MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 6

The probability distribution of a random variable

$X$ is given below

$X=x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P[X=X]$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

The variance of

$X$ is

0%
$1.6$
0%
$0.24$
0%
$0.84$
0%
$0.75$

Explanation

Now,

$Var(X) = (x-\mu)^2 \ P(X-x)$

Now, we have

$\mu =0\times 1+0.4 \times 1+0.3 \times 2+3\times 0.2 + 4 \times 0=1.6$

Thus, we have

$Var(X) = (0-1.6)^2 \times 0.2 + (1-1.6)^2 \times 0.4 + (2-1.6)^2 \times 0.3 + (3-1.6)^2 \times 0.2 + (4-1.6)^2 \times 0$

Thus,

$Var(X) = 0.256 + 0.144+ 0.048 + 0.392$

or,

$\Rightarrow Var(X) = 0.840$

The probability function of a binomial distribution is

$P(x) = \binom{6}{x} p^{x} q^{6 - x}, x = 0, 1, 2, ..., 6$ . If

$2P(2) = 3P(3)$ , then

$p =$ __________.

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

Explanation

Given

$P(x) = \binom{6}{x} p^{x} q^{6 - x}$ for

$x = 0, 1, 2, ..., 6$

Also,

$2P(2) = 3P(3)$

$2\times {}^{6}C_{2}p^{2}q^{4} = 3\times {}^{6}C_{3}p^{3}q^{3}$

$\Rightarrow 2\times \dfrac{6!}{2!4!}p^{2}q^{4}=3\times \dfrac{6!}{3!3!}p^{3}q^{3}$

$\Rightarrow q=2p$
(as

$p + q = 1)$ .

$\Rightarrow p = \dfrac {1}{3}$

The probability that an event A occurs in a single trial of an experiment is

$0.3$ . Six independent trials of the experiment are performed. What is the variance of probability distribution of occurrence of event A?

0%
$12.6$
0%
$0.18$
0%
$1.26$
0%
$1.8$

Explanation

X	0	1	2	3	4	5	6
P(X)	$^6C_{0}(0.7^6)(0.3^0)$	$^6C_{1}(0.7^5)(0.3^1)$	$^6C_{2}(0.7^4)(0.3^2)$	$^6C_{3}(0.7^3)(0.3^3)$	$^6C_{4}(0.7^2)(0.3^4)$	$^6C_{5}(0.7^1)(0.3^5)$	$^6C_{6}(0.7^0)(0.3^6)$

$\mu =\sum X_{i}P(X_{i})$

$= \sum_{0}^{6}i \times ^6C_{i}(0.7^{6-i})(0.3^i)$

$= 1.8$

$variance =\sum X_{i}^2P(X_{i})-\mu^2$

$= 4.5 -1.8^2$

$= 1.26$

Hence the answer is option (C).

What is the mean of

$f(x)=3x+2$ where x is a random variable with probability distribution.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$1_{/6}$	$1_{/3}$	$1_{/3}$	$1_{/6}$

0%
$\displaystyle\frac{19}{2}$
0%
$\displaystyle\frac{5}{2}$
0%
$\displaystyle\frac{15}{2}$
0%
$\displaystyle\frac{5}{3}$

Explanation

Given,

$f(x) = 3x+2$

applying expectation on both sides, we get

$E(f(x))=3E(x)+2$ (by the properties of expectation)

mean

$E(x)= \sum_{i=1}^{n} x_i P(x_i)$

Given, number of samples

$n=4$

Hence, mean =

$\sum_{i=1}^4x_iP(x_i)$

substituting the values from the table we get,

$=1\left(\dfrac{1}{6}\right)+2\left(\dfrac{1}{3}\right)+3\left(\dfrac{1}{3}\right)+4\left(\dfrac{1}{6}\right)$

$=\dfrac{1+4+6+4}{6} = \dfrac{15}{6}$

$E(x)=\dfrac{5}{2}$

Therefore,

$E(f(x))=3\left(\dfrac{5}{2}\right)+2$

$=\dfrac{19}{2}$

A random variable

$X$ has the probability distribution,

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P (x)$	$\dfrac{1}{10}$	$K$	$\dfrac{1}{5}$	$2K$	$\dfrac{3}{10}$	$K$

0%
$\dfrac{2}{10}$
0%
$\dfrac{1}{10}$
0%
$\dfrac{3}{10}$
0%
$\dfrac{7}{10}$

Explanation

The sum of probability of all the possible outcomes should be

$1$ .

