x)F\left( x \right) =\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ x } \right), −∞<x<∞-\infty < x < \infty is a distribution function of a continuous variable XX, then find P(0≤x≤1)P\left( 0\le x\le 1 \right)

Find the mean and variance of distributionf(x)={3e−x0<x<∞0elsewheref(x) = \left\{\begin{matrix}3e^{-x} & 0<x<\infty \\ 0 & \text{elsewhere} \end{matrix}\right.

The following is the p.d.f. (Probability Density Function) of a continuous random variable XX:F(x)=x32,0<x<8;=0F(x) = \dfrac {x}{32}, 0 < x < 8; = 0, otherwiseAlso find its value at x=0.5x = 0.5 and 99.

A random variable XX has the following probability distribution:X=xX=x00112233445566P(X=x)P(X=x)kk3k3k5k5k7k7k9k9k11k11k13k13k(a) Find kk(b) Find P(0<X<4)P(0< X< 4)(c) Obtain cumulative distribution function (c.d.f) of XX.

The table gives the probability distribution of a random variable XX.xx12345P(X=x)P(X=x)0.20.20.10.10.30.30.30.3pp(i) Find PP.(ii) Find the mean of XX(iii) Find the variance of XX.

X =12345P(X =)\dfrac{1}{2}\dfrac{1}{2}\dfrac{1}{4}\dfrac{1}{4}\dfrac{1}{8}\dfrac{1}{8}\dfrac{1}{16}\dfrac{1}{16}PThe probability distribution of a random variable X taking values 1, 2, 3, 4, 5 is given(a) Find the value of P.(b) Find the mean of X.(c) Find the variance of X.

The mean and variance of a binomial distribution are 44 and 33 respectively. Fix the distribution and find P(X \geq 1)P(X \geq 1).

Verify f(x)=\begin{cases} 30{ x }^{ 4 }{ e }^{ -6{ x }^{ 5 } };\quad \quad x>0 \\ 0;\quad otherwise \end{cases}f(x)=\begin{cases} 30{ x }^{ 4 }{ e }^{ -6{ x }^{ 5 } };\quad \quad x>0 \\ 0;\quad otherwise \end{cases} for p.d.f if f(x)f(x) is a p.d.f then find P(1)P(1).

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is draw at random with replacement.Find the probability distribution of the number defective bulbs.

From a lot of 66 items containing 22 defective items, a sample of 44 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

From a lot of 66 items containing 22 defective items, a sample of 44 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Mean of X.

A random variable xx has the following probability distribution:xx00112233445566P(X = x)P(X = x)kk3k3k5k5k7k7k9k9k11k11k13k13kFind 'k''k'.

The mean and standard deviation of a random variable X are 1010 and 55 respectively. Find.E\left(\left(\dfrac{x-10}{5}\right)^2\right)E\left(\left(\dfrac{x-10}{5}\right)^2\right).

The P.M.F of a r.v. XX is P\left( x \right) = \frac{1}{5}P\left( x \right) = \frac{1}{5} for x = 1,2,3, \ldots 14,15;x = 1,2,3, \ldots 14,15; P(x)=0P(x)=0 otherwise.Find (i) E(X)E(X) (ii) Var(X)Var(X)

The mean and standard deviation of a random variable X are 1010 and 55 respectively. Find.E\left(\left(\dfrac{x-10}{5}\right)\right)E\left(\left(\dfrac{x-10}{5}\right)\right).

The mean and standard deviation of a random variable X are 1010 and 55 respectively. Find E(X^2)E(X^2)

Find the mean of the probability distribution.

Find k, the probability distribution of X is as follows?X1234P(X)2k3k3k2k

Calculate the expected value and variance of xx, if xx denotes the number obtained on the uppermost face when a fair die is thrown.

A r.v. XX has the following probability distribution:X=xX=x-2-2-1-100112233P(x)P(x)0.10.1kk0.20.22k2k0.30.3kkFind the value of kk and calculate mean and variance of XX.

