SOURH7WK wrote:

In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256

B) 53/512

C) 105/512

D) 459/512

E) 47/256

We can compute either the probability of answering 3, 4, or 5 questions correctly, or subtract from 1 the probability of answering 0, 1, or 2 questions.

The probability of answering any question correctly is 1/4, and obviously, answering incorrectly is 3/4.

P(3 correct answers) =\(\frac{5*4}{2}(\frac{1}{4})^3(\frac{3}{4})^2=\frac{90}{4^5}\)

P(4 correct answers) =\(5(\frac{1}{4})^4(\frac{3}{4})=\frac{15}{4^5}\)

P(5 correct answers) =\((\frac{1}{4})^5=\frac{1}{4^5}\)

The sum of the above probabilities gives \(\frac{106}{4^5}=\frac{53}{512}\).

Answer B

Relevant topic for this question - Binomial probability distribution

Being a straightforward technical question, I question its importance for a real test.

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PhD in Applied Mathematics

Love GMAT Quant questions and running.