Step $- 1$: Simplify the problems to solve using probability

We know that we have 3 alternative answer but only 1 is correct

$\therefore$ The probability of getting a correct answer $\frac{1}{3}$

So the probability of getting an incorrect answer is $\left(1-\frac{1}{3}\right)=\frac{2}{3}$

Step $- 2$: Use binomial distribution

So the probability of getting $4$ or correct answer $=$

probability of $4$ or more correct $+$ probability of $5$ correct answer ………$(i)$

Now we can say, for probability of $4$ correct answer $x=4 \,and\, n=5$

By binomial distribution,Probability of getting $4$ correct answer $=^5 C _4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^{5-4}$

$=5 \times\left(\frac{1}{3}\right)^4 \times\left(\frac{2}{3}\right)$

$=5 \times \frac{2}{3^5}$

For probability of $5$ correct answer $x=5$ and $n=5$

So the probability of getting $5$ correct answers $={ }^5 C _5 \times\left(\frac{1}{3}\right)^5 \times\left(\frac{2}{3}\right)^{5-5}$

Probability of getting $5$ correct answers $=\left(\frac{1}{3}\right)^5$

Step $- 3:$ Put the values in $(i)$

So the probability of getting $4$ or more correct answers $=5 \times \frac{2}{3^5}+\frac{1}{3^5}$

$=\frac{10}{3^5}+\frac{1}{3^5}$

$=\frac{11}{3^5}$

Hence, the probability that a student will get 4 or more answer correct answer is $\frac{11}{3^5}$