Right.
Another way to approach the first problem: There are ways for the instructor to pick the 3 problems, all of which are equally likely. In of these cases, the student knows all 3. So...
An instructor gave her students 12 problems, telling them that 3 of the problems will be on a quiz and that passing the quiz requires solving all 3 of the problems.
a) Given that the instructor chooses the 3 problems at random, what is the probability for a student who knows only 10 problems to pass?
I calculate that the student has 10/12 chances to know the first problem, 9/11 to know the second, and 8/10 to know the third. Therefore the student's probability of passing the quiz is (10/12)*(9/11)*(8/10)= .545
b) What are the chances to fail for a student who knows only 8 problems.
That student has 1 - (8/12)*(7/11)*(6/10) = .745 probability of failing.
Are these answers correct?