An experiment consists of rolling a fair 6sided die two time

An experiment consists of rolling a fair 6-sided die two times. Define X as the larger of the two results. If the same number is obtained for both rolls of the die, then X is simply that value A. Determine the probability mass function (pmf) for this random variable, X. What is the sample space of two die rolls? Find the cumulative distribution function (cdf) for this random variable X

Solution

8. Sample space, S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), ..... (6,6)} = {(i,j) | 1<= i,j <= 6}

Total no. of elements sample space have = 6²= 36

A. The possible value X can take are {1, 2, 3, 4, 5, 6}.

For X = 1, possible outcomes are {(1,1)}

Thus, P(X=1) = 1/36

For X = 2, possible outcomes are {(1,2), (2,1) (2, 2)}

Thus, P(X=2) = 3/36

For X = 3, possible outcomes are {(1,3). (3,1), (2,3), (3,2), (3,3)}

Thus, P(X=3) = 5/36

For X = 4, possible outcomes are {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3), (4,4)}

Thus, P(X=4) = 7/36

For X = 5, possible outcomes are {(1,5), (5,1), (2,5), (5,2), (3,5), (5,3), (4,5), (5,4), (5,5)}

Thus, P(X=5) = 9/36

For X = 6, possible outcomes are {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5), (6,6)}

Thus, P(X=6) = 11/36

Thus, probablity mass function is given by,

P(X=1) = 1/36, P(X=2) = 3/36, P(X=3) = 5/36, P(X=4) = 7/36, P(X=5) = 9/36 & P(X=6) = 11/36.

B.

Cumulative distribution function (cdf) is given by,

F(x) = P(X<=x)

So for x=0,

F(x) = P(X=0) = 0

So for x=1,

F(x) = P(X=0) + P(X=1) = 0+ 1/36 = 1/36

So for x=2,

F(x) = P(X=0) + P(X=1) + P(X=2) = 0 + 1/36 + 3/36 = 4/36

So for x=3,

F(x) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0 + 1/36 + 3/36 + 5/36 = 9/36

So for x=4,

F(x) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)= 0 + 1/36 + 3/36 + 5/36 +7/36 = 16/36

So for x=5,

F(x) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0 + 1/36 + 3/36 + 5/36 +7/36 + 9/36 = 25/36

So for x=6,

F(x) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 0 + 1/36 + 3/36 + 5/36 +7/36 + 9/36 + 11/36 = 1

Thus, cdf for X is given by,

F(0) = 0, F(1) = 1/36, F(2) = 4/36, F(3) = 9/36, F(4) = 16/36, F(5) = 25/36 & F(6) = 1.