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 _________
f(x)=ae^{-2|x|}
The value of 'a' is _________
0.5 | |
1 | |
1.5 | |
2 |
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}
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
p/\left ( 1-p \right ) | |
\left ( 1-p \right )/p | |
1/p | |
1/p^{2} |
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}
\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 ________
0.1 | |
0.26 | |
0.65 | |
0.85 |
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
\frac{kT}{C} | |
\sqrt{\frac{kT}{C}} | |
\frac{C}{kT} | |
\frac{\sqrt{kT}}{C} |
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}}
\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
\frac{7}{8} | |
\frac{49}{64} | |
\frac{7}{64} | |
\frac{105}{64} |
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 ________.
0.4 | |
0.9 | |
1.5 | |
2.6 |
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}
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
\frac{1}{2} | |
\frac{4}{9} | |
\frac{5}{9} | |
\frac{6}{9} |
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
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
2,\frac{1}{2} | |
4,\frac{1}{2} | |
2,\frac{1}{4} | |
4,\frac{1}{4} |
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 _____.
0.1 | |
0.2 | |
0.4 | |
0.8 |
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
5/11 | |
1/2 | |
7/13 | |
6/11 |
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}
\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.