Advertisements
Advertisements
Question
In a multiple choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Sum
Advertisements
Solution
Let X denote the number of correct answers.
Since only one of the three options is correct,
P(getting correct answer by guessing) = p = `(1)/(3)`
∴ 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) = `""^5"C"_x (1/3)^x (2/3)^(5 - x), x` = 0, 1,...,5
P(getting 4 or more correct answers by guessing) = P(X ≥ 4) = P(X = 4 or X = 5)
= P(X = 4) + (X = 5)
= `""^5"C"_4(1/3)^4 (2/3) + ""^5"C"_5(1/3)^5`
= `5 xx (1)/3^4 xx (2)/(3) + (1)/3^5`
= `(10 + 1)/(3^5)`
= `(11)/(3^5)`
= `(11)/(243)`.
shaalaa.com
Is there an error in this question or solution?
