Probability and Statistics

Question 1
Let the probability density function of a random variable x be given as
f(x)=ae^{-2|x|}
The value of 'a' is _________
A
0.5
B
1
C
1.5
D
2
GATE EE 2022   Engineering Mathematics
Question 1 Explanation: 
f(x)=\left\{\begin{matrix} ae^{2x} &x \lt 0 \\ ae^{-2x} &x \gt 0 \end{matrix}\right.
Therefore,
\begin{aligned} \int_{-\infty }^{\infty }f(x)dx&=1\\ \int_{-\infty }^{0}ae^{2x}dx+\int_{0 }^{\infty }ae^{-2x}dx&=1\\ a\left [ \left [ \frac{e^{2x}}{2} \right ]_{-\infty }^0 +[ \left [ \frac{e^{-2x}}{-2} \right ]^{\infty }_0\right ]&=1\\ a\left [ \frac{1}{2}+\frac{1}{2} \right ]&=1\\ a&=1 \end{aligned}
Question 2
Suppose the probability that a coin toss shows "head" is p, where 0\lt p\lt 1. The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is
A
p/\left ( 1-p \right )
B
\left ( 1-p \right )/p
C
1/p
D
1/p^{2}
GATE EE 2021   Engineering Mathematics
Question 2 Explanation: 
F(H) = p, let x = Number of tosses
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \ldots \\ \hline P(x) & p & (1-p) p & (1-p)^{2} p & & \\ \hline \end{array}\\ &\begin{aligned} E(x) &=\sum x_{i} P(x)=p+2(1-p) p+3(p)(1-p)^{2}+\ldots \\ &=p\left[1+2(1-p)+3(1-p)^{2}+\ldots\right] \\ &=p[1-(1-p)]^{-2}=\frac{1}{p} \end{aligned} \end{aligned}
Question 3
The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places)is ________
A
0.1
B
0.26
C
0.65
D
0.85
GATE EE 2019   Engineering Mathematics
Question 3 Explanation: 
\begin{aligned} p&=0.02\\ n&=50\\ \lambda &=np=50(0.02)=1\\ p(x\geq 2)&=1-p(x \lt 2)\\ &=1-(p(x=0)+p(x=1))\\ &=1-\left ( \frac{e^{-\lambda }\lambda ^0}{0!}+\frac{e^{-\lambda }\lambda ^1}{1!} \right )\\ &=1-e^{-\lambda }(1+\lambda )\\ &=1-e^{-1}(1+1)=0.26 \end{aligned}
Question 4
The mean-square of a zero-mean random process is \frac{kT}{C}, where k is Boltzmann's constant, T is the absolute temperature, and C is a capacitance. The standard deviation of the random process is
A
\frac{kT}{C}
B
\sqrt{\frac{kT}{C}}
C
\frac{C}{kT}
D
\frac{\sqrt{kT}}{C}
GATE EE 2019   Engineering Mathematics
Question 4 Explanation: 
Given that,
\begin{aligned} E(x^2)&=\frac{kT}{C}\\ E(x)&=0\\ &=E(x^2)-(E(x))^2\\ Var(x)&=\frac{kT}{C}-0=\frac{kT}{C} \end{aligned}
Standard deviation =\sqrt{\frac{kT}{C}}
Question 5
A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var {Y}, where var {.} denotes the variance, equals
A
\frac{7}{8}
B
\frac{49}{64}
C
\frac{7}{64}
D
\frac{105}{64}
GATE EE 2017-SET-2   Engineering Mathematics
Question 5 Explanation: 


\begin{aligned} E(y)&=0\times \frac{1}{8}+1 \times \frac{7}{8}=\frac{7}{8} \\ E(y^2) &=0^2\left ( \frac{1}{8} \right )+1^2 \times \frac{7}{8}=\frac{7}{8} \\ \text{Variance(y)} &=E(y^2)-(E(y))^2 \\ &= \frac{7}{8}-\frac{49}{64}=\frac{56-49}{64}=\frac{7}{64} \end{aligned}
Question 6
Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green (vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is ________.
A
0.4
B
0.9
C
1.5
D
2.6
GATE EE 2017-SET-2   Engineering Mathematics
Question 6 Explanation: 
t be arrival time of vehicles of the junction is uniformaly distributed in [0,5].
Let y be the waiting time of the junction.