That is

$\sum P(x_i)=1$

$\implies \dfrac{1}{10}+K+\dfrac{1}{5}+2K+\dfrac{3}{10}+K=1$

$\implies 4K+\dfrac{6}{10}=1$

$\implies 4K=1-\dfrac{6}{10}=\dfrac{4}{10}$

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

An experiment is known to be random if the results of the experiment -

0%
Cannot be predicted
0%
Can be predicted
0%
Can be split into further experiments
0%
Can be selected at random

Let

$a={ i }^{ { k }_{ 1 } }+{ i }^{ { k }_{ 2 } }+{ i }^{ { k }_{ 3 } }+{ i }^{ { k }_{ 4 } },(i=\sqrt { -1 } )$ where each

${k}_{n}$ is randomly chosen from the set

${1,2,3,4}$ . The probability that

$a=0$ , is

0%
$\frac { 7 }{ 64 }$
0%
$\frac { 9 }{ 64 }$
0%
$\frac { 37 }{ 256 }$
0%
$\frac { 39 }{ 256 }$

Two cards are drawn successively with replacement from a well shuffled deck of

$52$ cards. Find the probability distribution of the number of aces.

0%
$\dfrac{1}{169}$
0%
$\dfrac{1}{221}$
0%
$\dfrac{1}{265}$
0%
$\dfrac{4}{663}$

Explanation

Total no. of cards = 52

Total no. of aces = 4

$P(ace) = \cfrac{4}{52} = \cfrac{1}{13}$

$P(non-ace) = \cfrac{48}{52} = \cfrac{12}{13}$

Let X be a random variable, such that

X = no. of aces obtained in 2 draws.

Case I:- X = 0 (No aces are obtained)

$P(X = 0) = P(\text{non-ace and non-ace})$

$= P(non-ace) \times P(non-ace)$

$= \cfrac{12}{13} \times \cfrac{12}{13}$

$= \cfrac{144}{169}$

Case II:- X = 1 (1 ace is obtained)

$P(X = 1) = P(\text{(ace and non-ace) or (non-ace and ace)})$

$= \left( P(non-ace) \times P(non-ace) \right) + \left( P(ace) \times P(non-ace) \right)$

$= \left( \cfrac{12}{13} \times \cfrac{1}{13} \right) + \left( \cfrac{1}{13} \times \cfrac{12}{13} \right)$

$= \cfrac{24}{169}$

Case III:- X = 2 (two aces are obtained)

$P(X = 2) = P(\text{ace and ace})$

$= P(ace) \times P(ace)$

$= \cfrac{1}{13} \times \cfrac{1}{13}$

$= \cfrac{1}{169}$

A box contains

$10$ items,

$3$ of which are defective. If

$4$ are selected at random without replacement, the probability that at least

$2$ are defective is?

0%
$50\%$
0%
$33.33\%$
0%
$67\%$
0%
$100\%$

Given two independent events A and B such that

$P\left( A \right) = 0.3,$

$P\left( A\cap B \right) = 0.1$ . Find

$P\left( {A \cap B'} \right)$

0%
$0.1$
0%
$0.2$
0%
$0.15$
0%
$0.3$

Explanation

Given

$P\left(A\right)=0.3,\,\,P\left(A\cap\,B\right)=0.1$

$A$ and

$B$ are independent events,

$P\left(A\cap\,B\right)=P\left(A\right)P\left(B\right)$

$\Rightarrow\,P\left(B\right)=\dfrac{P\left(A\cap\,B\right)}{P\left(A\right)}=\dfrac{0.1}{0.3}=\dfrac{1}{3}$

$A$ and

$B$ are independent events,

$P\left(A\cap{B}^{\prime}\right)=P\left(A\right)\times P\left({B}^{\prime}\right)$

$=P\left(A\right)\left(1-P\left(B\right)\right)$

$=0.3\left(1-\dfrac{1}{3}\right)$

$=0.3\times\dfrac{3-1}{3}=0.3\times\dfrac{2}{3}=0.1\times 2=0.2$

If the variance of a random variable X is

$\sigma^2$ , then the variance of the random variable X-

$5$ is?

