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  1. In some university $80\%$ of female students and $75\%$ of male students passed the mid-term exam.
  2. There are $60%$ of male and $40%$ of female students.
  3. If a randomly chosen student did not pass the exam, what is the probability she is a woman?

The answer is $0.35$. Could you explain how we get it?

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  • $\begingroup$ You're looking for $P(\text{woman}\mid \text{did not pass exam})$. What do you know about conditional probabilities? Do you know a formula for $P(A\mid B)$? $\endgroup$ – kccu Oct 23 '16 at 17:20
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    $\begingroup$ Are you just posting your homework here for us to do for you? This and your recent prior questions certainly look like that. $\endgroup$ – lulu Oct 23 '16 at 17:21
  • $\begingroup$ I know the formula for conditional probability P(A|B)=P(A and B)/P(B) $\endgroup$ – Mary Oct 23 '16 at 17:23
  • $\begingroup$ I have an exam tomorrow and I just want to clarify some questions $\endgroup$ – Mary Oct 23 '16 at 17:24
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    $\begingroup$ Next time please add what you tried. $\endgroup$ – suomynonA Oct 23 '16 at 17:27
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An easy way to do this is to just make up a number for the total number of students.

Here I am using $100$. Then, there would be $60$ male students and $40$ female students. Multiply $40$ by $0.8$ and $60$ by $0.75$, and you get that $32$ female students passed the exam and $45$ male students passed the exam, and $8$ female students and $15$ male students did not pass the exam.

The probability that a student that didn't pass the exam is female is $\frac{8}{23}$ which is about $0.35$

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Seeing the comments, you know the formula for conditional probability. Then the rest should follow.

Let $A$ denote the event that the person chosen is a woman. Let $B$ denote the event that the person chosen did not pass.

Then $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

Therefore we get $$ \frac{(\frac{40}{100})(\frac{20}{100})}{(\frac{40}{100})(\frac{20}{100})+(\frac{60}{100})(\frac{25}{100})} $$

which equals $\frac{8}{23}$ as the other answer.

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  • $\begingroup$ +1 I made a small change in your notation for better readability. Hope it is fine. $\endgroup$ – msm Oct 24 '16 at 0:38
  • $\begingroup$ There is a mistake in your edit: In the denominator, it should read $\frac{40}{100}$ not $\frac{40}{10}$ $\endgroup$ – Math_Enthusiast Oct 24 '16 at 13:23
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    $\begingroup$ Thank you for the edit though. $\endgroup$ – Math_Enthusiast Oct 24 '16 at 13:24

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