# On a multiple choice examination with three possible answers for each of the five questions,

Question:

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?

Solution:

The repeated guessing of correct answers from 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, $p=\frac{1}{3}$

$\therefore q=1-p=1-\frac{1}{3}=\frac{2}{3}$

Clearly, $\mathrm{X}$ has a binomial distribution with $n=5$ and $p=\frac{1}{3}$

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-\mathrm{x}} p^{x}$

$={ }^{5} \mathrm{C}_{x}\left(\frac{2}{3}\right)^{5-x} \cdot\left(\frac{1}{3}\right)^{x}$

$={ }^{5} C_{x}\left(\frac{2}{3}\right)^{5-x} \cdot\left(\frac{1}{3}\right)^{x}$

P (guessing more than 4 correct answers) = P(X ≥ 4)

$=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)$

$={ }^{5} \mathrm{C}_{4}\left(\frac{2}{3}\right) \cdot\left(\frac{1}{3}\right)^{4}+{ }^{5} \mathrm{C}_{5}\left(\frac{1}{3}\right)^{5}$

$=5 \cdot \frac{2}{3} \cdot \frac{1}{81}+1 \cdot \frac{1}{243}$

$=\frac{10}{243}+\frac{1}{243}$

$=\frac{11}{243}$