Probability Question 10
Question 10 - 2024 (31 Jan Shift 1)
Three rotten apples are accidently 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
(1) $\frac{37}{153}$
(2) $\frac{57}{153}$
(3) $\frac{47}{153}$
(4) $\frac{40}{153}$
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Answer (4)
Solution
3 bad apples, 15 good apples.
Let $\mathrm{X}$ be no of bad apples
Then $P(X=0)=\frac{{ }^{15} C_{2}}{{ }^{18} C_{2}}=\frac{105}{153}$
$P(X=1)=\frac{{ }^{3} C_{1} \times{ }^{15} C_{1}}{{ }^{18} C_{2}}=\frac{45}{153}$
$P(X=2)=\frac{{ }^{3} C_{2}}{{ }^{18} C_{2}}=\frac{3}{153}$
$E(X)=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153}$
$=\frac{1}{3}$
$\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}$
$=0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^{2}$
$=\frac{57}{153}-\frac{1}{9}=\frac{40}{153}$
