# Probability and Statistics

 Question 1
The mean and variance, respectively, of a binomial distribution for $n$ independent trials with the probability of success as $p$, are
 A $\sqrt{np},np(1-2p)$ B $\sqrt{np}, \sqrt{np(1-p)}$ C $np,np$ D $np,np(1-p)$
GATE ME 2021 SET-2   Engineering Mathematics
Question 1 Explanation:
Mean= np
Variance = npq = np(1 - p)
 Question 2
Consider a single machine workstation to which jobs arrive according to a Poisson distribution with a mean arrival rate of 12 jobs/hour. The process time of the workstation is exponentially distributed with a mean of 4 minutes. The expected number of jobs at the workstation at any given point of time is ________ (round off to the nearest integer).
 A 3 B 4 C 6 D 8
GATE ME 2021 SET-1   Engineering Mathematics
Question 2 Explanation:
$\lambda=12$ per hour, $\frac{1}{\mu}=4 \min / \mathrm{Jobs}, \mu=15 \mathrm{~Jobs} / \mathrm{hr}$
\begin{aligned} \rho &=\frac{12}{15}=\frac{4}{5} \\ L_{s} &=\frac{\rho}{1-\rho}=\frac{\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{4}{5}}{\frac{1}{5}} \\ \frac{4}{5} \times \frac{5}{1} &=4 \text { Jobs } \end{aligned}
 Question 3
Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour. The manager notes that no customer arrives for the first 3 minutes after the shop opens. The probability that a customer arrives within the next 3 minutes is
 A 0.39 B 0.86 C 0.5 D 0.61
GATE ME 2021 SET-1   Engineering Mathematics
Question 3 Explanation:
10 Customers arrived in 1 hour, $\lambda =10/hr$
For 3 minutes, $\lambda =\frac{1}{2}/hr$
$P(X=0)=\frac{e^{-\lambda }\lambda ^x}{x!}=\frac{e^{-\frac{1}{2}}\left ( \frac{1}{2} \right )^0}{0!}=0.606$
The probability that a customer arrives within the next 3 minutes,
$P = 1 – 0.606 = 0.39$
 Question 4
Robot Ltd. wishes to maintain enough safety stock during the lead time period between starting a new production run and its completion such that the probability of satisfying the customer demand during the lead time period is 95%. The lead time period is 5 days and daily customer demand can be assumed to follow the Gaussian (normal) distribution with mean 50 units and a standard deviation of 10 units. Using $\phi ^{-1}(0.95)=1.64$, where $\phi$ represents the cumulative distribution function of the standard normal random variable, the amount of safety stock that must be maintained by Robot Ltd. to achieve this demand fulfillment probability for the lead time period is _______ units (round off to two decimal places).
 A 12.25 B 58.36 C 36.67 D 18.62
GATE ME 2021 SET-1   Engineering Mathematics
Question 4 Explanation:
\begin{aligned} \text { Safety stock }&=Z \times \sigma \\ \sigma &=\sqrt{10^{2}+10^{2}+10^{2}+10^{2}+10^{2}} \\ &=10 \sqrt{5} \\ &=22.3606 \\ \text { Safety stock } &=1.64 \times 22.3606 \\ &=36.67 \end{aligned}
 Question 5
Consider a binomial random variable $X$. If $X_1,X_2,..., X_n$ are independent and identically distributed samples from the distribution of $X$ with sum $Y=\sum_{i=1}^{n}X_i$, then the distribution of $Y$ as $n\rightarrow \infty$ can be approximated as
 A Exponential B Bernoulli C Binomial D Normal
GATE ME 2021 SET-1   Engineering Mathematics
 Question 6
A fair coin is tossed 20 times. The probability that 'head' will appear exactly 4 times in the first ten tosses, and 'tail' will appear exactly 4 times in the next ten tosses is ______ (round off to 3 decimal places).
