A multiple choice test consists of 5 question | vedclass

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(B) Let $n = 5$ be the total number of questions. Let $p$ be the probability of answering a question incorrectly. Since there are $3$ incorrect respon

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$\frac{459}{512}$

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(B) Let $n = 5$ be the total number of questions. Let $p$ be the probability of answering a question incorrectly. Since there are $3$ incorrect responses out of $4$,$p = \frac{3}{4}$. Let $q$ be the probability of answering a question correctly,so $q = 1 - p = \frac{1}{4}$. We need to find the probability of answering at least $3$ questions incorrectly,which is $P(X \ge 3)$,where $X$ follows a binomial distribution $B(n, p)$. $P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$ $P(X=k) = \binom{n}{k} p^k q^{n-k}$ $P(X=3) = \binom{5}{3} (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 	imes \frac{27}{64} 	imes \frac{1}{16} = \frac{270}{1024}$ $P(X=4) = \binom{5}{4} (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 	imes \frac{81}{256} 	imes \frac{1}{4} = \frac{405}{1024}$ $P(X=5) = \binom{5}{5} (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 	imes \frac{243}{1024} 	imes 1 = \frac{243}{1024}$ $P(X \ge 3) = \frac{270 + 405 + 243}{1024} = \frac{918}{1024} = \frac{459}{512}$

$A$ multiple choice test consists of $5$ questions,each question having $4$ responses. There is only one correct response and the remaining $3$ are incorrect responses. If a candidate attempts all the $5$ questions,then the probability that he answers at least $3$ questions incorrectly is

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