Probability that the student can solve 5 of 7 problems on the exam 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.
 A: 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.
A: 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?
A: There are $\binom{12}{7}$ ways the instructor can choose $7$ questions from $12$.
As the student can solve $9$ of the $12$, unless the instructor picks the $3$ she can't solve (and $4$ others), then she will be able to solve at least $5$ of them.
There are $\binom{9}{4}$ ways to pick the questions like this.
Therefore the probability that the student can solve at least five of the questions is:
$$1-\frac{\binom{9}{4}}{\binom{12}{7}}$$
$$=1-\frac{126}{792}$$
$$=1-\frac{7}{44}$$
$$=\frac{37}{44}$$
