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Binomial Distribution — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsBinomial Distribution2 Marks
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?
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Answer

Let $X=$ number of correct answers.
$p=$ probability that a candidate gets a correct answer from three possible answers.
$\therefore p=\frac{1}{3}$ and $q=1-p=1-\frac{1}{3}=\frac{2}{3}$
Given: $\mathrm{n}=5$
$\therefore X \sim B\left(5, \frac{1}{3}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}, x=0,1,2,4,5$
i.e. $p(x)={ }^5 C_x\left(\frac{1}{3}\right)^x\left(\frac{2}{3}\right)^{5-x}$
i.e. $p(x)={ }^5 C_x\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{5-x} \quad, x=0,1,2,3,4,5$
$P$ (four or more correct answers) $=P[X \geq 4]$
$ =p(4)+p(5)$
$={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^{5-4}+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^{5-5}$
$=5 \times\left(\frac{1}{3}\right)^4 \times\left(\frac{2}{3}\right)^1+1 \times\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0$
$=\left(\frac{1}{3}\right)^4\left[5 \times \frac{2}{3}+\frac{1}{3}\right]$
$=\left(\frac{1}{3}\right)^4\left[\frac{10}{3}+\frac{1}{3}\right]=\frac{1}{81} \times \frac{11}{3}=\frac{11}{243} $
Hence, the probability of getting four or more correct answers $=\frac{11}{243}$.

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