MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 5

The counter part of probability mass function is:

0%
Probability distribution
0%
Probability density function
0%
Distribution function
0%
None of these

A variable that can assume any possible value between two points is called:

0%
Discrete sample space
0%
Discrete random variable
0%
Continuous random variable
0%
Random variable

Amit tosses a fair coin twice, and let

$X$ be defined as the number of heads he observe. Find probability mass function

$P_x$ .

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{4}$
0%
None off these

Explanation

Given Amit tosses a fair coin twice

$\therefore$ Sample space,

$S=[HH,HT,TH,TT]$

$\Rightarrow$ When

$P(x)=1$ it is a PMF

i.e,

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

$\therefore$ Probability mass function

$Px=\dfrac{1}{4}.$

Hence, the answer is

$\dfrac{1}{4}.$

Probability distribution of discrete random variable is classified as

0%
probability mass function.
0%
posterior mass function.
0%
interior mass function.
0%
continuous mass function.

Explanation

The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values.

It is also sometimes called the probability function or the probability mass function.

Thus option

$(A)$ is correct.

A random variable

$X$ has the following p.d.f.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^{2}$	$2k^{2}$	$7k^{2} + k$

The value of

$k$ is

0%
$\dfrac {1}{8}$
0%
$\dfrac {1}{10}$
0%
$0$
0%
$-1$

Explanation

$\displaystyle \sum_{x=0} ^{x=7} P(X)=1$

$\Rightarrow P(0)+P(1)+P(2)...P(7)=1$

$\Rightarrow 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1$

$\Rightarrow 10k^{2}+9k-1=0$

$\Rightarrow 10k^{2}+10k-k-1=0$

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

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

$k=-1$ or

$k=\dfrac{1}{10}$ .

But

$0\leq P(x)\leq 1$
Hence

$k=\dfrac{1}{10}$ .

$x$ is a continuous random variable, then

$P(a < x < b) =$

0%
$P(a \leq x \leq b)$
0%
$P(a < x \leq b)$
0%
$P(a \leq x < b)$
0%
all the three above

Explanation

Consider there are

$n$ number of points between

$a$ and

$b$ where

$n$ tends to infinite

$($ given

$x$ is a continuous random variable

$)$

$P(x=a)=P(x=b)=\dfrac{1}{n}$ . As

$n\rightarrow \infty$ ,

$\dfrac{1}{n}\rightarrow 0$

Therefore

$P(x=a)=p(x=b)\rightarrow 0$

$\Longrightarrow P(a<x<b)=P(a\le x\le b)=P(a<x\le b)=P(a\le x<b)$

Therefore the correct option is

$D$ .

A random variable X has the following probability distribution

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

Then

$E(x) =$

0%
$0.8$
0%
$0.9$
0%
$0.7$
0%
$1.1$

Explanation

$E(x) = \sum x. P(x)$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(x)$	$0.1$	$0.1$	$0.2$	$0.2$	$0.3$	$0.1$
$x. P(x)$	$-0.2$	$-0.1$	$0$	$0.2$	$0.6$	$0.3$

$E(x) = \sum x. P(x) = -0.2 - 0.1 + 0.2 + 0.6 + 0.3$

Thus

$E(x)= 0.8$

$E(X + C) = 8$ and

$E(X - C) = 12$ then the value of

$C$ is

0%
$-2$
0%
$4$
0%
$-4$
0%
$2$

Explanation

We know that

$E(X+C)=E(X)+C$ ...(i)
Therefore

$E(X+C)=E(X)+C=8$

$E(X-C)=E(X)-C=12$
Subtracting equation ii from i

$2C=-4$

$C=-2$ .

Let the random variable X follow B(6, p). If 16 P(X = 4) = P(X = 2), then what is the value of p?

