MCQ

MCQ 11 Mark

The $pdf$ of Poisson Distribution is given by __________

✓
$\frac{e^{-m} m^x}{x!}$
B
$\frac{e^{-m} x!}{m^x}$
C
$\frac{x!}{m^x e^{-m}}$
D
$\frac{e^m m^x}{x!}$

Answer

Correct option: A.

$\frac{e^{-m} m^x}{x!}$

(A) $\frac{e^{-m} m^x}{x!}$

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MCQ 21 Mark

If '$m$' is the mean of Poisson Distribution, the standard. deviation is given by __________.

✓
$\sqrt{m}$
B
$m^2$
C
$m$
D
$\frac{m}{2}$

Answer

Correct option: A.

$\sqrt{m}$

(A) $\sqrt{m}$
Explanation: Standard deviation $=\sqrt{\text { Variance }}$
$=\sqrt{m}$
( $\because$ For Poisson Distribution mean $=$ Variance $)$

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MCQ 31 Mark

If '$m$' is the mean of a Poisson Distribution, then variance is given by _________.

A
$m^2$
B
$m^{1 / 2}$
✓
$m$
D
$\frac{m}{2}$

Answer

Correct option: C.

$m$

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MCQ 41 Mark

For larger values of '$n$', Binomial Distribution __________.

A
loses its discreteness
✓
tends to Poisson's Distribution
C
Stays as it is
D
gives oscillatory values

Answer

Correct option: B.

tends to Poisson's Distribution

(B) tends to Poisson's Distribution
Explanation: $P(X)=\lim _{n \rightarrow \infty}^n C_x p^x q^{n-x}$
$=\frac{e^{-m} m_x}{x!}$
where $m = np$ is the mean of Poisson Distribution.

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MCQ 51 Mark

Binomial distribution is a _________

A
Continuous distribution
✓
Discrete distribution
C
Irregular distribution
D
Not a Probability distribution

Answer

Correct option: B.

Discrete distribution

(B) Discrete distribution
Explanation: It is applied to discrete random variable, hence it is a discrete distribution.

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MCQ 61 Mark

In a Binomial distribution, if '$n$' is the number of trials and '$p$' is the probability of success, then the mean value is given by __________.

A
np
B
n
C
p
D
np(1 - p)

Answer

(А) nр
Explanation: For a discrete probability function, the mean value or the expected value is given by
$\operatorname{Mean}(\mu)=\sum_{x=0}^\pi x p(x)$
For Binomial distribution $P(X)=n x p^x q^{n-x}$, substitute in above equation and solve to get $\mu=n p.$

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MCQ 71 Mark

$E(X)=\lambda$ is for which distribution?

A
Bernoulli's
B
Binomial
✓
Poisson's
D
None of these

Answer

Correct option: C.

Poisson's

(C) Poisson's
Explanation: In Poisson's distribution, there is a positive constant $\lambda$ which is the mean of the distribution and variance of the distribution.

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MCQ 81 Mark

$E(X) = npq$ is for which distribution?

A
Bernoulli's
✓
Binomial
C
Poisson's
D
None of these

Answer

Correct option: B.

Binomial

(B) Binomial
Explanation: In Binomial distribution, probability of success is given by $p$ and that of future is given by $q$ and the event is done n times. The mean of the distribution is given by $npq.$

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MCQ 91 Mark

A river passing near a town floods it on an average twice every 10 years. Assuming Poisson distribution find the probability that the town faces flooding at least once in 10 years.

A
0.0198
B
0.1353
C
0.5657
✓
0.8647

Answer

Correct option: D.

0.8647

(D) 0.8647
$\begin{aligned} P(X \geq 1) & =1-P(0) \\ & =1-\frac{e^{-2}(2)^0}{0!} \\ & =1-e^{-2} \\ & =0.86 47\end{aligned}$

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MCQ 101 Mark

During a pandemic, $10\%$ of the patients who have the disease get complications. If $100$ patients of a locality get infected by the disease, then the standard deviation of the number of patient getting complications is:

A
10
B
9
C
6
✓
3

Answer

Correct option: D.

