Question

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

Answer 1

Let X = number of correct answers.

p = probability that a candidate gets correct answer from three possible answers.

∴ `p = 1/3` and q = 1 – p = `1 - 1/3 = 2/3`

Given: n = 5

∴ `X ∼ B(5, 1/3)`

 The p.m.f. of X is given by

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

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

P(four correct answers) = P[X = 4] = p(4)

= `""^5C_4 (1/3)^4 (2/3)^(5  -  4)`

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

= `5 xx 1/81 xx 2/3`

= `10/243`

Hence, the probability of getting four or more correct answers = `10/243`.