JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 22)
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. If the variance of $$X$$ is $$\sigma^2$$, then $$96 \sigma^2$$ is equal to __________.
Answer
56
Explanation
| $$x$$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $$P(x)$$ | $$ \frac{{ }^7 C_5}{{ }^{10} C_5}=\frac{1}{12} $$ |
$$ \frac{C_4 \cdot{ }^3 C_1}{{ }^{10} C_5}=\frac{5}{12} $$ |
$$ \frac{{ }^7 C_3 \cdot{ }^3 C_2}{{ }^{10} C_5}=\frac{5}{12} $$ |
$$ \frac{{ }^7 C_2 \cdot{ }^3 C_3}{{ }^{10} C_5}=\frac{1}{12} $$ |
| $$xP(x)$$ | 0 | $$\frac{5}{12}$$ | $$\frac{10}{12}$$ | $$\frac{3}{12}$$ |
$$\begin{aligned} & \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\ & \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\ & =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12} \end{aligned}$$
$$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$$
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