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

We have been given that, there are 3 possible answers of the total 5 questions out of which only 1 answer is correct.

Let p be the probability of getting a correct answer out of the 3 alternative answers.

Then, q is the probability of not getting a correct answer out of 3 alternatives.

And, p + q = 1

⇒ q = 1 – p

Let X be any random variable representing a number of correct answers just be guessing out of 5 questions.

Then, the probability that the candidate would get r answers correct by just guessing out of 5 questions is given by this Binomial distribution.

P (X = r) = ^{n}C_{r}p^{r}q^{n-r}

Here, n = 5.

Substitute the value of n, p, and q in the above formula.

…(i)

We need to find the probability that a candidate would get four or more correct answers just by guessing.

The probability can be expressed as,

Probability = P (X ≥ 4)

This is in turn can be written as,

P (X ≥ 4) = P (X = 4) + P (X = 5)

Put r = 4, 5 subsequently in equation (i) and then substitute in the above formula.

⇒ P (X ≥ 4) = 0.0453

Hence, the probability that a candidate would get four or more correct answers just by guessing is 0.0453.

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