Problem statement -
We roll two dice. Find the conditional probability that at least one of the numbers is even, given that the sum is 8.
My attempt -
A = {at least one of the numbers is even}
B = {the sum is 8}
Want to find P(A|B)
P(A|B) = P(A and B)/P(B)
Find P(B). I know that
5+3=8 2+6=8 4+4=8
Are the only possible ways of obtaining a sum of 8. So there are 3 ways of rolling the dice such that the sum is 8. There are two die, so the number of total possibilities is 6*6 = 36. So
P(B) = 3/(6*6) = 3/36 = 1/12
Next, I want to find P(A and B)
There are only two possible ways of having at least one even number and, and the sum being 8 (2+6, and 4+4). So
P(A and B) = 2/36 = 1/18
So P(A|B) = (1/18)/(1/12) = 2/3
This answer seemed right to me, until I checked the back of the book for the answer, and it said 3/5.
What did I do wrong?
The following outcomes are equally probable: $$ (2,6), \quad (3,5) \quad (4,4) \quad (5,3) \quad (6,2) $$