Thread: Just checking

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

Hello, Brendan!

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

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

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

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

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


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

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

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