Answer:
[tex]Pr = \frac{2}{9}[/tex]
Step-by-step explanation:
Given
A roll of two dice
Required
Probability of getting a sum of 8
First, we list out the sample space
[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), (3, 1) (3, 2) (3, 3)[/tex][tex](3, 4) (3, 5) (3, 6), (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6),(5, 1) (5, 2) (5, 3) ,(5, 4) ,(5, 5) ,(5, 6),[/tex]
[tex](6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)\}[/tex]
The total is:
[tex]n(S) = 36[/tex]
From the sample space, the outcomes that give a sum of 8 are:
[tex]Sum =\{(2, 6), (3, 5),(4,4),(5,3),(6,2)\}[/tex]
The total is:
[tex]n(Sum) = 8[/tex]
So, the probability is:
[tex]Pr = \frac{n(Sum)}{n(S)}[/tex]
[tex]Pr = \frac{8}{36}[/tex]
[tex]Pr = \frac{2}{9}[/tex]