\begin{aligned} \text{Then }y&=\left\{\begin{matrix} 0 & t \lt 2 \\ 5-t & 2\leq t \lt 5 \end{matrix}\right.\\ y\rightarrow &[0,5]\\ f(y)&=\frac{1}{5-0}=\frac{1}{5}\\ E(y)&=\int_{-\infty }^{0}y(y)dy=\int_{0}^{5}yf(y)dy\\ &=\int_{2}^{5}y\left ( \frac{1}{5} \right )dy=\frac{1}{5}\int_{2}^{5}(5-t)dt\\ &=\frac{1}{5}\left ( 5t-\frac{t^2}{2} \right )|_2^5\\ &=\frac{1}{5}\left [ \left ( 25-\frac{25}{2} \right )-\left ( 10-\frac{4}{2} \right ) \right ]\\ &=\frac{1}{5}\left ( \frac{25}{2}-8 \right )=\frac{1}{5}\frac{9}{2}=0.9 \end{aligned}
Question 7
An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is
A
\frac{1}{2}
B
\frac{4}{9}
C
\frac{5}{9}
D
\frac{6}{9}
GATE EE 2017-SET-2   Engineering Mathematics
Question 7 Explanation: 


P(red)=\frac{5}{10}\frac{4}{9}+\frac{5}{10}\frac{5}{9}=\frac{45}{90}=\frac{1}{2}
Question 8
Let the probability density function of a random variable, X, be given as:
f_X(x)=\frac{3}{2}e^{-3x}u(x)+ae^{4x}u(-x)
where u(x) is the unit step function.
Then the value of 'a' and Prob\{X\leq 0\}, respectively, are
A
2,\frac{1}{2}
B
4,\frac{1}{2}
C
2,\frac{1}{4}
D
4,\frac{1}{4}
GATE EE 2016-SET-2   Engineering Mathematics
Question 8 Explanation: 
\begin{aligned} f_x(x)&=\left\{\begin{matrix} ax^{4x} &x \lt 0 \\ \frac{3}{2}e^{-3x}& x\geq 0 \end{matrix}\right.\\ \int_{-\infty }^{\infty }f_x(x)&=1\\ \int_{-\infty }^{0}ax^{4x}dx+\int_{0}^{\infty }\frac{3}{2}e^{-3x}dx&=1\\ \left [ \frac{ax^{4x}}{4} \right ]_{-\infty }^0+\left [ \frac{\frac{3}{2}e^{-3x}}{-3} \right ]_0^\infty &=1\\ \frac{a}{3}+\frac{3}{6}&=1\\ a&=2\\ P(x\leq 0)&=\int_{-\infty }^{0}2e^{4x}dx \\ &=\left [ \frac{e^{4x}}{2} \right ]_{-\infty }^0=\frac{1}{2} \end{aligned}
Question 9
Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all 3 pens having the same colour is _____.
A
0.1
B
0.2
C
0.4
D
0.8
GATE EE 2016-SET-1   Engineering Mathematics
Question 10
Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is
A
5/11
B
1/2
C
7/13
D
6/11
GATE EE 2015-SET-1   Engineering Mathematics
Question 10 Explanation: 
P(A wins) = P( 6 in first throw by A) + P(A not 6, B not 6,)+...
\begin{aligned} &=\frac{1}{6}+\frac{5}{6}\frac{5}{6}\frac{1}{6}+....\\ &=\frac{1}{6}\left ( 1+\left ( \frac{5}{6} \right )^2+\left ( \frac{5}{6} \right )^4+... \right )\\ &=\frac{1}{6}\left ( \frac{1}{1-\left (\frac{5}{6} \right )^2} \right )=\frac{6}{11} \end{aligned}
There are 10 questions to complete.