How do you calculate Pr(A | B, C, D, E) if you know Pr( A ∩ B ∩ C ∩ D ∩ E) and Pr(B ∩ C ∩ D ∩ E)? all!
Excuse me if the question sounds naive. I have searched on the Web but could not find the answer.
I have studied Chain Rule in my textbook as well as on the Web and understand the basics of it. And I see that it's very useful to calculate the probability of intersection of events when we're given conditional probabilities.
However, if I need to calculate Pr(A | B, C, D, E) and am able to calculate the probabilities Pr( A ∩ B ∩ C ∩ D ∩ E) and Pr(B ∩ C ∩ D ∩ E), can I simply divide the former by the latter? Here, A, B, C, D, and E are all events on the same sample space with non-zero probabilities.
If not, why not? And how to use the Chain Rule in such a case?
Many thanks for answering such a basic question.
Regards,
Quester.
 A: If you need to calculate $P(A\mid B,C,D,E)$ which is just another notation for $P(A\mid B\cap C\cap D\cap E)$, then your proposed calculation is merely using the definition of the conditional probability of $A$ given that $B, C, D, E$ all occurred.  The chain rule, on the other hand, says that $$P(A\cap B\cap C\cap D\cap E) = P(E)P(D\mid E)P(C\mid D,E)P(B\mid C,D,E)P(A\mid B,C,D,E),$$ that is, we use the conditional probability to find the unconditional probability that all the events occurred, not the other way around.
The chain rule can be "justified" as saying that in order for $A,B,C,D,E$ all
to occur, we must assume that $E$ occurred (which has probability $P(E)$), and 
having assumed that, we also need to assume that $D$ occurred which has
probability not $P(D)$ but rather $P(D\mid E)$ since we have already assumed
that $E$ occurred; and then we need to assume that $C$ occurred for which we
use the conditional probability $P(C\mid D,E) = P(C\mid D\cap E)$, and so on.
Note that the first multiplication on the right gives
$$P(E)P(D\mid E) = P(E) \frac{P(D\cap E)}{P(E)} = P(D\cap E)$$
and so the next one gives
$$P(D\cap E)P(C\mid D\cap E) = P(D\cap E)\frac{P(C\cap (D\cap E))}{P(D\cap E)}
= P(C\cap (D\cap E))$$
Do you see a pattern emerging?
