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

To solve the problem of finding the variance of the random variable \(x\), which represents the number of rotten apples in a draw of two apples from a mixture of rotten and good apples, we need to proceed with the following steps:

Step 1: Determine probabilities. 

• Total apples = 3 (rotten) + 15 (good) = 18 apples.
• Total ways to select 2 apples from 18 apples = \(\binom{18}{2} = \frac{18 \times 17}{2} = 153\) ways.

Step 2: Establish the probability distribution of \(x\).

• \(P(x = 0)\): We choose 0 rotten apples (only good ones) from the draw. This is given by selecting 2 out of 15 good apples: \(P(x=0)= \frac{\binom{15}{2}}{\binom{18}{2}} = \frac{105}{153}\)
• \(P(x = 1)\): We choose 1 rotten and 1 good apple. This is: \(P(x=1) = \frac{\binom{3}{1} \times \binom{15}{1}}{\binom{18}{2}} = \frac{45}{153}\)
• \(P(x = 2)\): We choose 2 rotten apples. This is given by selecting 2 out of the 3 rotten apples: \(P(x=2) = \frac{\binom{3}{2}}{\binom{18}{2}} = \frac{3}{153}\)

Step 3: Calculate the expected value \(E(x)\).

• \(E(x) = 0 \times P(x=0) + 1 \times P(x=1) + 2 \times P(x=2)\)
• \(E(x) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 2 \times \frac{3}{153} = \frac{51}{153}\)

Step 4: Calculate the expected value of \(x^2\), \(E(x^2)\).

• \(E(x^2) = 0^2 \times P(x=0) + 1^2 \times P(x=1) + 2^2 \times P(x=2)\)
• \(E(x^2) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 4 \times \frac{3}{153} = \frac{57}{153}\)

Step 5: Calculate the variance \(Var(x)\).

• Variance is given by \(Var(x) = E(x^2) - [E(x)]^2\)
• \(Var(x) = \frac{57}{153} - \left(\frac{51}{153}\right)^2 = \frac{57}{153} - \frac{2601}{23409} = \frac{6120}{23409} - \frac{2601}{23409} = \frac{3519}{23409} = \frac{40}{153}\)

Thus, the variance of \(x\) is \( \frac{40}{153} \), matching the given correct answer.

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:

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