Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable to be the number of rotten apples in a draw of two apples, the variance of is

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Solution

Q.
Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is [2024]

1 $\frac{37}{153}$

2 $\frac{57}{153}$

3 $\frac{40}{153}$

4 $\frac{47}{153}$

Ans.
(3)

We have,

Number of rotten apples = 3

Number of good apples = 15

$x$ is the number of rotten apples

$x$ 0 1 2

$p (x)$ $\frac{{}^{15}{C_{2}}}{{}^{18}{C_{2}}} = \frac{105}{153}$ $\frac{{}^{3}{C_{1} \times {}^{15}C_{1}}}{{}^{18}{C_{2}}}$ $= \frac{3 \times 15}{153} = \frac{45}{153}$ $\frac{{}^{3}{C_{2}}}{{}^{18}{C_{2}}} = \frac{3}{153}$

$E (x) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + \frac{6}{153} = \frac{51}{153} = \frac{1}{3}$

$E (x^{2}) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + \frac{12}{153} = \frac{57}{153}$

$∴$ $Var (x) = E (x^{2}) - {(E (x))}^{2} = \frac{57}{153} - {(\frac{1}{3})}^{2}$ $= \frac{57}{153} - \frac{1}{9} = \frac{40}{153}$

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