Q:

Conditional Probability that sum of dice is even

If two dice are rolled and let $ X and Y $ be the two random variables. What is the conditional probability that $X+Y$ is even when $X$ is odd. And when $X$ is odd? What is the total Probability of $X+Y$ to be even?
I am using the Bayes theorem but the answer is coming out 1. I am confused.

A:

Let $A$ be the event $X+Y$ is even, let $B$ be the event $X$ is odd, and let $C$ be the event $X$ is even. By the Law of Total Probability we have
$$\Pr(A)=\Pr(A\mid B)\Pr(B)+\Pr(A\mid C)\Pr(C).\tag{1}$$
It looks as if you calculated $\Pr(A\mid B)$ and $\Pr(A\mid C)$ correctly. They are both $1/2$. But $\Pr(B)=\Pr(C)=1/2$. Substituting in (1) we get $\Pr(A)=(1/2)(1/2)+(1/2)(1/2)=1/2$.
There are many other ways to compute the probability that $X+Y$ is even.  We can do it the long way, by counting the number of ordered pairs $(x,y)$  where $x+y$ is even, and $1\le x\le 6$, $1\le y\le 6$, and dividing by $36$. A much simpler way related to the answer above is that whatever the first roll is, the probability the second roll results in an even sum is $1/2$.

