Of all the people applying for a certain job, $89\%$ are qualified. The personnel manager claims that she approves qualified people $61\%$ of the time and she approves an unqualified person $12\%$ of the time.
Find the probability that a randomly chosen applicant who was approved by the manager is qualified. Round your answer to the nearest tenth of a percent.
I tried it many times but I always failed. I do not know how to solve this I need help.
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$\begingroup$Let $Q$ mean 'applicant is qualified', $Q'$ 'applicant is not qualified' and $A$ 'approved'. We are given
$$P(Q) = 0.89, P(A|Q) = 0.61, P(A| Q') = 0.12$$
What have you written down for $P(Q|A)$?
$\endgroup$ 3 $\begingroup$Denote with $Q$ the event that an applicant is qualified and with $A$ that he is approved. Using Bayes Rule and the law of total probability (to calculate the denominator) you obtain that $$\begin{align*}P(Q\mid A)&=\frac{P(A\mid Q)P(Q)}{P(A)}=\frac{0.61\cdot0.89}{P(A\mid Q)P(Q)+P(A\mid Q')P(Q')}\\[0.3cm]&=\frac{0.61\cdot0.89}{0.61\cdot0.89+0.12\cdot0.11}=0.976\end{align*}$$
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