Question

1. A die is rolled twice. Find the joint probability mass function of X andY if X denotes the value on the first roll and Y denotes the minimum of the values of the two rolls.

Answer 1

X is the value on the first roll and Y denotes the minimum of the 2 values of the two rolls.

For each combination of 2 dice values, the values of X and Y here are computed as:

	1	2	3	4	5	6
1	X = 1, Y = 1	X = 1, Y = 1	X = 1, Y = 1	X = 1, Y = 1	X = 1, Y = 1	X = 1, Y = 1
2	X = 2, Y = 1	X = 2, Y = 2	X = 2, Y = 2	X = 2, Y = 2	X = 2, Y = 2	X = 2, Y = 2
3	X = 3, Y = 1	X = 3, Y = 2	X = 3, Y = 3	X = 3, Y = 3	X = 3, Y = 3	X = 3, Y = 3
4	X = 4, Y = 1	X = 4, Y = 2	X = 4, Y = 3	X = 4, Y = 4	X = 4, Y = 4	X = 4, Y = 4
5	X = 5, Y = 1	X = 5, Y = 2	X = 5, Y = 3	X = 5, Y = 4	X = 5, Y = 5	X = 5, Y = 5
6	X = 6, Y = 1	X = 6, Y = 2	X = 6, Y = 3	X = 6, Y = 4	X = 6, Y = 5	X = 6, Y = 6

Therefore, the joint PDF here is obtained here as:

P(X = 1, Y = 1) = 6/36 = 1/6
P(X = 2, Y = 1) = 1/36
P(X = 2, Y = 2) = 5/36
P(X = 3, Y = 1) = 1/36
P(X = 3, Y = 2) = 1/36
P(X = 3, Y = 3) = 4/36
P(X = 4, Y = 1) = 1/36
P(X = 4, Y = 2) = 1/36
P(X = 4, Y = 3) = 1/36
P(X = 4, Y = 4) = 3/36
P(X = 5, Y = 1) = 1/36
P(X = 5, Y = 2) = 1/36
P(X = 5, Y = 3) = 1/36
P(X = 5, Y = 4) = 1/36
P(X = 5, Y = 5) = 2/36
P(X = 6, Y = 1) = 1/36
P(X = 6, Y = 2) = 1/36
P(X = 6, Y = 3) = 1/36
P(X = 6, Y = 4) = 1/36
P(X = 6, Y = 5) = 1/36
P(X = 6, Y = 6) = 1/36

This is the required joint PDF here.