Answer:
1/3
Step-by-step explanation:
The sample space for two dice is given as:
[tex](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) (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)\\(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)[/tex]
The Sample space for the total of each pair is:
[tex]2,3,4,5,6,7\\3,4,5,6,7,8\\4,5,6,7,8,9\\5,6,7,8,9,10\\6,7,8,9,10,11\\7,8,9,10,11,12[/tex]
Total Sample Space, n(S)=36
Factors of 18 are 1,2,3,6,9 and 18
Let Event A be the event that the total of each pair is a factor of 18.
Sample Space of A is:
[tex]2,3,6\\3,6,\\6,9\\6,9\\6,9\\9[/tex]
n(A)=12
Therefore, the probability of A
P(A)=n(A)/n(S)=12/36=1/3
The probability of getting a total that is a factor of 18 is One-Third.
