Question

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

• A. $\frac{11}{243}$
• B. $\frac{1}{243}$
• C. $\frac{11}{81}$
• D. $\frac{1}{81}$

Answer 1

The correct option is A $\frac{11}{243}$

In this question, we have the repeated correct answer guessing form the given multiple choice question are Bernoulli trials
Let us now assume, $X$ represents the number of correct answer by guessing in the multiple choice set.
Now, probability of getting a correct answer, $p=1/3$
Thus, $q=1-p=1-1/3=2/3$
Clearly, we have $X$ is a binomial distribution where $n=5$ and $p=1/3$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$
${=}^5{C}_x{\left(\frac{2}{3}\right)}^{5-x}\times {\left(\frac{1}{3}\right)}^x$
Hence, probability of guessing more than $4$ correct answer = $P(X\ge 4)=P(x=4)+P(x=5){=}^5{C}_4\left(\frac{2}{3}\right)\times {\left(\frac{1}{3}\right)}^4{+}^5{C}_5{\left(\frac{1}{3}\right)}^5=5\times \frac{2}{3}\times \frac{1}{81}+1\times \frac{1}{243}=\frac{11}{243}$