(D) $3$
Explanation:
$\begin{aligned} n & =100, \\ p & =\frac{1}{10} \\ q & =\frac{9}{10} \\ \sigma & =\sqrt{npq} \\ & =\sqrt{100 \times \frac{1}{10} \times \frac{9}{10}} \\ & =3\end{aligned}$

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MCQ 111 Mark

An insurance company has found that $50\%$ of its claims are for damages resulting from accidents. The probability that a random sample of 10 claims will contain fewer than 2 for accidents is:

A
$\frac{1}{1024}$
B
$\frac{5}{512}$
✓
$\frac{11}{1024}$
D
$\frac{15}{1024}$

Answer

Correct option: C.

$\frac{11}{1024}$

(C) $\frac{11}{1024}$
Explanation
$P(r < 2) = P(0$ or $1)$
$=C_0^{10}\left(\frac{1}{2}\right)^{10}+C_1^{10}\left(\frac{1}{2}\right)^{10}$
$=\frac{1+10}{1024}$
$=\frac{11}{1024}$

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MCQ 121 Mark

A TV manufacturer tests a random sample of 6 picture tubes to determine any defect. Past experience suggests the probability of defective picture tube is 0.05. The probability that there is at least one defective picture tube in the sample is:

A
$\left(\frac{19}{20}\right)^6$
✓
$1-\left(\frac{19}{20}\right)^6$
C
$1-\left[\left(\frac{19}{20}\right)^6+\frac{3}{10}\left(\frac{19}{20}\right)^5\right]$
D
$\left(\frac{1}{20}\right)^6$

Answer

Correct option: B.

$1-\left(\frac{19}{20}\right)^6$

(B) $1-\left(\frac{19}{20}\right)^6$
Explanation: $p = 0.05$
$=\frac{1}{20}$
$q=\frac{19}{20}$
$P(x\geq1)=1-P(0)$
$=1-6_{c_0}(\frac{1}{20})^{0}(\frac{19}{20})^{6}$
$=1-(\frac{19}{20})^{6}$

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MCQ 131 Mark

For a Poisson distribution with mean $\lambda_r \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!}$ is equal to

A
$-1$
B
$0$
C
$1/2$
✓
$1$

Answer

Correct option: D.

$1$

(D) $1$
Explanation:
$\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!}=$ Total probability
$= 1$

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MCQ 141 Mark

An automatic machine produces 20000 pins per day. On rare occasion it produces a perfect pin whose chance is $\frac{1}{10000}$. Assuming Poisson distribution, the mean and variance of the number of perfect pins are ___ and ___, respectively

A
$\sqrt{2}, \sqrt{2}$
✓
$2, 2$
C
$2,4$
D
$4, 2$

Answer

Correct option: B.

$2, 2$

(B) $2,2$
Explanation: For Poisson distribution
Mean $=$ variance
$= np$
$=20000 \times \frac{1}{10000}$
$= 2$

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MCQ 151 Mark

There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0-1. The probability that all the lines are busy is:

A
$\frac{5^0 e^{-5}}{0!}$
B
$1-\frac{5^0 e^{-5}}{0!}$
✓
$\frac{5^{50} e^{-5}}{50!}$
D
$1-\frac{5^{50} e^{-5}}{50!}$

Answer

Correct option: C.

$\frac{5^{50} e^{-5}}{50!}$

(C) $\frac{5^{50} e^{-5}}{50!}$
Explanation: Given, 50 telephone line in an exchange probability that any of them will be busy is 0.1.
Let X be the Poisson variable.
$\therefore \quad P(X=r)=\frac{e^{-m} m^r}{r!}, r=0,1,2, \ldots$
Here, $n = 50, p = 0.1$
$\therefore \quad m=n p=50 \times 0.1=5$
$\therefore P(X=r)=\frac{e^{-5}(5)^r}{r!}, r=0,1,2, \ldots, 50$
Thus, P(all lines are busy) $= P(X = 50)$
$\therefore \quad=\frac{e^{-5}(5)^r}{r!}$
For $r = 50,$ $P=\frac{e^{-5} 5^{50}}{50!}$

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MCQ 161 Mark

If a random variable X has the Poisson distribution with mean 2. Then, $P(X > 1.5)$ is:

A
$2 e^{-2}$
B
$3 e^{-2}$
C
$1-2 e^{-2}$
✓
$1-3 e^{-2}$

Answer

Correct option: D.

