Answer:
[tex]Pr =\frac{1}{3}[/tex]
Step-by-step explanation:
Given
[tex]S = \{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)[/tex]
[tex](3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)[/tex]
[tex](5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}[/tex] --- sample space
First, list out all outcomes whose sum is divisible by 5
[tex]A = \{(4,6), (5,5),(6,4)\}[/tex]
So, we have:
[tex]n(A) = 3[/tex]
Next, list out all outcomes that has an outcome of 5 in both rolls
[tex]B = \{(5,5)\}[/tex]
[tex]n(B) =1[/tex]
The required conditional probability is:
[tex]Pr =\frac{n(B)}{n(A)}[/tex]
[tex]Pr =\frac{1}{3}[/tex]
