Question

A student is writing a multiple choice test consisting of 40 questions, each of which provides 4 possible choices. He is certain that he has 16 questions correct. If he guesses for all of the remaining 24 questions, what is the probability that he will pass the test?

Hint: There are only 24 questions left to choose from.

Answer 1

Look in your text for "binomial distribution" to find the formula that has been shown. To find P(at least 4) it is quickest to find P(0), P(1), P(2), and P(3) and subtract from 1:
....P(at least 4) = 1 - P(less than 4) = 1 - [P(0) + P(1) + P(2) + P(3)]

Let p = probability of getting right answer = 0.25
.....q = probability of wrong answer = 1 - p = 0.75

P(0) = P(24 wrong) = q^24 = 0.001 <-- not likely to miss all 24

One way to get exactly 1 right is to be correct on the first and wrong on the next 23: p^1 q^23.
But that one right one could be any of the 24, so that probability gets multiplied by 24:
P(1) = 24 * (0.25)(0.75^23) = . . .

One way to get exactly 2 right is to be correct on the first 2 and wrong on the next 22: p^2 q^22.
But the 1st right one could be any of 24, and the 2nd could be any of 23 except don't count the same pair twice:
P(2) = (24)(23)/2 * (0.25^2)(0.75)^22 = . . .

One way to get exactly 3 right is . . .
But . . .
P(3) = . . .

The pattern is
P(n) = (24!)/(n! (24-n)!) * (0.25^n)(0.75^(24-n))