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

The Correct Option is D

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:

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