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{6}$

Explanation

$16P(X=4)=P(X=2)$

$\therefore 16(^6C_4p^4q^2)=^6C_2p^2q^4$

$\therefore 16(^6C_4p^4(1-p)^2)=^6C_2p^2(1-p)^4$

$\therefore 16p^2=(1-p)^2$

$\therefore 15p^2+2p-1=0$

$\therefore (3p+1)(5p-1)=0$

$\therefore p=-\dfrac{1}{3}$ or

$p=\dfrac{1}{5}$

Probability cannot be negative.

So,

$p=\dfrac{1}{5}$

Answer is option (C)

What is

$P(Z = 5)$ equal to ?

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$

Explanation

Given set

$=\{1,2,3,4,5,6,7\}$

$Z=X+Y$

Where

$X$ is randomly selected from set of odd numbers and

$Y$ is randomly selected from set of even numbers

That gives

$Z$ is odd and greater than one

Now, for

$Z=5$

Possible ordered pairs are

$\{(1,4),(2,3)\}$

Probability of getting these 2 pairs

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

$=\cfrac{1}{4}\times \cfrac{1}{3}+\cfrac{1}{4}\times \cfrac{1}{3}$

$=\cfrac{1}{6}$

What is

$P(Z = 10)$ equal to ?

0%
$0$
0%
$1/2$
0%
$1/3$
0%
$1/5$

Explanation

Given set

$=\{1,2,3,4,5,6,7\}$

$Z=X+Y$

Where

$X$ is randomly selected from set of odd numbers and

$Y$ is randomly selected from set of even numbers.

Which gives

$Z$ as odd and greater than one

Given :

$Z=10$

But

$10$ is even

So,

$P(Z=10)=0$

$x$ is a continuous random variable then

$P(x \geq a) =$

0%
$P (x < a)$
0%
$1 - P(x > a)$
0%
$P(x > a)$
0%
$1 - P (x \leq a - 1)$

Explanation

Given

$x$ is continuous random variable , Let us consider that there are

$n$ points in total, Since

$x$ is continuous,

$n\rightarrow \infty$

The probability of

$x=a$ is

$\dfrac{1}{n}\rightarrow 0$ , therefore

$P(x=a)=0$

Which implies

$P(x\ge a) = P(x>a)+P(x=a) = P(x>a)$

Therefore the correct option is

$C$

The probability distribution of a random variable is given below :

$X = x$	0	1	2	3	4	5	6	7
$P(X = x)$	$0$	$K$	$2K$	$2k$	$3K$	$K^2$	$2K^2$	$7K^2 + k$

Then

$P(0 < X < 5) =$

0%
$\dfrac{1}{10}$
0%
$\dfrac{3}{10}$
0%
$\dfrac{8}{10}$
0%
$\dfrac{7}{10}$

Explanation

$\sum p(X=x_i)=1$

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

$10K^{2}+9K-1=0$

$10K^{2}+10k-k-1=0$

$10K(K+1)-1(K+1)=0$

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

$P(0< X< 5)=p(x=1)+p(x=2)+p(x=3)+p(x=4)=K+2K+2K+3K=8K=\dfrac{8}{10}$

$\therefore P(0< X< 5)=\dfrac{8}{10}$

Let

$X$ and

$Y$ be two random variables. The relationship

$E(XY) = E(X) \cdot E(Y)$ holds

0%
Always
0%
If $E(X +Y)= E(X)+ E(Y)$ is true
0%
If $X$ and $Y$ are independent
0%
If $X$ can be obtained from $Y$ by a linear transformation

Explanation

We know that the relationship

$E(XY)=E(X) \cdot E(Y)$
holds, when

$X$ and

$Y$ are independent variables.

