2. (10 points) An exam is to consist of 10 problems, which will be randomly chosen out of a list of 100 problems.
(a) (3 points) Out of all the exam questions, 20 are difficult and 80 are easy. What is the probability that the exam will consist of 2 difficult problems and 8 easy ones?
$P(2 \text{ difficult problems}) = \frac{\binom{20}{2} \binom{80}{8}}{\binom{100}{10}}$
(b) (4 points) A student studies the list of problems ahead of time, but on the exam day, they realize they only know how to solve 60 of them. What is the probability that the student will know how to solve at least 8 problems on the exam? 8/10 QS
Binomial
$p(x \ge 8) = p(x=8) + p(x=9) + p(x=10)$
$p(x) = \binom{10}{8} (0.6)^8 (0.4)^2 + \binom{10}{9} (0.6)^9 (0.4)^1 + \binom{10}{10} (0.6)^{10} (0.4)^0$
(c) (3 points) If there are 20 difficult questions on the list and 80 easy ones, what is the probability that our student, who knows how to solve 60 questions total, knows how to solve all the difficult questions?
$P(S|D) = \frac{P(S \cap D)}{P(S)}$
$P(D|S) = $