# High School Probability and Contradiction

So I recently came across this question (2(a)) that my friend who teaches high school math posed to me. I thought the solution could be found by using the identities $P(B\,|\,A) = \dfrac{P(A\cap B)}{P(A)}$ and $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. However it apparently leads to the conclusion that $P(A \cap B) > P(B)$. Does anyone have any suggestions why?

It seemed to work out to $\dfrac{2}{3}P(A) = P(A \cap B)$ and thus $\dfrac{16}{25} = P(A) + \dfrac{2}{5} - \dfrac{2}{3}P(A)$. So $P(A) = \dfrac{18}{25}$. But that means that $P(A \cap B) = \dfrac{12}{25}$.

Your approach is correct. The problem is with the question. The values are not consistent. It can be proved that given $P(B)$ and $P(B \mid A)$, the value of $P(A \cup B)'$ should be at least $\frac{7}{15}$.