X 1 2 3 4 5 6 P(X=x) k 2k 3k 4k 5k 6kThe probability distribution of discrete r.v. XX is Then P(X\le 4)P(X\le 4) =

Two number are selected at random (without replacement ) from the first six positive integers. Let X denote the larger of the numbers obtained. Find E (X).

The p. m. f. X- number of major defects randomly selected appliance of a certain type is : X-x01234P(a)0.080.150.450.270.05Find the expected value and standard deviation of X.

Let a pair of dice be thrown and the random variable X X be the sum of the numbers that appear on the two dice. Find the mean of X X .

12 different objecs are to be distributed among 9 persons. Find the number of ways in which objects can be distributed . (All objects must be distributed and each person should get atleast 1 object)

Four red balls are numbered from 11 to 44 and three blue balls are numbered from 33 to 55. If these balls are placed in a bag and a ball is drawn from a bag, what is the probability of choosing a red ball and a blue ball numbered 55?

The probability distribution of a random variable X is given belowX=x_iX=x_iP(X=x_i)P(X=x_i)11k222k2k333k3k444k4k555k5kFind the value of k and the mean and variance of X.

Two unbiased dice are thrown together at random. Find the expected value of the total number of points shown up.

Two cards are drawn (without replacement) from a well shuffled deck of 5252 cards. Find probability distribution and mean of number of cards numbered 44

Find the probability distribution of the number of doublets in three throws of a pair of dice

Find expected value of the random variable XX whose probability mass function is :X=xX=x123P(X=x)P(X=x)\dfrac{1}{5}\dfrac{1}{5}\dfrac{2}{5}\dfrac{2}{5}\dfrac{2}{5}\dfrac{2}{5}

If f(x)=kx,0 < x < 2f(x)=kx,0 < x < 2 =0=0, otherwise,is a probability density function of a random variable XX, then find:(i) Value of kk,(ii) P(1 < X< 2)P(1 < X< 2)

There are 5 cards , numbered 1 to 5, one number of each card . two cards are drawn at random without replacement . Let X denote the sum of the numbers on the two cards drawn.find the mean and variance of X.

The probability distribution of XX is : X X 00112233 P(X)P(X) 0.20.2 K K K K 2K2K Write the value of K K .

A biased die is such that \mathrm{P}(4)=\dfrac{1}{10} \mathrm{P}(4)=\dfrac{1}{10} and other scores being equally likely. The die is tossed twice. If X X is the 'number of four seen', find the variance of the random variable \mathbf{X} \mathbf{X}

If \mathrm{X} \mathrm{X} is the number of tails in three tosses of a coin, determine the standard deviation of x x

A die is thrown three times. Let x x be 'the number of two seen'. Find the expectation of x x

Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

The probability distribution of a discrete random variable X is given as under:Calculate :(i) The value of A if E(X) = 2.94(ii) Variance of X.

Find the probability distribution of the maximum of the two scores obtained when die is thrown twice. Determine also the mean of the distribution

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.XX0011223344P(X)P(X)0.10.10.50.50.20.2-0.1-0.10.30.3

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.YY-1-10011P(Y)P(Y)0.60.60.10.10.20.2

Let X be a random variable taking values x_1, x_2,....... x_nx_1, x_2,....... x_n with probabilitiesp_1, p_2,......p_np_1, p_2,......p_n respectively. Then var(X) = __________.

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.ZZ33331100-1-1P(Z)P(Z)0.30.30.20.20.40.40.10.10.050.05

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.XX001122P(X)P(X)0.40.40.40.40.20.2

Find the probability distribution of XX, the number of heads in a simultaneous toss of two coins.

Complete the following table:

A random variable XX has the following probability distribution:XX0011223344556677P(X)P(X)00kk2k2k2k2k3k3kk^2k^22k^22k^27k^2+k7k^2+kDeterminationP(X>6)P(X>6)

A random variable XX has the following probability distribution:XX0011223344556677P(X)P(X)00kk2k2k2k2k3k3kk^2k^22k^22k^27k^2+k7k^2+kDeterminationP(X<3)P(X<3)

A random variable XX has the following probability distribution:XX0011223344556677P(X)P(X)00kk2k2k2k2k3k3kk^2k^22k^22k^27k^2+k7k^2+kDeterminationKK

A random variable XX has the following probability distribution:XX0011223344556677P(X)P(X)00kk2k2k2k2k3k3kk^2k^22k^22k^27k^2+k7k^2+kDeterminationP(0<X<3)P(0<X<3)

A fair coin is tossed until a head or five tails occur. If XX denotes the number of tosses of the coin, find the mean of XX.

