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From a lot of $12$ items containing $3$ defectives, a sample of $5$ items is drawn at random. Let the random variable $\mathrm{X}$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to..........
$71$
$34$
$72$
$76$
Solution
$ \mathrm{a}=1-\frac{{ }^3 \mathrm{C}_5}{{ }^{12} \mathrm{C}_5} $
$ \mathrm{~b}=3 \cdot \frac{{ }^9 \mathrm{C}_4}{{ }^{12} \mathrm{C}_5} $
$ \mathrm{c}=3 \cdot \frac{{ }^9 \mathrm{C}_3}{{ }^{12} \mathrm{C}_5} $
$ \mathrm{~d}=1 \cdot \frac{{ }^9 \mathrm{C}_2}{{ }^{12} \mathrm{C}_5} $
$ \mathrm{u}=0 \cdot \mathrm{a}+1 \cdot \mathrm{b}+2 \cdot \mathrm{c}+3 \cdot \mathrm{d}=1.25 $
$ \sigma^2=0 \cdot \mathrm{a}+1 \cdot b+4 \cdot c+9 \mathrm{~d}-\mathrm{u}^2 $
$ \sigma^2=\frac{105}{176}$
Ans. $176-105=71$
Similar Questions
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
| Height | Weight | |
| Mean | $162.6\,cm$ | $52.36\,kg$ |
| Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