If the range of random variable

$X = \left \{0, 1, 2, ...\right \}$ and

$P(X = k) = \dfrac {c(k + 1)}{2^{k}}$ for

$k = 0, 1, 2, ...$ then

$c =$

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$

Explanation

$P(X=0)=c$

$P(X=1)=\dfrac{2c}{2}$

$P(X=2)=\dfrac{3c}{2^{2}}$

$\sum P(X=r)=1$

$\Rightarrow \dfrac{c}{2^{0}}+\dfrac{2c}{2^1}+\dfrac{3c}{2^2}+........=1$

$\Rightarrow \dfrac{1c}{2}+\dfrac{2c}{2^{2}}+.......=\dfrac{1}{2}$

$\Rightarrow c+\dfrac{c}{2}+\dfrac{c}{2^{2}}+.......=\dfrac{1}{2}$

$\Rightarrow c\left[1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+.....\right]$

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

$\Rightarrow c[2]=\dfrac{1}{2}$

$\Rightarrow c=\dfrac{1}{4}$

$r. v. X\sim B(n, p)$ . If values of mean and variance of

$X$ are

$18$ and

$12$ respectively then total number of possible values of

$X$ are

0%
$54$
0%
$55$
0%
$12$
0%
$18$

Explanation

$Mean = np = 18$

$Variance = npq = 12$

$\dfrac {npq}{np} = \dfrac {12}{18}$

$\therefore q = \dfrac {2}{3}$
Now,

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

$\therefore p = \dfrac {1}{3}$
Also,

$np = 18$

$\implies n \left (\dfrac {1}{3}\right ) = 18$

$\implies n= 54$

$\therefore$ Possible values of

$X$ are

$0, 1, 2, ..... 54$

$\therefore$ No. of possible values are

$55$ .

$X$ is a binomial variate with the range

$\left\{ 0,1,2,3,4,5,6 \right\}$ and

$P(X=2)=4P(X=4)$ , then the parameter of

$X$ is

0%
$\cfrac{1}{3}$
0%
$\cfrac{1}{2}$
0%
$\cfrac{2}{3}$
0%
$\cfrac{3}{4}$

Explanation

Here n=6

According to the question,

$^6C_2p^2q^4=4\times ^6C_4p^4q^2\\\implies q^2=4p^2\\\implies (1-p)^2=4p^2\\\implies 3p^2+2p-1=0\\\implies (p+1)(3p-1)=0\\\implies p=\dfrac 13, p=-1$

p cannot be negative, hence

$p=\dfrac 13$

If the p.d.f of a r.v.

$X$ is given as

$xi$	$-2$	$-1$	$0$	$1$	$2$
$P(X=xi)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

then

$F(0)=$

0%
$P(X< 0)$
0%
$P(X> 0)$
0%
$1-P(X>0)$
0%
$1-P(X< 0)$

Explanation

We know that, probability function

$F$ for a random variable

$X$ is given by

$F(x)=\displaystyle \sum_{u\leq x} f(u)=P(X\leq x)$

$\therefore F(0)=P(X\leq 0)$

$=P(X=0)+P(X=-1)+P(X=-2)$

$=0.15+0.3+0.2$

$=0.65$

$=1-0.35$

$=1-(0.25+0.1)$

$=1-(P(X=1)+P(X=2))$

$=1-P(X>0)$

Hence,

$F(0)=1-P(X>0)$

The value of k when

$f(x) = \frac {1}{\sqrt{x}},0 < x < 4$ , is the p.d.f of r.v.x

0%
$-4$
0%
$\dfrac {-1}{4}$
0%
$4$
0%
$\dfrac{1}{4}$

Explanation

$f(x)=\dfrac {k}{\sqrt x}$ where

$x\in (0,4)$

So,

$\displaystyle \int_0^4{\dfrac k{\sqrt x}dx}=1$ (Summation of all probability distribution=1)

i.e,

$2k\sqrt x]_0^4=1\Rightarrow 4k=1$

So,

$k=\dfrac 14$

Hence, option D is correct.

If r.v

$X$ : waiting time in minutes for bus and p.d.f of

$X$ is given by

$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$
then probability of waiting time not more than

$4$ minutes is

$=$ ...........

