Official solution
Each question has 3 alternative and exactly one is correct.
\[\therefore\] Probability of giving correct answer P(correct) = \[{1 \over 3}\]
\[\therefore\] Probability of giving wrong answer P(wrong) = \[{2 \over 3}\]
Here student give 4 or more correct answer.
\[\therefore\] Student give either 4 correct answer or 5 correct answer.
Using Binomial Distribution,
Required probability = \[{}^5{C_4}{\left( {{1 \over 3}} \right)^4}{\left( {{2 \over 3}} \right)^1}\] + \[{}^5{C_5}{\left( {{1 \over 3}} \right)^5}{\left( {{2 \over 3}} \right)^0}\] = \[{{11} \over {{3^5}}}\]