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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,
- the student gets 4 or more correct answers.
- the student gets less than 4 correct answers.
Solution
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) = nCx, px. qn–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`
Notes
Let X denote the number of correct questions. Since only one of 4 suggested answers is correct
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