Suppose that in one of your classes you are given 20 questions, and are told that the final exam will consist of 8 of them. Suppose that the 8 that are chosen for the final are selected randomly.

a) If you work out how to do 12 of the problems, what is the probability that you will be able to work at least 7 of those on the final?

b) How many problems do you need to have worked out to have a 90% probability of being able to solve all 8 of those on the final?

a) We can calculate this probability by dividing the total amount of favourable cases with the total amount of cases (because all possible exams of 8 questions are equally probable).

From 12 questions we work out on and 8 we didnt, we will be fine is at least 7 questions from the 8 given are the ones that we worked out.

The total amount of exams with exactly 7 questions we can answer is

[tex] {12 \choose 7} * 8 = 6336 [/tex]

First we pick 7 from 12, and then we pick 1 of the 8 questions we dont know the answer.

The total amount of exams with 8 questions we know the answer is the total amount of ways to pick 8 form 12, thus it is [tex] {12 \choose 8} = 495 . [/tex] Therefore, there are a total of 6336+495 = 6831 favourable cases.

The total amount of exams, on the other hand, is the total amount of ways to pick 8 from 20, therefore it is [tex] {20 \choose 8} = 125970 . [/tex]

The probability to get 7 questions we work out to do is 6831/125970 = 0.054227. Thus, we are going to be fine only 5.42% of the time.

b) Combinatorial numbers grows pretty quickly. If we were to solve 19 out of 20 problems, then the probability of getting all 8 problems is

19/20 * 18/19 * 17/18 * 16/17 * 15/16 * 14/15 * 13/14 * 12/13 = 12/20 = 0.6

(for the first question of the exam, 19 out of 20 are fine, for the second question, of the 19 questions remaining, 18 can be answered, and so on)

This means that, if you want a 90% chance of answering all 8 questions, you need indeed be able to answer all 20 questions provided you are given.