0%
$5\sigma^2$
0%
$25\sigma^2$
0%
$\sigma^2$
0%
$2\sigma^2$

Explanation

${\textbf{Step -1: Solve using the variables given}}{\textbf{.}}$

${\text{If variance of }}X = k,{\text{ then variance of }}ax + b = {a^2}k.$

${\text{Here }}a = 1,b = - 5$

$\therefore {\text{ variance of }}x - 5 = {\sigma ^2}$

${\textbf{Hence, the value of variance of the random variable }}\mathbf{x - 5}{\textbf{ is }}\mathbf{{\sigma ^2}.}$

Consider the word

$W = MISSISSIPPI$ Number of ways in which the letters of the word W can be arranged if at least one vowel is separated from rest of the vowels

0%
$\dfrac { 8\ !.16\ ! }{ 4\ !.4\ !.2\ ! }$
0%
$\dfrac { 8\ !.16\ ! }{ 4.4\ !.2\ ! }$
0%
$\dfrac { 8\ !.16\ !}{ 4\ !.2\ ! }$
0%
$\dfrac { 8\ ! }{ 4\ !.2\ ! } .\dfrac { 165 }{ 4\ ! }$

Explanation

W = MISSISSIPPI , Letters= 11

1 M. 4I's, 4S's, 2P's

Total number of arranging these letters

$=\quad \cfrac { 11! }{ 4!4!2! }$

Suppose all I's are together , then the Number of ways

$=\quad \cfrac { \left( 1+1+4+2 \right) ! }{ 4!2! } \quad =\quad \cfrac { 8! }{ 4!2! }$

$\therefore$ Number of ways in which atleast one vowel is separated

$=\quad \cfrac { 11! }{ 4!4!2! } \quad -\quad \cfrac { 8! }{ 4!2! } \\ =\quad \cfrac { 8!16! }{ 4!4!2! }$

A biased die is tossed and the respective probabilities for various faces to turn up are given below:

Face:	$1$	$2$	$3$	$4$	$5$	$6$
Probability	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

If an even face has turned up, then probability that it is face

$2$ or face

$4$ , is

0%
$0.25$
0%
$0.42$
0%
$0.75$
0%
$0.9$

Explanation

Required probability =

$\dfrac{P(A \cap B)}{P(B)} = P$

$B =$ even face

$A =$ (face

$= 2$ or

$4$ )

$A \cap B =$ (face

$= 2$ or

$4$ )

$\therefore P = \dfrac{P (2 \, or \, 4)}{P (2 \, or \, 4 \, or \, 6)} = \dfrac{0.24 + 0.18}{0.24 + 0.18 + 0.14} = 0.75$

A special die with number

$1$ ,

$-1$ ,

$2$ ,

$-2$ ,

$0$ and

$3$ is thrown thrice. The probability that total is

$6$ is

0%
$\dfrac{1}{{108}}$
0%
$\dfrac{{25}}{{216}}$
0%
$\dfrac{5}{{216}}$
0%
$\dfrac{5}{{108}}$

Explanation

outcomes are 1,-1.2,-2,0,3

For Sum to be 6 -

2,2,2 1 case

3,3,0 3!/2= 3 cases

1,2,3 3!=6 cases
Total favourable outcomes=1+3+6=10

Total outcomes =6

$\times$ 6

$\times$ 6=216

Probability=10/216=5/108

A discrete random variable takes

0%
only a finite number of values
0%
all possible values between certain given limits
0%
infinite number of values
0%
a finite or countable number of values

A random variable

$X$ has the following probability distribution. Find the value of

$10k$ .

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

0%
1
0%
2
0%
3
0%
4

Explanation

As we know that

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

$\implies 0.1+k+0.2+2{k}+0.3+k=1\implies k=0.1\implies 10{k}=1$

A random variable

$X$ has the probability distribution:

$X$ :	1	2	3	4	5	6	7	8
$p(X)$ :	0.2	0.2	0.1	0.1	0.2	0.1	0.1	0.1

For the events

$E=\left\{ X\quad is\quad a\quad prime\quad number \right\}$ and

$F=\left\{ X<4 \right\}$ , the

$P(E\cup F)$ is

0%
$0.50$
0%
$0.77$
0%
$0.35$
0%
$0.80$

Explanation

given I =

$\{$ 'x' is a prime number

$\}$

P(E) = p(x=2 or x = 3 or x =5 or x =7)

