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) = $\overset{n}{C}_x \cdot q^{n-x} \cdot p^{x}$ = $\overset{5}{C}_x \cdot \left(\frac{2}{3}\right)^{5-x} \cdot \left(\frac{1}{3}\right)^{x}$ P (guessing more than 4 correct answers) = P (X ≥ 4) = P (X = 4) + P (X = 5) = $\overset{5}{C}_4 \cdot \left(\frac{2}{3}\right)^{5-4} \cdot \left(\frac{1}{3}\right)^{4}$ + $\overset{5}{C}_5 \cdot \left(\frac{2}{3}\right)^{5-5} \cdot \left(\frac{1}{3}\right)^{5}$ = 5 × $\left(\frac{2}{3}\right) \cdot \left(\frac{1}{81}\right)$ + 1 × 1 × $\left(\frac{1}{243}\right)$ [Using $\overset{n}{C}_r$ = $\frac{n!}{(n-r)! r!}$] = $\frac{11}{243}$