JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 12)
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
$\frac{11}{15}$
$\frac{2}{15}$
$\frac{3}{5}$
$\frac{28}{75}$
Explanation
| $X$ | $P(X)$ | $XP(X)$ | $\left(X_i-\mu\right)^2$ | $P_i X\left(X_i-\mu\right)^2$ |
| $X=0$ | $\frac{{ }^7 C_2}{{ }^{10} C_2}$ | 0 | $\left(0-\frac{3}{5}\right)^2$ | $\frac{7}{15}\left(\frac{9}{25}\right)$ |
| $X=1$ | $\frac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2}$ | $\frac{7}{15}$ | $\left(1-\frac{3}{5}\right)^2$ | $\frac{7}{15}\left(\frac{4}{25}\right)$ |
| $x=2$ | $\frac{{ }^7 C_0{ }^3 C_2}{{ }^{10} C_2}$ | $\frac{2}{15}$ | $\left(2-\frac{3}{5}\right)^2$ | $\frac{2}{15}\left(\frac{49}{25}\right)$ |
$$\begin{aligned} & \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(X)= \\ & \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75} \end{aligned}$$
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