One multiple choice examination with three possible answers for each of the five questions. Then what is the probability that a candidate would get four or more correct answers is

Let X represents the number of correct answers by guessing in the set of 5 multiple choice Questions.

Probability of getting  correct answer $p=\frac{1}{3}$

$\therefore \mathrm{q}=1-p=1-\frac{1}{3}=\frac{2}{3}$ and given $\mathrm{n}=5$

P(guessing more then 4 correct answers)

   $$\begin{gathered} =\mathrm{P}(X\ge 4) \\ =\mathrm{P}(X=4)+\mathrm{P}(X=5) \\ ={}^5{\mathrm{c}}_4 \left(\frac{2}{3}\right) {\left(\frac{1}{3}\right)}^4+{}^5{\mathrm{c}}_5 {\left(\frac{1}{3}\right)}^5 \\ =5 \left(\frac{2}{3}\right) \left(\frac{1}{81}\right)+1 \left(\frac{1}{243}\right) \\ =\frac{10}{243}+\frac{1}{243} \\ =\frac{11}{243} \end{gathered}$$