A multiple choice examination has 5 questions | vedclass

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Question

(C) Let $n = 5$ be the total number of questions. Since each question has $3$ alternatives with only $1$ correct,the probability of success $p = \frac

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$\frac{11}{243}$

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(C) Let $n = 5$ be the total number of questions. Since each question has $3$ alternatives with only $1$ correct,the probability of success $p = \frac{1}{3}$ and the probability of failure $q = 1 - \frac{1}{3} = \frac{2}{3}$. We use the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$. We need to find the probability of getting $4$ or more correct answers,which is $P(X \geq 4) = P(X = 4) + P(X = 5)$. $P(X = 4) = {}^5C_4 \left(\frac{1}{3}ight)^4 \left(\frac{2}{3}ight)^1 = 5 	imes \frac{1}{81} 	imes \frac{2}{3} = \frac{10}{243}$. $P(X = 5) = {}^5C_5 \left(\frac{1}{3}ight)^5 \left(\frac{2}{3}ight)^0 = 1 	imes \frac{1}{243} 	imes 1 = \frac{1}{243}$. Therefore,$P(X \geq 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243}$.

$A$ 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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