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I would appreciate it if someone could please help me understand what is being asked here and how to approach questions like the one below.

"At the college entrance exam, a candidate is admitted according to whether he has passed or failed the test. Of the candidates who are really capable, 80 % pass the test and of the incapable, 25 % pass the test. Given that 40 % of the candidates are really capable, then the proportion of the really capable students who can pass the test to the total students who can pass is?"

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    but I think your solution there makes a lot of sense. The final answer was 68% though.2011-09-07

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You don’t need any fancy notation to solve the problem.

You know that $40$% of the candidates are really capable and that $80$% of those $40$% pass the test; $0.8 \cdot 0.4 = 0.32$, so $32$% of all candidates both pass the test and are really capable. Of the remaining $60$% of the candidates, $25$% pass; $0.25 \cdot 0.6 = 0.15$; so $15$% of all candidates both pass the test and aren’t really capable. Altogether, then, $32+15=47$% of the candidates can pass the test, and the fraction of those who are really capable is $32/47$.

Alternatively, you can do as gary suggested and imagine that you’re working with a specific number of candidates. Choose the number so that all of the percentages work out to whole numbers of people; in this case $100$ works. Then you have $40$ who are really capable, of whom $32$ pass, and $60$ who aren’t capable, of whom $15$ pass anyway. Thus, $47$ pass, of whom $32$ are really capable, and the desired proportion is $32/47$. As you can see, this is just doing with specific numbers what I did with the percentages in the previous paragraph.

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    @user10695: Not really. I’m assuming that this is from some sort of admissions test. In my experience the analytical reasoning questions on such tests rarely require any technical tools beyond what in the U.S. is high school mathematics; the difficulty lies not in the mathematical tools themselves, but in seeing which tools to use and how to use them (and perhaps to some extent in reading carefully). If you’re studying for such an exam, probably the best thing that you can do is what you’re doing: go through as many sample problems as you can find.2011-09-07
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We have $p(P|C) = 0.8$ p(P|C') = 0.25 and $p(C) = 0.4$ and p(C') = 0.6

Note that $P$ denotes the event of passing and $C$ denotes the event that a person is capable. Thus $p(P \cap C) = p(P|C)p(C) = (0.25)(0.4)$

Also p(P) = p(P \cap C)+ p(P \cap C')

= p(P|C)p(C)+ p(P|C')p(C') $ = (0.8)(0.4)+(0.25)(0.6)$

So compute $\frac{p(P \cap C)}{p(P)}$

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    The % of capable is 40% and the percentage of those Capable who passed is 80%. So wouldn't P n C be (0.8 * 0.4)?2011-09-07
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Assume that there are 100x students as given 40% (40x)are capable imply that remaining 60% (60x)are incapable.

So 80% of capable pass the test means 40x*.8=32x passed which are capable ...(1) And 25% of incapable pass the test means 60x*.25=14x passed the test which are in capable

So total pass=40x+15x=47x

so proportion of the really capable students who can pass the test to the total students who can pass is 32x/47x= 32/47 ans