On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Explanation

Let probability of getting a correct answer is `P(A) `.

Here, `P(A) = 1/3`

`:. P(barA) = 1-1/3 = 2/3`

Here, `P(barA)` is the probability of getting an incorrect answer.

Now, there are two cases when a candidate can answer four or more answers correctly.

Case 1: When he answers 4 questions correcly and one incorrect.

In this case, probability will be `= C(5,4)(1/3)^4(2/3) = 5*1/81*1/3 = 10/243`

Case 2: When he answers all 5 questions correcly .

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