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?

Solution

Let X denote the number of questions answered correctly by guessing in multiple choice examinations.

Probability of getting a correct answer by guessing, p= $\frac{1}{3}$

Therefore, q, the probability of an incorrect answer by guessing = 1 - $\frac{1}{3} = \frac{2}{3}$

There are in 5 questions in all.

So X follows binomial distribution with n = 5, p = $\frac{1}{3} and q = \frac{2}{3}$

$p (X = x) =^{n} C_{x} . q^{n - x} . p^{x} =^{5} C_{x} . {(\frac{2}{3})}^{5 - x} . {(\frac{1}{3})}^{x} p (guessing more than 4 correct answers) = p (X \geq 4) = p (X = 4) + p (X = 5) =^{5} C_{4} . {(\frac{2}{3})}^{5 - 4} . {(\frac{1}{3})}^{4} +^{5} C_{5} . {(\frac{2}{3})}^{5 - 5} . {(\frac{1}{3})}^{5} = 5 x (\frac{2}{3}) . (\frac{1}{81}) + 1 x 1 (\frac{1}{243}) [Using^{n} C_{r} = \frac{n!}{(n - 1)! r!}] = \frac{11}{243}$

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Probability

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?

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