Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

An urn contains 4 white and 6 red balls. Four balls are drawn at randomfrom the urn. Find the probability distribution of the number of white balls.

Answer» Let X be the event of drawing white balls
possibilities of X =0,1,2,3,4
` P(X=0) = (.^6C_4)/.^(10C_4)`
`P(X=1) = ((.^9C_1)*(.^6C_3))/(.^10C_4)`
`P(X=2) = ((.^4C_2)*(.^6C_2))/(.^10C_2)`
`P(X=3) = ((.^4C_3)*(.^6C_3))/(.^10C_4)`
`P(X=4) = (.^4C_4)/(.^10C_4)`
Discussion
2.

Find the probability distribution of the number of green balls drawnwhen 3 balls are drawn , one by one, without replacement from a bagcontaining 3 green and 5 white balls.

Answer» let no of green balls to be drawn be = Xpossibilities of X:`0, 1 , 2,3`
`P(X=0)= (5/8)*(4/7)*(3/6)`
`P(X=1) = (3*4*5)/(8*7*6)`
`P(X=2)= 3*(2*3*5)/(8*7*6)`
`P(X=3)= (3/8)*(2/7)*(1/6)`
Discussion
3.

A die is loaded in such a way that an even number is twice likely tooccur as an odd number. If the die is tossed twice, find the probabilitydistribution of the random variable X representing the perfect squares in thetwo tosses.

Answer» let the probability of odd no be `p`
let the probability of even no be `2p`
`p+p+p+2p+2p+2p =1`
`9p = 1 `
`p = 1/9`
let x= perfect square
x = 4,9
`p(x)= P(x) or p(9) = p(4) + p(9)`
`p(x) = p.p + 2p.2p +p.p + p.2p + 2p.p + p.2p + 2p.p`
`p^2 + 4p^2 + p^2 + 8p^2 = 14p^2`
`p(x) = 14p^2`
`= 14* 1/9^2 = 14/81`
Discussion
4.

Two numbers are selected are random (withoutreplacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the largerof the two numbers obtained. Find the mean and variance of the  probability distribution of X.

Answer» Here are some positive integers `(2,3,4,5,6,7)` from which two numbers are selected at random (without replacement) and from those selected numbers larger one is selected.
Smallest pair that can be formed `(2,3)` and Largest pair `(6,7)` therefore minimum value for X is `3` and maximum is `7`.
`P(X=3) = P{(2,3),(3,2)} = P(2,3) + P(3,2)``=1/6*1/5+1/6*1/5`
`P(X=3) = 1/15`Similarly,
`P(X=4) = 2/15`
`P(X=5) = 1/5`
`P(X=6) = 4/15`
`P(X=7) = 1/3`
`Mean = Sigmax . P(X=x)= 3*(1)/15+4*(2)/15+5*(3)/15+6*(4)/15+7*(5)/15`
`=1/15[3+8+15+24+35]`
`=85/15=5.67`
`E(x^2)=3^2*1/15+4^2*2/15+5^2*3/15+144/15+245/15`
`=1/15[9+32+75+144+245]``=33.67`Therefore, Variance `(x)= E(E^2)-E(X)^2`
`=33.67-(5.67)^2``=1.52`
Discussion