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In a management trainee program, 80 percent of the trainees are female, 20 percent male. 90 percent of the females attended college, 78 percent of the males attended college. A management trainee is selected at random. What is the probability that the person selected is a female who did attend college?

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  • $\begingroup$ $$P(FC)\stackrel{ind}{=}P(F)P(C)$$$$P(FC)=0.9\times0.8$$ $\endgroup$ – Inquest Feb 12 '13 at 3:09
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It doesn't matter how many trainees there are, so let's suppose there are $100$. Then $80$ are female, and $72$ of the $80$ went to college, so the probability the person selected is a female who attended college is ... ?

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The probability of two independent events is $P(AB) = P(A)P(B)$. In this situation, P(A), where A is that the selected person is female is $.8$, and P(B), where B is that a given female went to college is $.9$, so the final answer is $$ P(AB) = P(A)P(B) = (.8)(.9) = .72 $$

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