$1-3 e^{-2}$

(D) $1-3 e^{-2}$
Explanation: We know that,
$P(X=x)=\frac{e^{-\lambda} \lambda^x}{x!}$
Given that mean $=2=\lambda$
$\therefore \quad P(X>1.5)=P(X=2)+P(X=3)+\ldots$
$\Rightarrow P(X>1.5)=1-[P(X=0)+P(X=1)]$
$=1-\left|e^{-2}+2 e^{-2}\right|=1-3 e^{-2}$

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MCQ 171 Mark

If the mean of a binomial distribution is 81, then the standard deviation lies in the interval:

✓
[0,9)
B
(0,9]
C
[0,3]
D
(0,3]

Answer

Correct option: A.

[0,9)

(A) [0,9)
Explanation: Standard deviation, $\sigma=\sqrt{n p q} \geq 0$
Now, mean $= np = 81$ and $q < 1$
So, $\sigma=\sqrt{n p q}<\sqrt{n p}=\sqrt{81}=9$
$\therefore$ $0 \leq \sigma<9$
Hence, $\sigma$ lies in $[0,9)$

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MCQ 181 Mark

The mean $E(x)$ of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is:

A
1
✓
2
C
5
D
$\frac{8}{3}$

Answer

Correct option: B.

(B) $2$
Explanation: Total no. of observations $= 6$
$\therefore \quad P(X=1)=\frac{3}{6}=\frac{1}{2}$
$P(X=2)=\frac{2}{6}=\frac{1}{3}$
$P(X=5)=\frac{1}{6}$
(where X be a random variable representing a number on die)
$\therefore \quad$ Mean, $E(X)=\Sigma p_i X_i$
$=\frac{1}{2} \times 1+\frac{1}{3} \times 2+\frac{1}{6} \times 5$
$=\frac{1}{2}+\frac{2}{3}+\frac{5}{6}$
$=\frac{3+4+5}{6}$
$=\frac{12}{6}=2$

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MCQ 191 Mark

In a box of 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?

✓
$\left(\frac{9}{10}\right)^5$
B
$\frac{9}{10}$
C
$10^{-5}$
D
$\left(\frac{1}{2}\right)^2$

Answer

Correct option: A.

$\left(\frac{9}{10}\right)^5$

(A) $\left(\frac{9}{10}\right)^5$
Explanation: Probability of getting a defective bulb,
$p=\frac{10}{100}=\frac{1}{10}$
$\therefore \quad q=1-p$
$=1-\frac{1}{10}=\frac{9}{10}$
Here, n = 5
By binomial distribution probability of getting no defective bulb
$=P(X=0)={ }^n C_0 p^0 q^{n-0}$
$={ }^5 C_0\left(\frac{9}{10}\right)^5=\left(\frac{9}{10}\right)^5$

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MCQ 201 Mark

If '$m$' is the mean of Poisson distribution, then its standard deviation is given by:

✓
$\sqrt{m}$
B
$m^2$
C
$m$
D
$\frac{m}{2}$

Answer

Correct option: A.

$\sqrt{m}$

(A) $\sqrt{m}$
Explanation: In case of Poisson distribution
$S . D =\sqrt{\text { mean }}$
Thus, $S , D =\sqrt{m}\quad$ [Given, mean $= m$]

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MCQ 211 Mark

A variable that can assume any value between two given points is called _________.

✓
Continuous Random Variable
B
Discrete Random Variable
C
Irregular Random Variable
D
Uncertain Random Variable

Answer

Correct option: A.

Continuous Random Variable

(A) Continuous Random Variable
Explanation: This is the definition of a continuous random variable.

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MCQ 221 Mark

A table which contains possible value of a random variable and its corresponding probabilities is called _________.

A
Probability Mass Function
B
Probability Density Function
C
Cumulative Distribution Function
✓
Probability Distribution

Answer

Correct option: D.

