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Question

(A) The repeated guessing of correct answers in multiple choice questions follows Bernoulli trials. Let $X$ represent the number of correct answers by

Vedclass · Accepted Answer

$\frac{11}{243}$

Vedclass · Answer

(A) The repeated guessing of correct answers in multiple choice questions follows Bernoulli trials. Let $X$ represent the number of correct answers by guessing in a set of $n=5$ questions.The probability of getting a correct answer is $p = \frac{1}{3}$.Therefore,the probability of an incorrect answer is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.$X$ follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{3}$.The probability mass function is $P(X=x) = ^{5}C_{x} \cdot p^{x} \cdot q^{5-x} = ^{5}C_{x} \cdot (\frac{1}{3})^{x} \cdot (\frac{2}{3})^{5-x}$.We need to find the probability of getting $4$ or more correct answers,which is $P(X \geq 4) = P(X=4) + P(X=5)$.$P(X=4) = ^{5}C_{4} \cdot (\frac{1}{3})^{4} \cdot (\frac{2}{3})^{1} = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{243}$.$P(X=5) = ^{5}C_{5} \cdot (\frac{1}{3})^{5} \cdot (\frac{2}{3})^{0} = 1 \cdot \frac{1}{243} \cdot 1 = \frac{1}{243}$.Thus,$P(X \geq 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243}$.

On a multiple choice examination with three possible answers for each of the five questions,what is the probability that a candidate would get four or more correct answers just by guessing?

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