Re: Simple (?) question on conditional probability

hagman wrote:

On 31 Jul., 20:55, Grant55 <grant.thomp...@xxxxxxxx> wrote:


I've searched the old posts and haven't come across the answer to this question. Notation: P is probability; A, B, C, D, and E are random variables; and "&" means "and".

Suppose I know: P(A), P(E|A), P(C|A&E), P(B|A&C), and P(D|C&E).

What is P(A&B&C&D&E)?

I realize that P(A&C&E)=P(A)*P(E|A)*P(C|A&E), but I start to wonder about "double counting" when I get to the more complex problem above.




You always have P(U&V) = P(U)*P(V|U).
Substitute U=A&B&C&D, V=E to get P(A&B&C&D&E) = P(A&BC&D)*P(E|A&B&C&D)
Substitute U=A&B&C, V=D to get P(A&B&C&D) = P(A&B&C)*P(D|A&B&C)
Substitute U=A&B, V=C to get P(A&B&C) = P(A&B)*P(C|A&B)
Substitute U=A, V=B to get P(A&B) = P(A)*P(B|A)
All in all you have
P(A&B&C&D&E) = P(A)*P(B|A)*P(C|A&B)*P(D|A&B&C)*P(E|A&B&C&D)

No doubel counting involved, just repeated use of the same equation
for two events


But the OP is not given all the conditional probabilities on the right-hand side of your last equation.

--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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