Just checking

**Beebs** · Oct 27th 2010, 08:17 AM

I have P(B|A) and P(A), there isn't a quick way to get P(B) off this is there?

A and B are not independent.

Thanks,

Brendan

**Soroban** · Oct 27th 2010, 10:05 AM

Hello, Brendan!

I have $P(B|A)$ and $P(A).$

There isn't a quick way to get $P(B)$ off this, is there? . no

$\,A$ and $\,B$ are not independent.

From Bayes' Theorem, we have: . $P(B|A) \;=\;\dfrac{P(A \cap B)}{P(A)}$

. . Hence: . $P(A \cap B) \;=\;P(B|A)\cdot P(A)$

But without more information, such as $P(A \cup B)$ ,

. . we cannot isolate $P(B).$

**Beebs** · Oct 27th 2010, 10:09 AM

Thanks, I assumed that was the case but just wanted to be sure.

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