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chopasticks
Jun20-09, 10:57 AM
how can i proof that >>>>>> p(A | B') = p(A | B)
well i tried but i think there is a hole in the solution T.T
if we said that
p(A|B)=p(A&B)/p(B)

p(A|B')= p(A&B')/p(b')

= p(A) - p(A&B)/1-p(B)


so0o0o how can i complete it ??

arildno
Jun20-09, 11:33 AM
how can i proof that >>>>>> p(A | B') = p(A | B)


In general, you can't prove that.

That condition holds ONLY for independent events.

So, in a particular case, you may demonstrate whether that condition holds or not.

If it holds, A and B are independent events, if not, A and B are not independent events.

chopasticks
Jun20-09, 04:08 PM
sorry my bad ,, i forgot to mention that A,B are independent

HallsofIvy
Jun20-09, 05:37 PM
In that case, by definition of "independent", P(A|B)= P(A)= P(A|B').