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

.: Let $p$ be the probability of success that the answer is correct

$= \cfrac{1}{3}$ .

If $p = \cfrac{1}{3},q = 1 - \cfrac{1}{3} = \cfrac{2}{3}$

$\therefore$ $X$ has a binomial distribution with $n = 5,p = \cfrac{1}{3},q = \cfrac{2}{3}$

$\therefore$ $P(X = r){ = ^n}{C_r}{(q)^{n - r}}{p^r}$

$\therefore$ Required probability $= P$ (four or more correct answers)

$= P(X \ge 4) = P(X = 4) + P(X = 5)$

$= 5{C_4}{\left( {\cfrac{2}{3}} \right)^1}{\left( {\cfrac{1}{3}} \right)^4}{ + ^5}{C_5}{\left( {\cfrac{2}{3}} \right)^0}{\left( {\cfrac{1}{3}} \right)^5} = \cfrac{{(5)(2)}}{{{{(3)}^5}}} + \cfrac{{(1)(1)}}{{{{(3)}^5}}} = \cfrac{{11}}{{{3^5}}} = \cfrac{{11}}{{243}}$