Probability Distribution

(D) Probability Distribution
Explanation: The given statement is the definition of a probability distribution.

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MCQ 231 Mark

Find the expectation of a random variable X.

$X$	$0$	$1$	$2$	$3$
$f(X)$	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{2}{6}$	$\frac{1}{6}$

A
0.5
✓
1.5
C
2.5
D
3.5

Answer

Correct option: B.

1.5

(B) 1.5
Explanation:
$E(X)=0\left(\frac{1}{6}\right)+1\left(\frac{2}{6}\right)+2\left(\frac{2}{6}\right)+3\left(\frac{1}{6}\right)$
$=0+\frac{2}{6}+\frac{4}{6}+\frac{3}{6}$
$=\frac{9}{6}=\frac{3}{2}=1.5$

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MCQ 241 Mark

Find the mean and variance of X?

$X =x$	$0$	$1$	$2$	$3$	$4$
$f(X)$	$\frac{1}{9}$	$\frac{2}{9}$	$\frac{3}{9}$	$\frac{2}{9}$	$\frac{1}{9}$

✓
$2, \frac{4}{3}$
B
$3, \frac{4}{3}$
C
$2, \frac{2}{3}$
D
$3, \frac{2}{3}$

Answer

Correct option: A.

$2, \frac{4}{3}$

(A) $2, \frac{4}{3}$
Explanation:
Mean $= E(X) = ΣΧf(X)$
$=0\left(\frac{1}{9}\right)+1\left(\frac{2}{9}\right)+2\left(\frac{3}{9}\right)+3\left(\frac{2}{9}\right)+4\left(\frac{1}{9}\right)$
$= 2$
Variance $=E\left(X^2\right)-(E(X))^2$
$=\left(0+\frac{2}{9}+\frac{12}{9}+\frac{18}{9}+\frac{16}{9}\right)-4$
$=\frac{4}{3}$

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MCQ 251 Mark

Mean of a constant 'a' is _________

A
$0$
✓
$a$
C
$a/2$
D
$1$

Answer

Correct option: B.

$a$

(B) $a$
Explanation: Let $f(x)$ be the pdf of the random variable $X.$
Now, $E(a)=\int a f(x)=a \int f(x)=a .1=a$

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MCQ 261 Mark

The expectation of a random variable X (continuous or discrete) is given by __________.

✓
$\sum X f(x), \int X f(X)$
B
$\Sigma X^2 f(X), \int X^2 f(X)$
C
$\Sigma f(X), \int f(X)$
D
$\sum X f\left(X^2\right), \int X f\left(X^2\right)$

Answer

Correct option: A.

$\sum X f(x), \int X f(X)$

(A) $\sum X f(x), \int X f(X)$
Explanation: The expectation of a random variable X is given by the summation (integral) of X times the function in its interval. If it is a continuous random variable, then summation is used and if it is discrete random variable, then integral is used.

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MCQ 271 Mark

A sales promotion company sells tickets for ₹ 100 each to win a prize of ₹ 5 lakhs. If a person buys one of the 10,000 tickets sold, then his expected gain in rupees is

✓
$-50$
B
$0$
C
$50$
D
$100$

Answer

Correct option: A.

$-50$

(A) $-50$
Explanation:

Prize $\left(x_i\right)$	$p_i$	$x_i p_i$
$500000$	$\frac{1}{10000}$	$50$
$0$	$\frac{9999}{10000}$	$0$

So, $\sum x_i p_i=50$
Net expected gain $= 50-100$
$= -50$
$\therefore$ No gain
$\therefore$ Loss is ₹ 50

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MCQ 281 Mark

Let $X$ denotes the number of hours a student devotes to self-study during a randomly selected school day. The probability that $X$ takes the value $x,$ where $k$ is some unknown constant is:
$P(X=x)=\left\{\begin{array}{ccc}k & \text { if } & x=0 \\ k x & \text { if } & x=1 \text { or } 2 \\ k(5-x) & \text { if } & x=3 \text { or } 4 \\ 0 & & \text { otherwise }\end{array}\right.$
The probability that a student studies at least 3 hours on a particular day is:

A
$\frac{1}{7}$
B
$\frac{2}{7}$
✓
$\frac{3}{7}$
D
$\frac{1}{2}$

Answer

Correct option: C.

