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
.$$\begin{gathered} p\mathit{}=\mathrm{probability}\mathrm{of}\mathrm{guessing}\mathrm{right}\mathrm{answer}=\frac{1}{3} \\ q=\mathrm{probability}\mathrm{of}\mathrm{guessing}\mathrm{wrong}\mathrm{answer}=\frac{2}{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{5}C_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{5-r},r=0,1,2,...5 \\ \therefore P(\mathrm{The}\mathrm{student}\mathrm{guesses}4\mathrm{or}\mathrm{more}\mathrm{correct}\mathrm{answers})=P(\mathrm{X}\ge 4) \\ =P(X=4)+P(X=5) \\ ={}^{5}C_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1+{}^{5}C_5{\left(\frac{1}{3}\right)}^5{\left(\frac{2}{3}\right)}^0 \\ =\frac{10+1}{3^5} \\ =\frac{11}{243} \end{gathered}$$