Finding probability of $P(A \cap B)$

Question

Suppose that $$P(A|B) = P(B|A')$$ $$P(B'|A) = 0.75$$ $$P(B') = 0.4$$ Find $P(A \cap B)$.

Here is the my answer which seems to be wrong: $$P(B'|A) = 1 - P(B|A) \implies P(B|A) = 0.25$$ $$P(B' ) = 0.4 \implies P(B) = 0.6$$ $$P(A|B)P(B) = P(B|A)P(A) \implies 0.6P(A|B) = 0.25P(A) $$ $$P(B|A')P(A') = P(A'|B)P(B) \implies P(A|B)(1-P(A)) = 0.6(1-P(A|B))$$

Let $P(A|B) = y$ and $P(A) = x$, then $$0.6y = 0.25x$$ $$y(1-x) = 0.6(1-y)$$ And this leads to $$\frac{x}{2.4}(1-x) = 0.6(1-\frac{x}{2.4})$$ This equation doesn't have real solution. So what's wrong about my answer?

Answer 1

I don't see mistake in your reasoning.

My way:

Let $P(A)=x.$

Thus, $$P(A\cap B')=0.75x,$$ $$P(A\cap B)=0.25x,$$ $$P(A'\cap B)=0.6-0.25x$$ and $$P(A'\cap B')=0.4-0.75x.$$ Thus, from the first condition we obtain: $$\frac{0.25x}{0.6}=\frac{0.6-0.25x}{1-x},$$ which also has no real roots.

Answer 2

For all $A$ and $B$ we have $$P(A\cap B)+P(A'\cap B)= P(B) = 0,6$$

and

$$P(A\cap B)+P(A\cap B')= P(A) = x$$

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Now, from the first equation we have $$P(B\cap A') = {5-5x\over 3}P(A\cap B)$$ and from the second $$ P(B'\cap A) = {3\over 4}x$$

Let $p= P(A\cap B)$, then we have $$p+{5-5x\over 3}p ={3\over 5}$$

and $$p+{3\over 4}x =x\implies p = {x\over 4}$$

and thus $$p+{5-20p\over 3}p ={3\over 5}$$ Now solve that...and if you plug in your equation $x=4p$ you will get the same formula as I did. So nothing wrong with your calculation.