$\frac{3}{7}$

(C) $\frac{3}{7}$
Explanation: $\Sigma p_i=1$
$\Rightarrow \quad 7 k=1$
$\Rightarrow \quad k=\frac{1}{7}$
Now, $P(x \geq 3)=3 k=3 \times \frac{1}{7}$
$=\frac{3}{7}$

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MCQ 291 Mark

A candidate claims $70\%$ of the people in her constituency would vote for her. If $120000$ valid votes are polled, then the number of votes she expects from her constituency is:

A
100000
✓
84000
C
56000
D
36000

Answer

Correct option: B.

84000

(B) 84000
Explanation: Expected number of votes
$=np$
$=\frac{70}{100} \times 120000$
$= 84000$

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MCQ 301 Mark

The mean of the probability distribution of the number of doublets in 4 throws of a pair of dice, is:

A
$1$
✓
$\frac{2}{3}$
C
$1 \frac{3}{5}$
D
$2 \frac{2}{3}$

Answer

Correct option: B.

$\frac{2}{3}$

(B) $\frac{2}{3}$
Explanation: Let X: Number of doublets X = 0, 1, 2, 3, 4
Total number of possible outcomes = 36
Doublets = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
$P($Doublet$)=\frac{6}{36}=\frac{1}{6}$
P(Not a Doublet) $=\frac{5}{6}$

X	P(X)	X.P(X)
0	$\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{625}{1296}$	0
1	$4 \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{500}{1296}$	$\frac{500}{1296}$
2	$6 \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{150}{1296}$	$\frac{300}{1296}$
3	$4 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}=\frac{20}{1296}$	$\frac{60}{1296}$
4	$\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{1296}$	$\frac{4}{1296}$

$\therefore \quad$ Mean $=\sum XP(X)$
$=0+\frac{500}{1296}+\frac{300}{1296}+\frac{60}{1296}+\frac{4}{1296}$
$=\frac{864}{1296}=\frac{2}{3}$

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MCQ 311 Mark

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. Then the possible values of X are

A
0,1,3,5
✓
0, 2, 4, 6
C
0,2,5,6
D
1, 3, 4, 5

Answer

Correct option: B.

0, 2, 4, 6

(B) 0, 2, 4, 6
Explanation: The coin is tossed six times and X represents the difference between the number of heads and the number of tails.
$\therefore X(6H,0T)=|6-0|=6$
$X(5 H, 1 T)=|5-1|=4$
$X(4 H, 2 T)=|4-2|=2$
$X(3 H, 3 T)=|3-3|=0$
$X(2 H, 4 T)=|2-4|=2$
$X(1 H, 5 T)=|1-5|=4$
$X(0 H, 6 T)=|0-6|=6$
Thus, possible values are $0,2,4,6.$

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MCQ 321 Mark

Let X be a discrete random variable whose probability distribution is given below:

$X=x_i$	0	1	2	3	4	5	6	7
$P\left(X=x_i\right)$	0	2K	2K	3K	$K^2$	2$K^2$	7$K^2$	2K

The value of K is:

✓
$\frac{1}{10}$
B
$-1$
C
$-\frac{1}{10}$
D
$\frac{1}{5}$

Answer

Correct option: A.

$\frac{1}{10}$

(A) $\frac{1}{10}$
Explanation: Here, $\Sigma P_i=1$
$\therefore 0+2 K+2 K+3 K+K^2+2 K^2+7 K^2+2 K=1$
$\Rightarrow \quad 9 K+10 K^2=1$
$\Rightarrow \quad 10 K^2+9 K-1=0$
$\Rightarrow \quad 10 K^2+10 K-K-1=0$
$\Rightarrow 10 K(K+1)-1(K+1)=0$
$\Rightarrow \quad(K+1)(10 K-1)=0$
$\Rightarrow \quad K=-1$ and $K=\frac{1}{10}$
Since, $K \geq 0$ therefore $K=\frac{1}{10}$

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Applied Maths MCQ

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