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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 $P(A \cap B) = \dfrac{12}{25}$.

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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 $\frac{7}{15}$.

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  • $\begingroup$ I did the problem over and over again and couldn't figure out why my numbers didn't make sense (our numbers). I think this has to be a problem with the problem. Can you show how to prove that? $\endgroup$ – randomgirl May 3 '15 at 5:20
  • $\begingroup$ I was wondering if there was a theoretical explanation like it fails because of collinearity or whatever. $\endgroup$ – Eugene May 3 '15 at 5:21

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