Question 1 |

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)

3.2 | |

4 | |

6.4 | |

5 |

Question 1 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

Mean = Average of the 50th & 51th matches = 4

Question 2 |

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.

1.25 | |

3.32 | |

5.45 | |

6.55 |

Question 2 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

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 3 |

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

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

\frac{2}{5} | |

\frac{1}{3} | |

\frac{1}{2} | |

\frac{2}{3} |

Question 3 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}

2. Both head coin :

\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}

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}

2. Both head coin :

\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 4 |

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 _________.

0.15 | |

0.75 | |

0.55 | |

0.25 |

Question 4 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

\thereforeP(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

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

\thereforeP(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

Question 5 |

Let Z be an exponential random variable with mean 1. That is, the cumulative distribution function of Z is given by

F_z(x)=\left\{\begin{matrix} 1-e^{-x} & if & x\geq 0\\ 0 & if & x \lt 0 \end{matrix}\right.

Then Pr(Z\gt 2|Z \gt 1), rounded off to two decimal places, is equal to ______

F_z(x)=\left\{\begin{matrix} 1-e^{-x} & if & x\geq 0\\ 0 & if & x \lt 0 \end{matrix}\right.

Then Pr(Z\gt 2|Z \gt 1), rounded off to two decimal places, is equal to ______

0.37 | |

0.18 | |

0.54 | |

0.9 |

Question 5 Explanation:

Required probability

\begin{aligned} =&P(2 \gt 2) \cap(z \gt 1)]=\frac{P(z \gt 2]}{P[z \gt 1]} \\ P(z \gt 1] &=1-e^{-2} \Rightarrow P(z \gt 2)=e^{-2} \\ P(z \leq 2)=&1-e^{-1} \Rightarrow P(z \gt 1)=e^{-1} \end{aligned}

So.Required probability

=\frac{e^{-2}}{e^{-1}}=e^{-1} \simeq 0.37

\begin{aligned} =&P(2 \gt 2) \cap(z \gt 1)]=\frac{P(z \gt 2]}{P[z \gt 1]} \\ P(z \gt 1] &=1-e^{-2} \Rightarrow P(z \gt 2)=e^{-2} \\ P(z \leq 2)=&1-e^{-1} \Rightarrow P(z \gt 1)=e^{-1} \end{aligned}

So.Required probability

=\frac{e^{-2}}{e^{-1}}=e^{-1} \simeq 0.37

Question 6 |

If X and Y are random variables such that E[2X+Y]=0 and E[X+2Y]=33, then E[X]+E[Y]=________.

8 | |

10 | |

11 | |

13 |

Question 6 Explanation:

\begin{aligned} E[2 X+Y] &=0 \text { and } E[X+2 Y]=33 \\ \text { then, } \quad 2 E[X]+E[Y] &=0 \text { and } E[X]+2 E[Y]=33 \\ 3 E[X]+3 E[Y] &=0+33=33 \\ E[X]+E[Y] &=11 \end{aligned}

Question 7 |

Let X_{1} , X_{2} , X_{3} \; and \; X_{4} be independent normal random variables with zero mean and unit variance. The probability that X_{4} is the smallest among the four is _______.

1 | |

0.25 | |

0.45 | |

0.75 |

Question 7 Explanation:

P\left(X_{4} \text { is smallest }\right)=\frac{3 !}{4 !}=\frac{1}{4}=0.25

Question 8 |

Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is _________.

1 | |

1.5 | |

2 | |

2.5 |

Question 8 Explanation:

Protuatility of getting success =\frac{4}{10}

Propatiility of failure =\frac{6}{10}

\begin{aligned} &E[X]=\sum x_{i} P\left(x_{i}\right) \\ E[X]&=1 \times \frac{4}{10}+2 \times \frac{4}{10} \times \frac{6}{10}\\ &+3 \times \frac{4}{10}\left(\frac{6}{10}\right)^{2}+\ldots \; \; \; \ldots(i) \\ \frac{6}{10} E[X]&=1 \times \frac{4}{10} \times \frac{6}{10}\\ &+2 \times \frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \; \; \; \ldots(ii) \end{aligned}

subtracting (ii) from (i), we get

\begin{aligned} \frac{4}{10} E[X] &=\frac{4}{10}+\frac{4}{10} \times \frac{6}{10}+\frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \\ \frac{4}{10} E[X] &=\frac{\frac{4}{10}}{1-\frac{6}{10}}=1 \\ E[X] &=\frac{10}{4}=2.5 \end{aligned}

Propatiility of failure =\frac{6}{10}

\begin{aligned} &E[X]=\sum x_{i} P\left(x_{i}\right) \\ E[X]&=1 \times \frac{4}{10}+2 \times \frac{4}{10} \times \frac{6}{10}\\ &+3 \times \frac{4}{10}\left(\frac{6}{10}\right)^{2}+\ldots \; \; \; \ldots(i) \\ \frac{6}{10} E[X]&=1 \times \frac{4}{10} \times \frac{6}{10}\\ &+2 \times \frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \; \; \; \ldots(ii) \end{aligned}

subtracting (ii) from (i), we get

\begin{aligned} \frac{4}{10} E[X] &=\frac{4}{10}+\frac{4}{10} \times \frac{6}{10}+\frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \\ \frac{4}{10} E[X] &=\frac{\frac{4}{10}}{1-\frac{6}{10}}=1 \\ E[X] &=\frac{10}{4}=2.5 \end{aligned}

Question 9 |

Three fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place) __________.

0.018 | |

0.038 | |

0.028 | |

0.08 |

Question 9 Explanation:

When three dice are thrown

Total number of possible cases =6 \times 6 \times 6=216

Favourable cases of all three dice have same

number are,

\left\{\begin{array}{lll} (1,1,1) & (2,2,2)&(3,3,3) \\ (4,4,4) & (5,5,5) & (6,6,6) \end{array}\right\}

Number of favourable cases =6

\text { Required probability }=\frac{6}{216}=\frac{1}{36}=0.028

Total number of possible cases =6 \times 6 \times 6=216

Favourable cases of all three dice have same

number are,

\left\{\begin{array}{lll} (1,1,1) & (2,2,2)&(3,3,3) \\ (4,4,4) & (5,5,5) & (6,6,6) \end{array}\right\}

Number of favourable cases =6

\text { Required probability }=\frac{6}{216}=\frac{1}{36}=0.028

Question 10 |

The probability of getting a "head" in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a "head" is obtained. If the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is __________

0.01 | |

0.04 | |

0.07 | |

0.1 |

Question 10 Explanation:

\begin{aligned} P(H)&=0.3 \\ P(T)&=0.7 \end{aligned}

since all tosses are independent

so, probability of getting head for the first time in 5^{\text {th }} toss is

\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}

since all tosses are independent

so, probability of getting head for the first time in 5^{\text {th }} toss is

\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}

There are 10 questions to complete.