0%
$0.3$
0%
$0.8$
0%
$0.2$
0%
$0.5$

Explanation

Given :

$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$

Probability of waiting time not more than

$4$ is same as probability of waiting time less than

$4$ minutes and it is given by

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

$=\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}$ .... [From the definition of

$f$ ]

$=4\times \dfrac { 1 }{ 5 } =0.8$

The probability distribution of

$X$ is

$X$	0	1	2	3
$P(x)$	0.3	k	2k	3k

The value of

$k$ is

0%
$0.116$
0%
$0.7$
0%
$1$
0%
$0.3$

Explanation

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

$\displaystyle \sum _{ x=0 }^{ 3 }{ P(x) } =1$

$P\left( 0 \right) +P\left( 1 \right) +P\left( 2 \right) +P\left( 3 \right) =1$

$\Rightarrow0.3+k+2k+3k=1$

$\Rightarrow 6k=1-0.3$

$\Rightarrow 6k=0.7$

$k=\cfrac { 0.7 }{ 6 }; \$

$k=0.116$

For the following distribution function

$F(x)$ of a r.v

$X$ is given

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$F(x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

Then

$P(3 < x\leq 5) =$

0%
$0.48$
0%
$0.37$
0%
$0.27$
0%
$1.47$

Explanation

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$f(x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$
$p(x)$	$0.2$	$0.17$	$0.11$	$0.14$	$0.23$	$0.15$

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

$= 0.14 + 0.23$

$= 0.37$

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

0%
$\frac{14}{3}$
0%
$\frac{13}{3}$
0%
$\frac{12}{3}$
0%
$\frac{11}{3}$

Explanation

E(X) is nothing but mean of X

$\rightarrow$ first sic natural numbers are

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

$\rightarrow$ We can select two positive no.s in

$6\times 5=30$ ways

$\rightarrow 2$ numbers are selected at random among which X denotes the larger of the two no.s

$\rightarrow$ But as X is largest among the two ,

it can be

$2,3,4,5$ or

$6$

$\therefore P(X=2)=P$ (larger no. is

$2$ )

${(1,2),(2,1)}$

$=\cfrac { 2 }{ 30 }$

$\rightarrow P\left( X=3 \right) =P$ (Larger no. is

$3$ )

$=\left\{ \left( 1,3 \right) ,\left( 3,1 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) \right\}$

$=\cfrac { 4 }{ 30 }$

$\rightarrow P\left( X=4 \right) =P$ (larger no. is

$4$ )

$=\left\{ \left( 1,4 \right) ,\left( 4,1 \right) ,\left( 2,4 \right) ,\left( 4,2 \right) ,\left( 3,4 \right) ,\left( 4,3 \right) \right\}$

$=\cfrac { 6 }{ 30 }$

$\rightarrow P\left( X=5 \right) =P$ (Larger no. is

$5$ )

$=\left\{ \left( 1,5 \right) ,\left( 5,1 \right) ,\left( 2,5 \right) ,\left( 5,2 \right) ,\left( 3,5 \right) ,\left( 5,3 \right) ,\left( 4,5 \right) ,\left( 5,4 \right) \right\}$

$=\cfrac { 8 }{ 30 }$

$\rightarrow P\left( X=6 \right) =P$ (Larger no. is

$6$ )

$=\left\{ \left( 1,6 \right) ,\left( 6,1 \right) ,\left( 2,6 \right) ,\left( 6,2 \right) ,\left( 3,6 \right) ,\left( 6,3 \right) ,\left( 4,6 \right) ,\left( 6,4 \right) ,\left( 5,6 \right) ,\left( 6,50 \right) \right\}$

$=\cfrac { 10 }{ 30 }$

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

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

$=2\left( \cfrac { 2 }{ 30 } \right) +3\left( \cfrac { 4 }{ 30 } \right) +4\left( \cfrac { 6 }{ 30 } \right) +5\left( \cfrac { 8 }{ 30 } \right) +6\left( \cfrac { 10 }{ 30 } \right)$

$=\cfrac { 140 }{ 30 } =\cfrac { 14 }{ 3 }$

The expected value of the number of points, obtained in a single throw of die, is

0%
$\dfrac{3}{2}$
0%
$\dfrac{5}{2}$
0%
$\dfrac{7}{2}$
0%
$\dfrac{9}{2}$

Explanation

Considering that the dice is a fair dice, the probability of all outcomes is equal.

