Answer:
a. The possible values of X are 2,3,4,5,6
The possible values of Y are 1,2,3,4,5
b. X 1 2 3 4 5 6 Total
n 0 2 4 6 8 10 30
p(x) 0 1/15 2/15 3/15 4/15 1/3 1
c. X 1 2 3 4 5 6 Total
n 10 8 6 4 2 0 30
p(x) 1/3 4/15 3/15 2/15 1/15 0 1
Step-by-step explanation:
Rolling two fair dice, we obtain the following tables :
1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6
Let X be the maximum of the two numbers on the dice, then
X = {2,3,4,5,6,2,3,4,5,6,3,3,4,5,6,4,4,4,5,6,5,5,5,5,6,6,6,6,6,6}
Let Y be the minimum of the two numbers on the dice, then
Y = {1,1,1,1,1,1,2,2,2,2,1,2,3,3,3,1,2,3,4,4,1,2,3,4,5,1,2,3,4,5}
Finding the probability mass functions of X and Y which contains the possible values of X and Y and their associated probabilities, we have
X 1 2 3 4 5 6 Total
n 0 2 4 6 8 10 30
p(x) 0 1/15 2/15 3/15 4/15 1/3 1
and
X 1 2 3 4 5 6 Total
n 10 8 6 4 2 0 30
p(x) 1/3 4/15 3/15 2/15 1/15 0 1