In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0X = 0 if he opposed, and X = 1X = 1 if he is in favour. Find E\left(X\right)E\left(X\right) and Var\left(X\right)Var\left(X\right).

The numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X and hence find the means of the distribution.

Two the numbers are selected at random (without replacement) from first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of X. Find the mean and variance of this distribution.

From a lot of 1515 bulbs which include 55 defectives, a sample of 44 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.

Three numbers are selected at random (without replacement) from first six positive integers. Let XX denote the largest of the three numbers obtained. Find the probability distribution of XX. Also, find the mean and variance of the distribution.

A random variable 'XX' has the following probability distribution:X=x01234567P(X=x)00kk2k2k2k2k3k3k{k}^{2}{k}^{2}2{k}^{2}2{k}^{2}7{k}^{2}+k7{k}^{2}+kFind:(i) kk(ii) The Mean and(iii) P(0< X< 5)P(0< X< 5)

The probability mass function (p.m.f) of $X$ is given below:
$X=x$ $1$ $2$ $3$
$P(X=x)$ $\cfrac{1}{5}$ $\cfrac{2}{5}$ $\cfrac{2}{5}$

The mean and variance of a Binomial distribution is $4$ and $3$ respectively. Find its parameters.

The random variable x has a probability distribution $P(x)$ of the following form where $k$ is some number:
$P(x)=\begin{cases}k & if\, x=0\\2k&if\, x=1 \\3k&if\, x=2\\0&otherwise \end{cases}$
Determine the value of $k$ and $P(X \le 2).$

Find the variance and standard deviation of the random variable $X$ whose probability distribution is given below:
$x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\dfrac {1}{8}$ $\dfrac {3}{8}$ $\dfrac {3}{8}$ $\dfrac {1}{8}$

The probability that a doctor successfully performs an operation is $80\%$ . What is the value of $625$ times the probability that at least $3$ operations out of $4$ conducted by him will be successful?

Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Find the Probability that both are king.

Find $k$ such that the function
$P(x)=\quad \begin{cases} k\begin{pmatrix} 4 \\ x \end{pmatrix};\quad x=0,1,2,3,4,k>0 \\ 0;\quad otherwise \end{cases}$ is a probability mass function.

Given $X \sim B(n, p)$ , If $E(X)=6, \text{ Var. }(X)=4.2$ , find $n$ and $p$ .

In a dice game , a player pays a stake of Re $1$ for each throw of a die. She receives Rs. $5$ if the die shows a $3$ , Rs $2$ if the die shows a $1$ or $6$ , and nothing otherwise . What is the player’s expected profit per throw over a long series of throws ?

A random variable X has the following probability distribution
X 0 1 2 3 4
P(X) 0 k 2k 3k 1/2
Determine P(0 < X < 3)

A random variable has the following probability distribution :
Find P

Given below is the probability distribution of $X$ :
$x$ $0$ $1$ $2$ $3$ $4$
$P[X=x]$ $k$ $2k$ $4k$ $2k$ $k$
Find:
$P(X\ge 2), P(X < 3), P(X\le 1)$ .

A pair of dice is thrown $5$ times. If getting a doublet is considered as a success, then find the mean and variance of successes.

There are $100$ tickets in a raffle (Lottery). There is $1$ prize each of Rs. $1000/-,$ Rs. $500/-$ and Rs. $200/-.$ Remaining tickets are blank. Find the expected price of one such ticket.

A pair of dice is thrown $4$ times. If getting a total of $9$ in a single throw is considered as a success then find the mean and variance of successes.

