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

• A. $\frac{37}{153}$
• B. $\frac{57}{153}$
• C. $\frac{47}{153}$
• D. $\frac{40}{153}$

Answer 1

Answer: (D)

$\frac{40}{153}$

Answer 2

Consider 3 bad apples and 15 good apples. Let $X$ be the number of bad apples drawn. The probabilities are:

$P(X = 0) = \frac{\binom{15}{2}}{\binom{18}{2}} = \frac{105}{153}$ $P(X = 1) = \frac{\binom{3}{1} \cdot \binom{15}{1}}{\binom{18}{2}} = \frac{45}{153}$ $P(X = 2) = \frac{\binom{3}{2}}{\binom{18}{2}} = \frac{3}{153}$

Calculate the expected value $E(X)$:

$E(X) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 2 \times \frac{3}{153} = \frac{51}{153} = \frac{1}{3}$

Compute the variance:

$\text{Var}(X) = E(X^2) - (E(X))^2$ $E(X^2) = 0^2 \times \frac{105}{153} + 1^2 \times \frac{45}{153} + 2^2 \times \frac{3}{153} = \frac{57}{153}$ $\text{Var}(X) = \frac{57}{153} - \left( \frac{1}{3} \right)^2 = \frac{57}{153} - \frac{1}{9} = \frac{40}{153}$