If $P(A) = p$ and $P(B|A) = P(B^{c}|A^{c}) = 1 - p$, find the value of $P(A|B)$.

Question

I am trying to solve this question where I have to find P(A|B) given, $P(A) = p$ and, $P(B|A) = P(B’|A’) = 1-p$

Since, I don’t have P(B), direct Bayes theorem isn’t applicable but I have this hunch that the equality between the conditional probabilities can somehow be used to prove that events are independent and thus the answer comes out to p, but that’s just a hunch.

(edit) P(B’) means probability of B not happening. Same for A

Answer 1

According to the given information, one has that \begin{align*} \textbf{P}(B|A) = \textbf{P}(B^{c}|A^{c}) = 1 - p & \Longleftrightarrow \frac{\textbf{P}(A\cap B)}{\textbf{P}(A)} = \frac{\textbf{P}(A^{c}\cap B^{c})}{\textbf{P}(A^{c})} = 1-p\\\\ & \Longleftrightarrow \frac{\textbf{P}(A\cap B)}{p} = \frac{1 - \textbf{P}(A\cup B)}{1 - p} = 1 - p \end{align*}

Thus we conclude that \begin{align*} \begin{cases} \textbf{P}(A\cap B) = p(1-p)\\\\ \textbf{P}(A\cup B) = 2p - p^{2} \end{cases} & \Longrightarrow \textbf{P}(A) + \textbf{P}(B) - \textbf{P}(A\cap B) = p + \textbf{P}(B) - p(1-p) = 2p - p^{2}\\\\ & \Longrightarrow \textbf{P}(B) = 2p(1 - p) \Longrightarrow \textbf{P}(A|B) = \frac{\textbf{P}(A\cap B)}{\textbf{P}(B)} = \frac{p(1-p)}{2p(1-p)} = \frac{1}{2} \end{align*} and we are done.

Hopefully this helps.