It doesn't seem like it should be so hard... in fact, it makes perfect sense that P(B|B) = 1... but I really don't know how to show it. Please help!

- Thread starter hollysti
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It doesn't seem like it should be so hard... in fact, it makes perfect sense that P(B|B) = 1... but I really don't know how to show it. Please help!

Ok so using the theorem P(A|B) = [(P(B|A))(P(A)) / P(B)],

since P(B) does not equal 0, then that also implies that P(B|A) does not equal 0, right? If so, then P(A|B) would be greater than or equal to 0, but I am not sure how to show that...

I think using that theorem helps a little, but I still don't have a clear picure in my mind. Could you maybe help a little more? Thanks!

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\(\displaystyle P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P(B)}}\) or \(\displaystyle P\left( {A \cap B} \right) = P\left( {A|B} \right)P(B)\).

In either form the proofs follow at once if you note \(\displaystyle P\left( {A \cap B} \right) \ge 0\).