$=P(x=2)+p(x=3)+p(x=5)+p(x=7)$

$= 0.2+0.1 +0.2+0.1$

$p(B) = 0.6$

$p(F) = P(x = 1)+p(x=2)+p(x =3)$

$= 0.2+0.2+0.1$

$= 0.5$

$p(EAF) = p(x=2)+p(x=3)$

$= 0.2+0.1$

$0.3$

$p(E\vee F)=p(E)+p(F)-p(E\wedge F)$

$= 0.6+0.5 = 0.3$

$P(E\vee F) = 0.80$

1074833_1183729_ans_46ae65bbe970434a85dc8e51d8a9d826.png

$P ( A ) = 0.8,$

$P ( B ) = 0.5$ and

$P \left( \dfrac { B } { A } \right) = 0.4$ then

$P \left( \dfrac { A } { B } \right) =?$

0%
$0.32$
0%
$0.64$
0%
$0.16$
0%
$0.25$

Explanation

$P(A)=0.8 ; P(B)=0.5$

$P(\dfrac{B}{A})=0.4$

$P(A\cap B)=P(A)P(\dfrac{B}{A})$

$P(A\cap B)=P(B)P(\frac{A}{B})$

$\dfrac{P(A\cap B)}{P(A\cap B)}=\dfrac{P(A)\times P(B/A)}{P(B)\times P(A/B)}$

$\Rightarrow 1=\dfrac{0.8\times 0.4}{0.5\times P(A/B)}$

$\Rightarrow 0.5\times P(A/B)=0.32$

$P(A/B)=\dfrac{0.32}{0.5}\Rightarrow 0.64$

1205532_1297302_ans_22978b1d71ff4de4adf5d396e2c6a64e.jpg

Two cards are drawn successively with replacement from a well-shuffled deck of

$52$ cards. Let

$X$ denote the random variable of number of aces obtained in the two drawn cards. Then

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

0%
$\dfrac {52}{169}$
0%
$\dfrac {25}{169}$
0%
$\dfrac {49}{169}$
0%
$\dfrac {24}{169}$

Explanation

Two cards are drawn successively with replacement

$4\ Aces\ 48$ Non Aces

$P(x = 1) = \dfrac {^{4}C_{1}}{^{52}C_{1}}\times \dfrac {48C_{1}}{52C_{1}} + \dfrac {48C_{1}}{52C_{1}} \times \dfrac {4C_{1}}{52C_{1}} = \dfrac {24}{169}$

$P(x = 2) = \dfrac {^{4}C_{1}}{^{52}C_{1}}\times \dfrac {^{4}C_{1}}{^{52}C_{1}} = \dfrac {1}{169}$

$P(x = 1) + P(x = 2) = \dfrac {25}{169}$ .

In the following, find the value of k and find mean and variance of X:

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

Find the value of:

1) k

2) E(X)

3) V(X)

0%
$0.1 , 0.8, 2.16$
0%
$0.8 , 2.16, 2.16$
0%
$0.1 , 2, 3$
0%
$2.16, 0.8, 0.1$

The outcome of each of

$30$ items was observed;

$10$ items gave an outcome

$\dfrac{1}{2}$ - d each,

$10$ items gave outcome

$\dfrac{1}{2}$ each and the remaining

$10$ items gave outcome

$\dfrac{1}{2}+$ d each. If the variance of this outcome data is

$\dfrac{4}{3}$ then

$|d|$ equals:-

0%
$2$
0%
$\dfrac{\sqrt5}{2}$
0%
$\dfrac{2}{3}$
0%
$\sqrt2$

Explanation

Variance is independent of origin. So we shift the given data by

$\dfrac{1}{2}$ .

so,

$\dfrac{10d^2+10\times0^2+10d^2}{30}-(0)^2=\dfrac{4}{3}$

$\Rightarrow d^2=2\Rightarrow|d|=\sqrt2$

A perfect die is thrown twice. The expected value of the product of the number of point obtained in two thrown is

0%
$7/2$
0%
$7$
0%
$49/2$
0%
$none of these$

A random variable

$X$ has the following probability mass function:

$X$	$-2$	$3$	$1$
$P(X = x)$	$\dfrac{\lambda}{6}$	$\dfrac{\lambda}{4}$	$\dfrac{\lambda}{12}$

Then the value of

$\lambda$ is:

0%
$3$
0%
$1$
0%
$4$
0%
$2$

Explanation

We know that sum of Probability mass function

$=1$

$\sum{P\left(X=x\right)}=P\left(-2\right)+P\left(3\right)+P\left(1\right)$

$\Rightarrow 1=\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}$

$\Rightarrow 1=\dfrac{2\lambda}{6}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}$

$\Rightarrow \dfrac{2\lambda+3\lambda+\lambda}{12}=1$

$\Rightarrow 6\lambda=12$

$\therefore \lambda=\dfrac{12}{6}=2$

For the probability distribution given by

$\left.\begin{matrix} X=x_i & 0 \\ P. & \dfrac{25}{36}\end{matrix}\right|$

$\begin{matrix} 1 \\ 5 \\ 18\end{matrix}$

$\begin{vmatrix} 2 \\ 1 \\ 36\end{vmatrix}$ the standard deviation

$(\sigma)$ is?

