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 answer just by guessing, is

The total number of questions is $n = 5$. Each question has 3 alternatives with exactly one correct answer, so the probability of guessing a correct answer is $p = \dfrac{1}{3}$ and the probability of guessing an incorrect answer is $q = 1 - p = \dfrac{2}{3}$.

The number of correct answers $X$ follows a binomial distribution $B(n, p)$, where $P(X = k) = ^{n}C_{k} p^{k} q^{n-k}$.

We need to find the probability of getting 4 or more correct answers, which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.

$P(X = 4) = ^{5}C_{4} \left(\dfrac{1}{3}\right)^{4} \left(\dfrac{2}{3}\right)^{5-4} = 5 \times \dfrac{1}{81} \times \dfrac{2}{3} = \dfrac{10}{243}$.

$P(X = 5) = ^{5}C_{5} \left(\dfrac{1}{3}\right)^{5} \left(\dfrac{2}{3}\right)^{5-5} = 1 \times \dfrac{1}{243} \times 1 = \dfrac{1}{243}$.

$P(X \ge 4) = \dfrac{10}{243} + \dfrac{1}{243} = \dfrac{11}{243}$.

Since $243 = 3^{5}$, the probability is $\dfrac{11}{3^{5}}$.

Answer: $\dfrac{11}{3^5}$