A fair die is tossed. If $2, 3$ or $5$ occurs, the player wins that number of rupees, but if $1, 4$ or $6$ occurs, the player loses that number of rupees. Then find the possible payoffs for the player.

Find the mean and variance for the first $10$ multiples of $3$

Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X$ .

If $F\left( x \right) =\dfrac { 1 }{ \pi } \left( \dfrac { \pi }{ 2 } +\tan ^{ -1 }{ x } \right)$ , $-\infty < x < \infty$ is a distribution function of a continuous variable $X$ , then find $P\left( 0\le x\le 1 \right)$

Find the mean and variance of distribution
$f(x) = \left\{\begin{matrix}3e^{-x} & 0<x<\infty \\ 0 & \text{elsewhere} \end{matrix}\right.$

The following is the p.d.f. (Probability Density Function) of a continuous random variable $X$ :
$F(x) = \dfrac {x}{32}, 0 < x < 8; = 0$ , otherwise
Also find its value at $x = 0.5$ and $9$ .

A random variable $X$ has the following probability distribution:
$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$
(a) Find $k$
(b) Find $P(0< X< 4)$
(c) Obtain cumulative distribution function (c.d.f) of $X$ .

The table gives the probability distribution of a random variable $X$ .
$x$ 1 2 3 4 5
$P(X=x)$ $0.2$ $0.1$ $0.3$ $0.3$ $p$
(i) Find $P$ .
(ii) Find the mean of $X$
(iii) Find the variance of $X$ .

X = 1 2 3 4 5
P(X =) $\dfrac{1}{2}$
$\dfrac{1}{4}$ $\dfrac{1}{8}$ $\dfrac{1}{16}$ P
The probability distribution of a random variable X taking values 1, 2, 3, 4, 5 is given
(a) Find the value of P.
(b) Find the mean of X.
(c) Find the variance of X.

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Fix the distribution and find $P(X \geq 1)$ .

Verify $f(x)=\begin{cases} 30{ x }^{ 4 }{ e }^{ -6{ x }^{ 5 } };\quad \quad x>0 \\ 0;\quad otherwise \end{cases}$ for p.d.f if $f(x)$ is a p.d.f then find $P(1)$ .

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Mean of X.

A random variable $x$ has the following probability distribution:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$
Find $'k'$ .

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find.
$E\left(\left(\dfrac{x-10}{5}\right)^2\right)$ .

The P.M.F of a r.v. $X$ is $P\left( x \right) = \frac{1}{5}$ for $x = 1,2,3, \ldots 14,15;$ $P(x)=0$ otherwise.
Find (i) $E(X)$ (ii) $Var(X)$

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find.
$E\left(\left(\dfrac{x-10}{5}\right)\right)$ .

The mean and standard deviation of a random variable X are $10$ and $5$ respectively. Find $E(X^2)$

Find k, the probability distribution of X is as follows?
X 1 2 3 4
P(X) 2k 3k 3k 2k

Calculate the expected value and variance of $x$ , if $x$ denotes the number obtained on the uppermost face when a fair die is thrown.

A r.v. $X$ has the following probability distribution:
$X=x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(x)$ $0.1$ $k$ $0.2$ $2k$ $0.3$ $k$
Find the value of $k$ and calculate mean and variance of $X$ .

X 1 2 3 4 5 6
P(X=x) k 2k 3k 4k 5k 6k
The probability distribution of discrete r.v. $X$ is
Then $P(X\le 4)$ =

The p. m. f. X- number of major defects randomly selected appliance of a certain type is :
X-x 0 1 2 3 4
P(a) 0.08 0.15 0.45 0.27 0.05
Find the expected value and standard deviation of X.

Let a pair of dice be thrown and the random variable $X$ be the sum of the numbers that appear on the two dice. Find the mean of $X$ .

Four red balls are numbered from $1$ to $4$ and three blue balls are numbered from $3$ to $5$ . If these balls are placed in a bag and a ball is drawn from a bag, what is the probability of choosing a red ball and a blue ball numbered $5$ ?

The probability distribution of a random variable X is given below
$X=x_i$ $P(X=x_i)$
$1$ k
$2$ $2k$
$3$ $3k$
$4$ $4k$
$5$ $5k$
Find the value of k and the mean and variance of X.

