If P(A)>0, P(B)>0 and P(A)<P(A|B), show that P(B)<P(B|A)

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Thanks in advance!

Hi, roadapples-ga:

Thanks for asking this question... it has a fairly succinct answer!

The definition of conditional probability is that:

P(A|B) = P(A&B)/P(B)

and so of course also:

P(B|A) = P(A&B)/P(A)

We are given P(A),P(B) positive and:

P(A) < P(A|B) = P(A&B)/P(B)

So multiplying both sides by P(B)/P(A) gives the result:

P(B) < P(A&B)/P(A) = P(B|A)

Note that we use P(B)/P(A) positive to preserve the direction of the
inequality when we multiply.

regards, mathtalk-ga

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Request for Answer Clarification by roadapples-ga on 09 Feb 2003 14:15 PST

Just wondering - is the '&' meant to be A intersection B?

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Clarification of Answer by mathtalk-ga on 09 Feb 2003 14:27 PST

Hi, roadapples-ga:

I was thinking "and" in the logical sense -- event A and event B both
occur.  But in a set-theoretic sense, yes, the events A,B are subsets
of a probability space and the intersection models the logical "and"
between events.

regards, mathtalk-ga

Thanks for your help!