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(C) This is a binomial distribution problem where $n = 5$,$p = \frac{1}{3}$ (probability of success),and $q = \frac{2}{3}$ (probability of failure).We

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

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(C) This is a binomial distribution problem where $n = 5$,$p = \frac{1}{3}$ (probability of success),and $q = \frac{2}{3}$ (probability of failure).We need to find the probability of getting $4$ or more correct answers,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.Using the formula $P(X = k) = {^nC_k} \cdot p^k \cdot q^{n-k}$:$P(X = 4) = {^5C_4} \cdot (\frac{1}{3})^4 \cdot (\frac{2}{3})^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{3^5}$.$P(X = 5) = {^5C_5} \cdot (\frac{1}{3})^5 \cdot (\frac{2}{3})^0 = 1 \cdot \frac{1}{243} \cdot 1 = \frac{1}{3^5}$.Total probability = $\frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$.

$A$ 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:

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