Question

Solve the following problem:

An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer completely randomly. Find the probability that,

1. the student gets 4 or more correct answers.
2. the student gets less than 4 correct answers.

Answer 1

Let x = Number of correct answer

p = Probability of guessing correct answer

∴ p = `(1)/(4)`

q = 1 – p

= `1 - (1)/(4)`

= `(3)/(4)` 

Here, n = 5

∴ X ~ B(n, p)

∴ X ~ B`(5","1/4)`

For binomial distribution

p(x) = ⁿC_x, p^x. q^n–x

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

`"P"("X" = x) = ""^5"C"_x (1/4)^x (3/4)^(5-x)`, x = 0, 1, ...., 5

a. Probability that student gets 4 or more correct answers

= P(X ≥ 4)

= P(X = 4) or P(X = 5)

= ⁵C₄.p⁴.q¹ + ⁵C₅ p⁵.q⁰

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

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

= `15/1024 + 1/1024`

= `16/1024`

= `1/64`

b. Probability that student gets less than 4 correct answer:

= P(X < 4)

= 1 – P(X ≥ 4)

= `1 - 1/64`

= `63/64`