# Probability and Statistics

 Question 1
A random variable $X$, distributed normally as $N(0,1)$, undergoes the transformation $Y=h(X)$, given in the figure. The form of the probability density function of $Y$ is (In the options given below, $a, b, c$ are non-zero constants and $g(y)$ is piece-wise continuous function) A $a \delta(y-1)+b \delta(y+1)+g(y)$ B $a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)$ C $a \delta(y+2)+b \delta(y)+c \delta(y-2)+g(y)$ D $a \delta(y+1)+b \delta(y-2)+g(y)$
GATE EC 2023   Engineering Mathematics
Question 1 Explanation:
$X=N(0,1)$ $f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ \begin{aligned} Y= & -1 ; x \leq-2 \\ & 0 ;-1 \leq x \leq 1 \\ & 1 ; x \geq 2 \\ & x+1 ;-2 \leq x \leq-1 \\ & x-1 ; 1 \leq x \leq 2 \end{aligned}
$Y$ is taking discrete set of values and a continuous range of values, so it is mixed random variable.
From the given options, density function of '$Y$' will be.
$f_{Y}(y)=a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)$
 Question 2
The bar graph shows the frequency of the number of wickets taken in a match by a bowler in her career. For example, in 17 of her matches, the bowler has taken 5 wickets each. The median number of wickets taken by the bowler in a match is __________ (rounded off to one decimal place) A 3.2 B 4 C 6.4 D 5
GATE EC 2022   Engineering Mathematics
Question 2 Explanation:
Sum of frequency of No. of wickets = 5 +7 + 8 + 25 + 20 +17 + 8 + 4 + 3 + 2 + 1 = 100
Mean = Average of the 50th & 51th matches = 4

 Question 3
In a high school having equal number of boy students and girl students, $75\%$ of the students study Science and the remaining $25\%$ students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is ______ bits.
 A 1.25 B 3.32 C 5.45 D 6.55
GATE EC 2021   Engineering Mathematics
Question 3 Explanation:
\begin{aligned} \text{Given,}\qquad P(G)&=\frac{1}{2} \text { and } P(B)=\frac{1}{2} \\ P(C)&=\frac{1}{4} \text { and } P(S)=\frac{3}{4} \end{aligned}
Let probability of selected science student is a boy is $P(B/S) = x$
Given that Commerce students are two time more likely to be a boy than are Science students.
Then, probability of selected Commerce student is a boy
$P\left(\frac{B}{C}\right)=2 x$
We have to find probability of randomly selected girl studies Commerce i.e.
\begin{aligned} P\left(\frac{C}{G}\right)&=\frac{P(C \cap G)}{P(G)}=\frac{P(C) P(G / C)}{P(G)} \\ P\left(\frac{C}{G}\right)&=\frac{\frac{1}{4} \times P\left(\frac{G}{C}\right)}{\frac{1}{2}} &\ldots(i) \end{aligned}
To find $P(G / C)$, first we have to find $P(B/C).$
The probability of selected student is a boy
\begin{aligned} P(B) &=P(S) P(B / S)+P(C) \times P(B / C) \\ \frac{1}{2} &=\frac{3}{4} \times x+\frac{1}{4} \times 2 x \\ \Rightarrow\qquad\qquad x &=\frac{2}{5}\\ \text{Then},\qquad \qquad P(B / C)&=2 x=\frac{4}{5}\\ \text{We know that},\qquad \\ P\left(\frac{B}{C}\right)+P\left(\frac{G}{C}\right)&=1 \\ P\left(\frac{G}{C}\right) =1- \frac{4}{5}&=\frac{1}{5} \\ \text{From equation (i),} \\ P\left(\frac{G}{C}\right) = \frac{1}{4} \times \frac{1}{\frac{5}{1/2}} &= \frac{1}{10} \end{aligned}
The amount of information gained in knowing that a randomly selected girl student studies Commerce
$I\left ( \frac{C}{G} \right )=\log _2 \frac{1}{P\left ( \frac{C}{G} \right )}= \log _2 10=3.32$
 Question 4
Box contains the following three coins.

i. A fair coin with head on one face and tail on the other face.
ii. A coin with heads on both the faces.
iii. A coin with tails on both the faces.

A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is
 A $\frac{2}{5}$ B $\frac{1}{3}$ C $\frac{1}{2}$ D $\frac{2}{3}$
GATE EC 2021   Engineering Mathematics
Question 4 Explanation:
Let $\quad P\left(H_{2}\right)=$ Probability of getting head in second toss
$P\left(H_{1}\right)=$ Probability of getting head in first toss
\begin{aligned} P\left(\frac{H_{2}}{H_{1}}\right) &=\frac{P\left(H_{2} \cap H_{1}\right)}{P\left(H_{1}\right)} \\ P\left(H_{1}\right) &=\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times 1+\frac{1}{3} \times 0=\frac{1}{2} \end{aligned}
To get head in second toss when head came in first toss, following cases can be made
1. Fair coin :
\begin{aligned} \text { Both head coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 1=\frac{1}{12} \\ \text { Both tail coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 0=0 \end{aligned}
\begin{aligned} \text { Fair coin }&=\frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{12}\\ \text{Both tail coin }&=0\\ \therefore \qquad P\left(H_{2} \cap H_{1}\right)&=\frac{1}{12}+\frac{1}{12}=\frac{1}{6} \\ \therefore \qquad P\left(H_{2} / H_{1}\right)&=\frac{1 / 6}{1 / 2}=\frac{1}{3} \end{aligned}
 Question 5
The two sides of a fair coin are labelled as 0 and 1. The coin is tossed two times independently. Let M and N denote the labels corresponding to the outcomes of those tosses. For a random variable X, defined as X = min(M, N), the expected value E(X) (rounded off to two decimal places) is _________.
 A 0.15 B 0.75 C 0.55 D 0.25
GATE EC 2020   Engineering Mathematics
Question 5 Explanation:
s={(H,H),(H,T),(T,H),(T,T)}
$M=\begin{bmatrix} 1 & 1 & 0 & 0 \\ H & H & T & T \end{bmatrix}$ of first toss

$N=\begin{bmatrix} H & T & H& T \\ 1 & 0 & 1 & 0 \end{bmatrix}$ of second toss
Now,X=Min{M,N}
$\therefore$
X=Min{H,H}=Min{1,1}=1
X=Min{H,T}=Min{1,0}=0
X=Min{T,H}=Min{0,1}=0
X=Min(T,T)=Min(0,0)=0
$\therefore$P(X=1)=$\frac{1}{4}$,P(x=0)=$\frac{3}{4}$
We Know that, $E(X)=\sum_{i}X_{i}P(x_{i})=1\times \frac{1}{4}+0\times \frac{3}{4}=\frac{1}{4}=0.25$

There are 5 questions to complete.