Three rotten apples are accidently mixed with fifteen good apples.

Question

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

Accepted Answer

3 bad apples, 15 good apples.

Let $$\mathrm{X}$$ be no of bad apples

$$\begin{aligned}
& \text { Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \
& \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \
& \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \
& \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \
& =\frac{1}{3} \
& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \
& =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \
& =\frac{57}{153}-\frac{1}{9}=\frac{40}{153}
\end{aligned}$$

Solutions · 1