Given $P(A|B), P(A|B^c)$ and $P(B|A)$ how do I find $P(A)$ and $P(B)$

Question

For reference sake:

$P(A|B) = 1/3$,

$P(A|B^c) = 2/3$

$P(B|A) = 1/2$

Now, I know $P(A∩B) = P(A|B)*P(B)$ and that $P(A|B^c) = P(A∩B^c)/P(B^c)$

But where to go next I do not know.

I think I need to find either $P(A∩B)$ or $P(A∩B^c)$ which I would be able to sub into the earlier equations to find $P(A)$ and $P(B)$ with ease, but how to do that I don't know.

Am I missing an equation or am I just missing something that's staring me in the face?

Thank You

Answer 1

If not the Bayes' theorem then consider that from the first equation $$P(B)=\frac{P(A\cap B)} {\frac13}$$

and from the second one $$1-P(B)=\frac{P(A\cap B)} {\frac23}.$$

The quotient of the two is $$\frac{P(B)}{1-P(B)}=2.$$ From the third equation we learn that

$$P(A)=\frac{P(A\cap B)}{\frac12}.$$ We can do the trick above again and we already know $P(B)$.

Answer 2

$$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}$$ and rearrange.

With regards to finding $P(B)$, swap $A$ with $B$ in the formula wherever either occurs. $$P(B|A^c)=1-P(B|A)=\frac{1}{2}$$