two fair dice are rolled two fair dice are rolled, what is the probability their sum is a 10 given that the first roll is a 6?
not sure why i can't solve this simple problem. i presumed the following: A: event that sum is 10
B: event that first roll is 6
$$Pr(B) = \frac{1}{6}$$
$$Pr(A) = \frac{3}{36} =  \frac{1}{12}$$
$$Pr(B|A) = \frac{1}{3}$$
$$Pr(A \cap B) = Pr(A) * Pr(B|A) = \frac{1}{12} * \frac{1}{3} = \frac{1}{36}$$
$$Pr(A|B) = \frac{Pr(A \cap B)}{Pr(B)} = \frac{1/36}{1/6} = \frac{1}{36}* \frac{6}{1} =  \frac{1}{6} $$
but alas, this doesn't seem to be correct.
A: Your answer is not incorrect, but it is not at all efficient and relies on the fact that $P(B|A)=1/3$, which you don't seem to prove anywhere - and which is odd especially since you use only use this fact to show that $P(A\cap B)$ is $1/36$ which is much easier to see by noting that $A\cap B$ is the event "the first roll was $6$ and the rolls summed to $10$" - i.e. "the first roll was $6$ and the second was $4$". There's no need here even to compute $P(A)$, in fact.
Better is to use your intuition: if you rolled a $6$, the only way to get $10$ is to roll a $4$. If you were in this situation in real life rather than in a hypothetical mathematical question, you would likely realize that since there is only one out of six possible rolls that give success having already rolled a $6$, your chance of success at that moment is $1/6$.
How can you express that intuition? Well, let's let the rolls be given as random variables $X$ and $Y$. You are trying to compute $P(X+Y=10|X=6)$. Of course, this is the same as $P(Y=4|X=6)$ because we can substitute $X=6$ into the left side and cancel. However $Y$ and $X$ are independent, so this is just $P(Y=4)$, which is $1/6$ because $Y$ is uniform on a set of size $6$.
It's worth also thinking about this in terms of events - what the argument says is that if we define $C$ to be the event that the second roll is $4$, then $A\cap B = A \cap C$ which implies that $P(B|A)=P(C|A)$. The independence of $C$ and $A$ implies then that the final answer is equal to $P(C)$.
