A multiple choice examination has 5 questions | vedclass

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Question

(D) Let $p$ be the probability of guessing the correct answer for a single question. Since there are $3$ alternatives and only $1$ is correct,$p = \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $p$ be the probability of guessing the correct answer for a single question. Since there are $3$ alternatives and only $1$ is correct,$p = \frac{1}{3}$.Consequently,the probability of guessing incorrectly is $q = 1 - p = \frac{2}{3}$.Let $X$ be the random variable representing the number of correct answers in $n = 5$ questions. $X$ follows a binomial distribution $X \sim B(n, p) = B(5, \frac{1}{3})$.The probability of getting $k$ correct answers is given by $P(X = k) = {}^nC_k p^k q^{n-k}$.We need to find the probability of getting $4$ or more correct answers,which is $P(X \geq 4) = P(X = 4) + P(X = 5)$.$P(X = 4) = {}^5C_4 (\frac{1}{3})^4 (\frac{2}{3})^1 = 5 	imes \frac{1}{81} 	imes \frac{2}{3} = \frac{10}{3^5}$.$P(X = 5) = {}^5C_5 (\frac{1}{3})^5 (\frac{2}{3})^0 = 1 	imes \frac{1}{243} 	imes 1 = \frac{1}{3^5}$.Therefore,$P(X \geq 4) = \frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$.

$A$ 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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