Question

A student has to sit for an examination consisting of 3 questions selected randomly from a list of 100 questions. To pass, he needs to answer all three questions. What is the probability that the student will pass the examination if he knows the answers to 90 questions on the list?

Answer 1

The total number of ways to select 3 questions from 100 questions is given by the combination formula:

nCk = n! / (k! * (n-k)!)

Where n is the total number of questions and k is the number of questions we need to select. In this case, n=100 and k=3, so:

100C3 = 100! / (3! * (100-3)!) = 161700

This means there are 161700 possible combinations of 3 questions that can be selected.

If the student knows the answers to 90 of the questions, then he has a pool of 90 questions from which to select the 3 questions for the examination. The number of ways to select 3 questions from the pool of 90 is:

90C3 = 90! / (3! * (90-3)!) = 117480

Therefore, the probability that the student will pass the examination is the number of ways to select 3 questions from the pool of 90 that he knows divided by the total number of possible combinations:

P(pass) = 117480 / 161700 ≈ 0.726

So the probability that the student will pass the examination is approximately 0.726 or 72.6%