Question

one professor grades homework by randomly choosing 5 out of 12homework problems to grade.

a.) how many different groups of 5 problems can be chosen fromthe 12 problems?

b.) probability extension: jerry did only 5 problems of oneassignment. what is the probability that the problems he didcomprised the group that was selected to be graded?

c.) silvia did 7 problems. how many different groups of 5 didshe complete? what is the probability that one of the groups of 5she completed comprised the group selected to be graded?

Answer 1

(a) This is a binomial coefficient for 5 out of 12,written = 792

(b) There are 792 ways to choose 5 out of 12 problems andJerry chose 5 out of 12 problems in a single way, so his chancewould only be 1/792

(c) This is again like part (a), this is how many ways you canchoose 5 problems out of 7, which is

= 42/2 = 21

So there are 21 different combinations of 5 in the problemsSilvia. This is 21 of the 792 possible that the professorcould grade, so her probability of doing the 5 that were graded is21/792 = 7/264