# 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 ?

$\begin{array}{1 1}\large \frac{11}{243}\\\large \frac{10}{243} \\\large \frac{1}{243} \\\large \frac{5}{243} \end{array}$

Toolbox:
• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
• Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
The experiment is to get 4 or more correct answers by guessing given $n=5$ questions. It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
Given 3 possible answers, P (getting a correct answer) $= p = \large\frac{1}{3}$ and P (not getting a correct answer) $= q = 1 -p = \large \frac{2}{3}$
Let X be the number of correct answers. X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
We need to calculate P (getting 4 or more correct answers) = P (X = 4) + P (X = 5).
$P (X = 4) = \large^{5}C_4\large\frac{1}{3}^4$$\times$$\large\frac{2}{3}^{5-4} $$= 5 \times \large\frac{1}{81}$$\times\large\frac{2}{3} = \frac{10}{243}$
$P (X = 5) = \large^{5}C_5\large\frac{1}{3}^5$$\times$$\large\frac{2}{3}^{5-5}$$= \large\frac{1}{3}^5$$= \large\frac{1}{243}$
Therefore $P (X \geq 4) = \large\frac{10+1}{243} = \frac{11}{243}$