$\therefore$ sample space of outcomes of dice is

$\{1,2,3,4,5,6\}$

Probability of each outcome =

$\dfrac{1}{6}$

$\therefore$

$E[\text{outcome of a single throw}]=\dfrac{1}{6}\times(1+2+3+4+5+6)=\dfrac{21}{6}=\dfrac{7}{2}$

This is the required answer.

$X$ has three children in his family. What is the probability that all the three children are boys?

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{3}{8}$

Explanation

X has

$3$ children in his family.

All possible outcomes are if

$G$ represents girl child and B represent boy child,

$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$ , hence

$n(S)=8$

Out of which favorable outcomes for all the three chilren being boy,

$n(E)=1$

Hence, the probability that all the three children are boys is

$\dfrac {n(E)}{n(S)}=\dfrac 18$

$X$ has three children in his family. Probability of atleast two girls in the family is.....

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{4}$

Explanation

$X$ has

$3$ children in his family.

All possible outcomes are if

$G$ represents girl child and

$B$ represent boy child,

$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$ , hence

$n(S)=8$

Out of which favorable outcomes for at-least two girls are,

$n(E)=4$

Hence, the probability that there are at-least two girls is

$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$

$A$ and

$B$ are two non empty sets then

${ \left( A\cup B \right) }^{ C }=$

0%
${ A }^{ C }\cup { B }^{ C }$
0%
${ A }^{ C }\cap { B }^{ C }$
0%
$A\cup { B }^{ C }$
0%
${ A }^{ C }\cup B$

Explanation

${ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }$

The probability distribution of a random variable

$X$ is given below:

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

If mean of

$X$ is

$4.2$ , then

$a$ and

$b$ are respectively equal to

0%
$0.3,0.2$
0%
$0.1,0.4$
0%
$0.1,0.2$
0%
$0.2,0.1$

Explanation

Given, mean of

$x$ is

$4.2$

Thus that

$a+2a+3a+4b+5b+1.8=4.2$

$\Rightarrow 6a+9b=2.4$ .....(1)

Also

$a+a+a+b+b+0.3=1$

$\Rightarrow 3a+2b=0.7$ .....(2)

Multiply equation (2) by

$2$ ,

$6a+4b=1.4$ .....(3)

Subtracting equations (1) and (3), we get

$5b=1.0\Rightarrow b=0.2$

Substituting this value in equation (1), we get

$6a+9(0.2)=2.4$

$\Rightarrow 6a+1.8=2.4\Rightarrow a=0.1$
Therefore,

$a=0.1,b=0.2$

$X$ has three children in his family. What is the probability of two or more boys in the family?

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{8}$

Explanation

$X$ has

$3$ children in his family.

All possible outcomes are if

$G$ represents girl child and

$B$ represent boy child,

$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$ , hence

$n(S)=8$

Out of which favorable outcomes for two or more boys are,

$n(E)=4$

Hence, the probability that there are two or more boys in the family is

$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$

$X$ has three children in his family. The probability of one girl and two boys is......

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{8}$

Explanation

$X$ has

$3$ children in his family.

All possible outcomes are if G represents girl child and

$B$ represent boy child,

$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$ , hence

$n(S)=8$

Out of which favorable outcomes for one girl and two boys are,

$n(E)=3$

Hence, the probability that there are one girl and two boys is

$\dfrac {n(E)}{n(S)}=\dfrac 38$

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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