Using Bayes Law to find $P(A)$

Question

Let $A$ and $B$ be the events with $0\lt P(B)\lt 1, P(A|B)=P(A|B^c)=1/3.$ Is it possible to calculate $P(A)$ with this information?

For my answer, after using Bayes Law I come up with... $$P(B|A)=\frac{P(AB)}{P(A)}= \frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|B^c)(P(B^c)}=\frac{(1/3)P(B)}{(1/3)(P(B))+(1/3)(1-P(B))}=P(B)$$ And thus $P(A)=1$, but this doesn't seem to correct to me...

Answer 1

Simply $P(A)=\frac{1}{3}$ since this is the conditional probability in the new cases that cover the entire sample space ($B$ and $B^c$.

Mathematically:

$P(A)=P(A\cap B)+P(A\cap B^c)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)=\frac13 P(B)+\frac13 P(B^c)=\frac13$

Answer 2

Since $A\cap B$ and $A\cap B^{c}$ are disjoint sets and $(A\cap B) \cup (A\cap B^{c})=A$ you have $$P(A)=P(A\cap B)+ P(A\cap B^{c})=P(A|B)P(B)+P(A|B^{c})P(B^{c})$$ $$P(A|B)P(B)+P(A|B^{c})(1-P(B))=\frac{1}{3}$$