Question

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?

Answer 1

Let X be the number of right answers in the 5 questions. 
X can take values 0,1,2...5.
X follows a binomial distribution with n =5

\[p = \text{ probability of guessing right answer }  = \frac{1}{3} \]
\[q = \text{ probability of guessing wrong answer }  = \frac{2}{3}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) = ^{5}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{5 - r} , r = 0, 1, 2, . . . 5\]
\[ \therefore P(\text{ The student guesses 4 or more correct answers} ) = P(X \geq 4) \]
\[ = P(X = 4) + P(X = 5)\]
\[ =^{5}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^1 + ^{5}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^0 \]
\[ = \frac{10 + 1}{3^5}\]
\[ = \frac{11}{243}\]