Question

Two identical fair 6-sided dice are rolled simultaneously. Each die that shows a number less than...

Two identical fair 6-sided dice are rolled simultaneously. Each die that shows a number less than or equal to 4 is rolled once again. Let X be the number of dice that show a number less than or equal to 4 on the first roll, and let Y be the total number of dice that show a number greater than 4 at the end.

(a) Find the joint PMF of X and Y . (Show your final answer in a table.)

(b) Find the marginal PMF of X.

(c) Compute the mean and variance of X.

(d) Find the conditional PMF of Y given X = 1.

(e) Compute the conditional mean and variance of Y given X = 1.

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Answer #1

a) The joint PDF for X, Y is obtained here as:

X = 0 X = 1 X = 2
Y = 0 0 0 (1/3)4 = 1/81
Y = 1 0 (2c1)*(2/3)*(1/3)*(1/3) = 4/27 (1/3)3(2/3)*2 = 4/81
Y = 2 (2/3)2 = 4/9 (2c1)*(2/3)*(1/3)*(2/3) = 8/27 (1/3)2*(2/3)2 = 4/81

b) The marginal PDF from above joint PDF now could be obtained as:

P(X = 0) = 4/9
P(X = 1) = 12/27 = 4/9
P(X = 2) = 1/9

P(Y = 0) = 1/81
P(Y = 1) = 16/81
P(Y = 2) = 64/81

c) The mean of X is computed as:

E(X) = 1*4/9 + 2/9 = 6/9 = 2/3
E(X2) = 1*4/9 + 22/9 = 8/9

Var(X) = E(X2) - [E(X)]2 = (8/9) - (2/3)2 = 4/9

d) P(Y = 0 | X = 1) = 0
P(Y = 1 | X = 1) = 4/12 = 1/3
P(Y = 2 | X = 2) = 2/3

e) E(Y | X = 1) = 1*(1/3) + 2*(2/3) = 5/3
E(Y2 | X = 1) = 1*(1/3) + 22*(2/3) = 3
Var(Y | X = 1) = 3 - (5/3)2 = 2/9

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