Consider the population of students attending a university. Suppose that 64% of these students are from Town O, 15% are from Town K, and 21% are from Town T. Of all the students from Town O, 8% eventually withdraw from their program. The withdrawal rates for students from Town K and Town T are 22% and 13%, respectively.
Suppose a student from the university does withdraw from their program. What is the probability that the student is from Town O?

I solved for the probability that a randomly selected student will withdraw, which is 0.1115 and it is correct. I tried P(O|W)= P*(O∩W) / P(O) = 0.08 / 0.1115 = 0.7175 , but this does not match the answer of 0.4592.

Any help is much appreciated!

EDIT: Should be P(O|W)= P*(O∩W) / P(W), sorry for the typo.

  • 2
    $\begingroup$ Your formula is incorrect. $P(O|W) = P(O \cap W)/\mathbf{P(W)}$. Note that you'll have to calculate $P(W)$ from the data given. For this, use the total probability law $\endgroup$ – stochasticboy321 Nov 5 '15 at 19:57

Imagine 10000 students. .64(10000)= 6400 of them are from town O and .08(6400)= 512 withdraw. .15(10000)= 1500 are from town K and .22(1500)= 330 withdraw. .21(10000)= 2100 are from town T and .13(2100)= 273 a withdraw.

So a total of 512+ 330+ 273= 1115 students withdraw (and 1115 out of 10000 is, as you say 0.0115) and 512 of them, or 512/1115= 0.4592, are from town O.

You used .08 which is the percentage of students from town O without regard that to the number of students from town O- the number of students from the various towns is not the same.

  • $\begingroup$ Thank you very much, it makes sense to me now. $\endgroup$ – Astag Nov 5 '15 at 20:13

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