 A 0.012 B 0.052 C 0.042 D 0.068
GATE ME 2020 SET-2   Engineering Mathematics
Question 6 Explanation:
Fair coin tossed 20 times
First 10 times probability that head will appear exactly 4 times
P[4 heads in 10 tosses] ? P[4 tails in 10 tosses]
\begin{aligned} &=10_{C_{4}}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{6} \cdot 10_{C_{4}}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{4} \\ &=\frac{10_{C_{4}} \cdot 10_{C_{4}}}{2^{10} \cdot 2^{10}}=\frac{210 \times 210}{2^{10} \times 2^{10}}=0.042 \end{aligned}
 Question 7
The sum of two normally distributed random variables X and Y is
 A always normally distributed B normally distributed, only if X and Y are independent C normally distributed, only if X and Y have the same standard deviation D normally distributed, only if X and Y have the same mean
GATE ME 2020 SET-2   Engineering Mathematics
Question 7 Explanation:
\begin{aligned} X_{1} &\sim N\left(\mu_{1}, \sigma_{1}\right)\\ \text{and }\quad X_{2} &\sim N\left(\mu_{2}, \sigma_{2}\right)\\ \text{then }\quad X_{1}+X_{2} &\sim N\left(\mu_{1}+\mu_{2}, \sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}\right) \end{aligned}
Always normally distributed.
 Question 8
Consider two exponentially distributed random variables X and Y, both having a mean of 0.50. Let Z = X + Y and r be the correlation coefficient between X and Y. If the variance of Z equals 0, then the value of r is ___________ (round off to 2 decimal places).
 A 1 B -1 C -2 D 2
GATE ME 2020 SET-1   Engineering Mathematics
Question 8 Explanation:
\begin{aligned} x \sim E\left(\lambda_{1}\right) ; \quad \text { mean } &=\frac{1}{\lambda_{1}}=0.5 \\ \Rightarrow \qquad \lambda_{1} &=2 \\ \text { Variance, } x &=\frac{1}{\lambda_{1}^{2}}=\frac{1}{4}=0.25 \\ Y \sim E\left(\lambda_{2}\right) ; \quad \text { Mean } &=\frac{1}{\lambda_{2}}=0.5 \\ \Rightarrow \quad \lambda_{2} &=2 \\ \text { Variance, } y &=0.25 \\ \text { Given Var }(\mathrm{Z})=\text{Var}(x)+& \text{Var}(\mathrm{y})+2 \mathrm{COV}(x, y) \\ 0 &=0.25+0.25+2 \mathrm{COV}(x, y) \\ COV(x,y)&=-\frac{0.5}{2}=-0.25\\ \text { Correlation, } \qquad \rho&=\frac{\text{COV}(x, y)}{\sigma_{x} \sigma_{y}} \\ &=\frac{-0.25}{\sqrt{(0.25) \sqrt{(0.25)}}}=-1 \end{aligned}
 Question 9
A company is hiring to fill four managerial vacancies. The candidates are five men and three women. If every candidate is equally likely to be chosen then the probability that at least one women will be selected is _______ (round off to 2 decimal places).
 A 0.21 B 0.48 C 0.93 D 0.78
GATE ME 2020 SET-1   Engineering Mathematics
Question 9 Explanation:
5 men, 3 women
P[atleast one women selected for 4 vacancies]
$\begin{array}{l} =1-P[\text { none }] \\ =1-\frac{5_{C_{4}} 3_{C_{0}}}{8_{C_{4}}}=1-\frac{5}{70}=1-\frac{1}{14}=\frac{13}{14}=0.93 \end{array}$
 Question 10
The probability that a part manufactured by a company will be defective is 0.05. If 15 such parts are selected randomly and inspected, then the probability that at least two parts will be defective is ______ (round off to two decimal places).
 A 0.25 B 0.17 C 0.08 D 0.55
GATE ME 2019 SET-2   Engineering Mathematics
Question 10 Explanation:
Let p= probability of making defective part
$\begin{array}{c} =0.05 \\ \Rightarrow \mathrm{q}=1-\mathrm{p}=0.95 \end{array}$
given n=15
Let X be number of defective parts be a random variable.
\begin{aligned} \mathrm{P}(\mathrm{X} \geq 2) &=1-\mathrm{P}(\mathrm{X} \lt 2) \\ &=1-\{\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)\} \\ &=1-\left(^{15} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{\mathrm{n}}+^{15} \mathrm{C}_{1} \mathrm{p}^{1} \mathrm{q}^{14}\right) \\ &\left.=1-\left\{(0.95)^{15}+15(0.05)(0.95)^{14}\right)\right\} \\ &=0.1709 \end{aligned}
There are 10 questions to complete.