In a multiple-choice examination with three possible answers for … - Vidyadip

Question

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

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Answer

The repeated guessing of correct answers form multiple choice questions are bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.Probability of getting a correct answer is, $\text{p}=\frac{1}{3}$
$\therefore\text{q}=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Clearly, X has a binomial distribution with $\text{n = 5}$ and $\text{p}=\frac{1}{3}$
$\therefore\text{P(X = x)}=\text{ }^{\text{n}}\text{C}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}$
$=\text{ }^5\text{C}_{\text{x}}\big(\frac{2}{3}\big)^{5-\text{x}}.\big(\frac{1}{3}\big)^{\text{x}}$
P(guessing more than 4 correct answer) $=\text{P(X}\geq4)$
$=\text{P(X}=4)+\text{P(X}=5)$
$=\text{ }^5\text{C}_4\big(\frac{2}{3}\big).\big(\frac{1}{3}\big)^4+\text{ }^5\text{C}_5\big(\frac{1}{3}\big)^5$
$=5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}$
$=\frac{10}{243}+\frac{1}{243}$
$=\frac{11}{243}$

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