Two cards are drawn (without replacement) from a well shuffled deck of $52$ cards. Find probability distribution and mean of number of cards numbered $4$

Find expected value of the random variable $X$ whose probability mass function is :
$X=x$ 1 2 3
$P(X=x)$ $\dfrac{1}{5}$ $\dfrac{2}{5}$ $\dfrac{2}{5}$

If $f(x)=kx,0 < x < 2$
$=0$ , otherwise,

is a probability density function of a random variable $X$ , then find:
(i) Value of $k$ ,
(ii) $P(1 < X< 2)$

The probability distribution of $X$ is :
$X$ $0$ $1$ $2$ $3$
$P(X)$ $0.2$ $K$ $K$ $2K$

Write the value of $K$ .

A biased die is such that $\mathrm{P}(4)=\dfrac{1}{10}$ and other scores being equally likely. The die is tossed twice. If $X$ is the 'number of four seen', find the variance of the random variable $\mathbf{X}$

If $\mathrm{X}$ is the number of tails in three tosses of a coin, determine the standard deviation of $x$

A die is thrown three times. Let $x$ be 'the number of two seen'. Find the expectation of $x$

The probability distribution of a discrete random variable X is given as under:
Calculate :
(i) The value of A if E(X) = 2.94
(ii) Variance of X.

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$X$ $0$ $1$ $2$ $3$ $4$
$P(X)$ $0.1$ $0.5$ $0.2$ $-0.1$ $0.3$

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$Y$ $-1$ $0$ $1$
$P(Y)$ $0.6$ $0.1$ $0.2$

Let X be a random variable taking values $x_1, x_2,....... x_n$ with probabilities
$p_1, p_2,......p_n$ respectively. Then var(X) = __________.

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$Z$ $3$ $3$ $1$ $0$ $-1$
$P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
$X$ $0$ $1$ $2$
$P(X)$ $0.4$ $0.4$ $0.2$

Find the probability distribution of $X$ , the number of heads in a simultaneous toss of two coins.

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(X>6)$

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(X<3)$

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$K$

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determination
$P(0<X<3)$

A fair coin is tossed until a head or five tails occur. If $X$ denotes the number of tosses of the coin, find the mean of $X$ .

In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take $X = 0$ if he opposed, and $X = 1$ if he is in favour. Find $E\left(X\right)$ and $Var\left(X\right)$ .

From a lot of $15$ bulbs which include $5$ defectives, a sample of $4$ bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.

Three numbers are selected at random (without replacement) from first six positive integers. Let $X$ denote the largest of the three numbers obtained. Find the probability distribution of $X$ . Also, find the mean and variance of the distribution.

A random variable ' $X$ ' has the following probability distribution:
X=x 0 1 2 3 4 5 6 7
P(X=x) $0$ $k$ $2k$ $2k$ $3k$ ${k}^{2}$ $2{k}^{2}$ $7{k}^{2}+k$
Find:
(i) $k$
(ii) The Mean and
(iii) $P(0< X< 5)$

Find the mean and variance for the probability density function : $f(x)=\begin{cases} \alpha { e }^{ -\alpha x }\,\,\, \text{if}\, x>0 \\ 0\,\,\, \text{otherwise} \end{cases}$

A random variable $X$ has the following probability distribution.
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^{2}$ $2k^{2}$ $7k^{2} + k$
Find (i) $k$ (ii) The mean and (iii) $P(0 < X < 5)$

Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.

A player tosses $2$ fair coins. He wins Rs $5$ if $2$ heads appear, Rs $2$ if head appears and Rs $1$ if no head appears. Find his expected winning amount and variance of winning amount.

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(X\leq 3/X > 0)$ .

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(-1 \leq X < 2)$ .

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find $P(X \geq 5)$ .

The following is the c.d.f. of a discrete r.v. X:
X $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Find the p.m.f. of X.

A die is thrown $120$ times and getting $1$ or $5$ is considered a success. Find the mean and variance of the number of sucess.