Re: Conditional Probability




Le 17/06/2005 23:00, dans
1119042013.577383.21700@xxxxxxxxxxxxxxxxxxxxxxxxxxxx, « Mathias »
<news.abc42@xxxxxxxx> a écrit :

> Hi,
>
>>> I am pretty sure the answer is no. You need more data about the
>>> probabilities of joint events. Play with a spread*** to come up
>>> with two numerical examples where P(B | A1) and P(B | A2) maintain
>>> their values but P(B | A1 A2) differ.
>
> Why would it proof that the answer is no? The idea and the way it is
> going to show it is not clear to me. Could you possibly explain it a
> bit more detailed?
>

If you can find particular events B X1 X2 Y1 Y2 such that

P(B | Y1 and Y2) != P(B | X1 and X2)

P(B | Y1) = P(B | X1)

P(B | Y2) = P(B | X2)

Then P(B | A1 and A2) is NOT a function
of only P(B | A1) and P(B | A2): for
same input, you should have same output !

.

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