A box contains 10 pens of which 3 are defective.

Question

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

Accepted Answer

$X$
    $P(X)$
    $XP(X)$
    $\left(X_i-\mu\right)^2$
    $P_i X\left(X_i-\mu\right)^2$

$X=0$
    $\frac{{ }^7 C_2}{{ }^{10} C_2}$
    0
    $\left(0-\frac{3}{5}\right)^2$
    $\frac{7}{15}\left(\frac{9}{25}\right)$

$X=1$
    $\frac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2}$
    $\frac{7}{15}$
    $\left(1-\frac{3}{5}\right)^2$
    $\frac{7}{15}\left(\frac{4}{25}\right)$

$x=2$
    $\frac{{ }^7 C_0{ }^3 C_2}{{ }^{10} C_2}$
    $\frac{2}{15}$
    $\left(2-\frac{3}{5}\right)^2$
    $\frac{2}{15}\left(\frac{49}{25}\right)$

$$\begin{aligned}
& \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \
& \text { Variance }(X)= \
& \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75}
\end{aligned}$$

Solutions · 1