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
Solution
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<td class="tg-0pky">$X$</td>
<td class="tg-0pky">$P(X)$</td>
<td class="tg-0pky">$XP(X)$</td>
<td class="tg-0pky">$\left(X_i-\mu\right)^2$</td>
<td class="tg-0pky">$P_i X\left(X_i-\mu\right)^2$</td>
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<td class="tg-0pky">$X=0$</td>
<td class="tg-0pky">$\frac{{ }^7 C_2}{{ }^{10} C_2}$</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">$\left(0-\frac{3}{5}\right)^2$</td>
<td class="tg-0pky">$\frac{7}{15}\left(\frac{9}{25}\right)$</td>
</tr>
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<td class="tg-0pky">$X=1$</td>
<td class="tg-0pky">$\frac{{ }^7 C_1{ }^3 C_1}{{ }^{10} C_2}$</td>
<td class="tg-0pky">$\frac{7}{15}$</td>
<td class="tg-0pky">$\left(1-\frac{3}{5}\right)^2$</td>
<td class="tg-0pky">$\frac{7}{15}\left(\frac{4}{25}\right)$</td>
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<td class="tg-0pky">$x=2$</td>
<td class="tg-0pky">$\frac{{ }^7 C_0{ }^3 C_2}{{ }^{10} C_2}$</td>
<td class="tg-0pky">$\frac{2}{15}$</td>
<td class="tg-0pky">$\left(2-\frac{3}{5}\right)^2$</td>
<td class="tg-0pky">$\frac{2}{15}\left(\frac{49}{25}\right)$</td>
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<p>$$\begin{aligned}
& \mu=\sum X_i P\left(X_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\
& \text { Variance }(X)= \\
& \sum P_i\left(X_i-\mu\right)^2=\frac{7}{15}\left(\frac{9}{25}\right)+\frac{7}{15}\left(\frac{4}{25}\right)+\frac{2}{15}\left(\frac{49}{25}\right)=\frac{28}{75}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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