A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

Probability of getting correct answer \((p)=\frac{1}{3}\)
\(\therefore \quad q=1-\frac{1}{3}=\frac{2}{3}\)
Also, \(\mathrm{n}=5\)
\(\therefore \quad\) Required probability
\(\begin{aligned}
& =\mathrm{P}(\mathrm{X} \geq 4) \\
& =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& ={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^1+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0 \\
& =5\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+1\left(\frac{1}{3}\right)^5 \\
& =\left(\frac{1}{3}\right)^4\left(\frac{10}{3}+\frac{1}{3}\right)=\frac{11}{3^5}=\frac{11}{243}
\end{aligned}\)