On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

#### Solution

The repeated guessing of correct answers from multiple-choice questions is Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple-choice questions.

Probability of getting a correct answer is, p = `1/3`

` therefore q = 1 - p = 1 -1/3 = 2/3`

Clearly, X has a binomial distribution with n = 5 and p = `1/3`.

The p.m.f. of X is given by

P(X = x) = `"^nC_x p^x q^(n - x)`, x = 0, 1, 2, 4, 5

i.e. p(x) = `"^nC_x (1/3)^x (2/3)^(5-x)` x = 0, 1, 2, 3, 4, 5

P(four or more correct answers) = P[X ≥ 4] = p(4) + p(5)

`= ""^5C_4 (1/3)^4 (2/3)^(5 - 4) + "^5C_5 (1/3)^5 (2/3)^(5 - 5)`

`= 5xx(1/3)^4 xx (2/3)^1 + 1xx (1/3)^5 (2/3)^0`

`= (1/3)^4 [5 xx 2/3 + 1/3]`

`= (1/3)^4 [10/3 +1/3] = 1/81 xx 11/3 = 11/243`

Hence, the probability of getting four or more correct answers `11/243`.