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At a certain university, 28% of students major in zoology. Of all the students majoring in zoology, 68% are males. It is also known that 56% of all students at the university are male. Let Z represent the event that a randomly chosen student is majoring in zoology. Let M represent the event that a randomly chosen student is male. What proportion of males at the university are majoring in zoology?

Previously solved for P(M∩Z)=0.1904 and this is correct. I've tried P(Z|M) = P(M∩Z) / P(Z), and get the answer 0.68 which is the probability that a student majoring in zoology is a male. The correct answer is 0.34, which is exactly half, but I am not sure how to get this answer.

Any help is appreciated!

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    $\begingroup$ The equation $P(Z\vert M) = P(M\cap Z) / P(Z)$ is equivalent to $P(Z)P(Z\vert M) = P(M\cap Z)$, which is not correct. stattrek.com/probability/probability-rules.aspx $\endgroup$ – mathmandan Nov 5 '15 at 16:00
  • $\begingroup$ Thank you, I took a look at the site and found where I went wrong. $\endgroup$ – Astag Nov 5 '15 at 16:10
  • $\begingroup$ Great! Glad you found the solution! $\endgroup$ – mathmandan Nov 5 '15 at 16:11
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Let it be that there are $10000$ students in total.

Then $2800$ of them major in zoology and $68\times28=1904$ of these are male.

Also it is known that $5600$ of all students are male.

So the proportion of males majoring in zoology is: $$\frac{1904}{5600}$$

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  • $\begingroup$ Thank you very much for the further explanation. $\endgroup$ – Astag Nov 5 '15 at 16:24
  • $\begingroup$ You are welcome. Nice to see that you found it yourself. $\endgroup$ – drhab Nov 5 '15 at 16:26
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Thanks to mathmandan I found my error and got the right answer.

P(Z|M) = P(M∩Z) / P(M) = 0.1904/0.56 = 0.34

Should have been divided by P(M), not P(Z).

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