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

$\frac{11}{243}$

The correct answer is Option (3) → $\frac{11}{243}$

As each multiple choice question has three possible answers (out of which only one is correct),

the probability of correct answer = $p =\frac{1}{3}$,

so the probability of wrong answer = $q =\frac{2}{3}$

Here, the number of questions to be answered = $n = 5$.

As the questions to be answered are independent, it is a problem of binomial distribution with $p=\frac{1}{3},q=\frac{2}{3}$ and $n = 5$.

∴ The probability of getting four or more correct answers

$=P(4)+P(5)$

$={^5C}_4(\frac{1}{3})^4.(\frac{2}{3})^1+{^5C}_5(\frac{1}{3})^5$

$=5.\frac{2}{3^5}+1.\frac{1}{3^5}=\frac{10+1}{3^5}=\frac{11}{243}$