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 is = 1 - $\frac{1}{3}$ = $\frac{2}{3}$
P (X = x) = $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{x}}.{\mathit{q}}^{\mathit{n}-\mathit{x}}.{\mathit{p}}^{\mathit{x}}$ = $\stackrel{5}{}{\mathit{C}}_{\mathit{x}}.{\left(\frac{2}{3}\right)}^{5-\mathit{x}}.{\left(\frac{1}{3}\right)}^{\mathit{x}}$
P (guessing more than 4 correct answers) = P (X ≥ 4)
= P (X = 4) + P (X = 5)
= $\stackrel{5}{}{\mathit{C}}_{4.}{\left(\frac{2}{3}\right)}^{5-4}.{\left(\frac{1}{3}\right)}^4$ + $\stackrel{5}{}{\mathit{C}}_{5.}{\left(\frac{2}{3}\right)}^{5-5}.{\left(\frac{1}{3}\right)}^5$
= 5 × $\left(\frac{2}{3}\right).\left(\frac{1}{81}\right)$ + 1 × 1 × $\left(\frac{1}{243}\right)$ [Using $\stackrel{\mathit{n}}{}{\mathit{C}}_{\mathit{r}}$ = $\frac{\mathit{n}!}{\left(\mathit{n}-\mathit{r}\right)!\mathit{r}!}$]
= $\frac{11}{243}$