Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 Explanation:
P\left(X_{4} \text { is smallest }\right)=\frac{3 !}{4 !}=\frac{1}{4}=0.25
Question 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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}
Question 8 |
Two random variables X and Y are distributed according to
f_{x,y}(x,y)=\left\{\begin{matrix} (x+y) &0 \leq x\leq 1,0\leq y\leq 1 \\ 0 & otherwise. \end{matrix}\right.
The probability p(x+y\leq 1) is ________
f_{x,y}(x,y)=\left\{\begin{matrix} (x+y) &0 \leq x\leq 1,0\leq y\leq 1 \\ 0 & otherwise. \end{matrix}\right.
The probability p(x+y\leq 1) is ________
0.14 | |
0.33 | |
0.55 | |
0.68 |
Question 8 Explanation:
\begin{aligned} P(X+Y \leq 1) &=\int_{x=0}^{1} \int_{y=0}^{(1-x)} f_{x y}(x, y) d x d y \\ &=\int_{x=0}^{1} \int_{y=0}^{1-x}(x+y) d x d y \\ &=\int_{x=0}^{1}\left(x y+\frac{y^{2}}{2}\right)_{0}^{1-x}\\ &=\int_{x=0}^{1}\left(x(1-x)+\frac{(1-x)^{2}}{2}\right) d x \\ &=\int_{x=0}^{1}\left(\frac{1}{2}-\frac{x^{2}}{2}\right) d x=\left(\frac{x}{2}-\frac{x^{3}}{6}\right)_{0}^{1} \\ &=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}=0.33 \end{aligned}
Question 9 |
The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is _______
0.5 | |
1 | |
2 | |
3 |
Question 9 Explanation:
In Poisson distribution,
Mean = First moment =\lambda
secondmoment =\lambda^{2}+\lambda
Given, that second moment is 2
\begin{array}{r} \lambda^{2}+\lambda=2 \\ \lambda^{2}+\lambda-2=0 \\ (\lambda+2)(\lambda-1)=0 \\ \lambda=1 \end{array}
Mean = First moment =\lambda
secondmoment =\lambda^{2}+\lambda
Given, that second moment is 2
\begin{array}{r} \lambda^{2}+\lambda=2 \\ \lambda^{2}+\lambda-2=0 \\ (\lambda+2)(\lambda-1)=0 \\ \lambda=1 \end{array}
Question 10 |
A fair die with faces {1, 2, 3, 4, 5, 6} is thrown repeatedly till '3' is observed for the first time. Let X denote the number of times the die is thrown. The expected value of X is ____.
3 | |
4 | |
5 | |
6 |
There are 10 questions to complete.