0%
$\sqrt{\dfrac{1}{3}}$
0%
$\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$
0%
$\sqrt{\dfrac{5}{36}}$
0%
None of the above

Explanation

$M^2+\sigma^2=\sum x^2_iP(x=x_i)$

$M=\sum x_iP(x=x_i)=\dfrac{5}{18}+\dfrac{2}{36}=\dfrac{6}{18}=\dfrac{1}{3}$

$\therefore\dfrac{1}{9}+\sigma^2=\dfrac{5}{18}+\dfrac{4}{36}=\dfrac{7}{18}$

$\therefore \sigma^2=\dfrac{7}{18}-\dfrac{2}{18}=\dfrac{5}{18}$

$\therefore S.D=\dfrac{1}{3}\sqrt{\dfrac{5}{2}}$ .

A coin is rolled n times. If the probability of getting head at least once is greater than

$90\%$ then the minimum value of n is?

0%
$4$
0%
$3$
0%
$5$
0%
$6$

Explanation

$1-\dfrac{1}{2^n} > \dfrac{9}{10}\Rightarrow \dfrac{1}{10} > \dfrac{1}{2^n}\Rightarrow 2^n > 10$

$\therefore$ minimum value of n is

$4$ .

A random variable X has following probability distribution.

$X=x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	K	$3K$	$5K$	$7K$	$8K$	K

Then

$P(2\leq x < 5)=$ _______.

0%
$\dfrac{3}{5}$
0%
$\dfrac{7}{25}$
0%
$\dfrac{23}{25}$
0%
$\dfrac{24}{25}$

Explanation

$since\quad \sum { P(x)=1 } \\ k+3k+5k+7k+8k+k=1\\ k=1/25\\ P(2\le x<5)=P(2)+P(3)+P(4)\\ \qquad \qquad \quad \quad =3k+5k+7k=15k=15/25=3/5$

A random variable

$X$ has the following probability distribution:

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^{2}$	$2k$	$k$	$2k$	$5k^{2}$

Then

$P(X>2)$ is equal to

0%
$\dfrac{1}{6}$
0%
$\dfrac{7}{12}$
0%
$\dfrac{1}{36}$
0%
$\dfrac{23}{36}$

Explanation

$\displaystyle \sum Pi=1$

$\Rightarrow 6k^2+5k=1$

$6k^2+5k-1=0$

$6k^2+6k-k-1=0$

$6k(k+1)-1(k+1)=0$

$(k+1)(6k-1)=0$

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

$\therefore\ k=\dfrac 16$ as k can't be negative

$p(x>2)=k+2k+5k^2$

$=\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{5}{36}$

$=\dfrac{23}{36}$

If the c.d.f.(cumulative distribution function) is given by

$F(x)=\dfrac{x-25}{10}$ , then

$P(27\leq x\leq 33)=$ ________.

0%
$\dfrac{3}{5}$
0%
$\dfrac{3}{10}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{10}$

Explanation

By definition,

$F(x) = P(X \leq x)$ .

Therefore,

$P(27 \leq X \leq 33) = P(X \leq 33) - P(X \leq 27)$

$= F(33) - F(27)$

$= \dfrac{33-25}{10} - \dfrac{27 - 25}{10}$

$= \dfrac{6}{10} = \dfrac{3}{5}$ .

That is,

$P(27 \leq X \leq 33) = \dfrac{3}{5}$ .

A fair die is tossed repeatedly until a 6 is obtained.Let X denote the number of tosses required.

The probability that X=3 equals

0%
25/216
0%
25/36
0%
5/36
0%
125.216

For the following probability distribution.
E(X) is equal to:

0%
0
0%
-1
0%
-2
0%
-1.8

Explanation

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

$=4\times (0.1)+(-3\times0.2)+(-2\times0.3)+(-1\times0.2)+(0\times0.2)$

$=-0.4-0.6-0.6-0.2=-1.8$

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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