There is a question my friends and me have been going on in circles about. We're trying to understand how we can add up probabilities of mutually exclusive events, when there are conditional probabilities involved. Assume I have a die. If I roll this, I have an equal probability of getting anything from between 1 to 6. Now let me define 3 events $A =$ Rolling a 2, $B =$ rolling an event number, $X =$ rolling a prime. Now let's define $Y$ as $A\mid B$. $P(Y) = P(A|B) = 1/3$.
Now
P(Y|X) P(X) + P(Y|X')(1-P(X)) = P(Y).
$P((A|B)|X) =$ probability of rolling a 2 if I rolled an even number ... if Ive rolled a prime. Working backwards, the primes are 2,3,5. The probability of rolling a 2 if the rolled an even number is 1. $P(X) = 1/2$.
P((A|B)|X') = probability of rolling a 2 if I rolled an even number ... if I've rolled a non prime. Working backwards, the non-primes are 1,4,6. The probability of rolling a 2 if the rolled an even number is 0.
Hence $P(Y)$ should be P((A|B)|X)\cdot 1/2 + P((A|B)|X')\cdot 1/2, which should be $1/2$.
what are we missing?