A box contains 10 pens of which 3 are defective. A sample of 2 pe… — Solution | JEE Mathematics

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A box contains $10$ pens of which $3$ are defective. A sample of $2$ pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

A
$\frac{11}{15}$
B
$\frac{28}{75}$
C
$\frac{2}{15}$
D
$\frac{3}{5}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: There are $10$ pens, of which $3$ are defective and $7$ are non-defective. A sample of $2$ pens is drawn without replacement, and $X$ denotes the number of defective pens.

Find: The variance of $X$ .

Using the probability distribution of $X$ :

P(X=0)=\frac{{}^7C_2}{{}^{10}C_2}=\frac{21}{45}=\frac{7}{15}, \qquad P(X=1)=\frac{{}^7C_1 \cdot {}^3C_1}{{}^{10}C_2}=\frac{21}{45}=\frac{7}{15}, \qquad P(X=2)=\frac{{}^3C_2}{{}^{10}C_2}=\frac{3}{45}=\frac{1}{15}

Hypergeometric Formula

Since sampling is without replacement, $X$ follows a hypergeometric distribution with $N=10$ , $K=3$ , and $n=2$ .

Therefore,

E(X)=n\frac{K}{N}=2\cdot\frac{3}{10}=\frac{3}{5}

and

\operatorname{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}

Substituting the values,

\operatorname{Var}(X)=2\cdot\frac{3}{10}\cdot\frac{7}{10}\cdot\frac{8}{9}=\frac{336}{900}=\frac{28}{75}

Therefore, the correct option is B.

Common mistakes

Using the binomial variance formula as if the draws were independent is incorrect because the pens are drawn without replacement. Use the hypergeometric model or compute variance from the exact distribution.
Confusing the mean $E(X)=\frac{3}{5}$ with the variance is wrong. The question asks for variance, not expected value.
Computing probabilities for $X=0,1,2$ incorrectly by ignoring combinations can lead to a wrong distribution. Count selections using combinations such as $\binom{3}{x}\binom{7}{2-x}$ .

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Answer

Step-by-step solution

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