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A student prepares for an exam by studying a list of 12 problems. She can solve 9 of them. For the exam, the instructor selects 7 problems at random from the 12 on the list given to the students. What is the probability that the student can solve 5 of the 7 problems on the exam?

I am really confused by this problem. My first initial thought was to do:

(5 chose 7)(7 chose 0)/ (12 chose 7)

I am not 100% sure about this though.

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I believe it should be (7 chose 5)(7 choose 2)/ (12 chose 7) is this correct? – user125627 Feb 13 '14 at 19:24

Your denominator is right. That's the total number of possible exams. In the numerator you want the total number of exams where the student can solve 5/7 problems. So you want to chose the set of solvable problems and then chose the set of not solvable problems.

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so you're saying it should just be (5 choose 7) / (12 choose 7)? – user125627 Feb 13 '14 at 6:35
That corresponds to choosing the solvable problems. You still have to chose the not solvable problems. – Kai Sikorski Feb 13 '14 at 6:40
Okay. hm.. would that be (2 choose 0) because there will be 2 other problems that you can't solve? – user125627 Feb 13 '14 at 6:43

In order for the student to solve exactly $5$ problems out of the $7$, the exam must contain exactly $5$ out of the $7$ problems the student knows how to solve, and exactly $2$ of the other three problems.

How many different combinations are there?

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Could you explain why you said "exactly 2 of the other three problems"? I understand the 5 out of 7 but I feel like I am missing the other half. My professor told me that the top if summed together should equal the denominator. – user125627 Feb 13 '14 at 17:15
@user125627 Is the student gets exactly $5$ of $7$ questions right, what does he do on the other two? – N. S. Feb 14 '14 at 2:24

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