1. Key Concepts and Formulas

• Discrete Random Variable: A variable whose value can only take a finite number of distinct values. In this problem, $X$ represents the number of rotten apples drawn, which can be 0, 1, or 2.
• Probability Distribution: For a discrete random variable $X$, its probability distribution lists all possible values $x_i$ and their associated probabilities $P(X=x_i)$. The sum of all probabilities must equal 1.
• Combinations ($^n C_r$): Used to count the number of ways to choose $r$ items from a set of $n$ distinct items without regard to the order of selection. The formula is $^n C_r = \frac{n!}{r!(n-r)!}$. This is essential for calculating the probabilities of drawing specific combinations of apples.
• Expected Value (Mean) $E(X)$: Represents the average value of $X$ over many trials. For a discrete random variable, it is calculated as: $E(X) = \sum_{i} x_i P(X=x_i)$
• Expected Value of $X^2$, $E(X^2)$: This is needed to calculate the variance and is computed similarly: $E(X^2) = \sum_{i} x_i^2 P(X=x_i)$
• Variance $Var(X)$: A measure of how much the values of a random variable deviate from its mean. The primary formula for variance is: $Var(X) = E(X^2) - (E(X))^2$
• Hypergeometric Distribution: This problem describes a scenario that follows a Hypergeometric Distribution, which models the probability of $k$ successes in $n$ draws without replacement from a finite population. The variance for a Hypergeometric distribution is given by: $Var(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ where $N$ is the total population size, $K$ is the number of successes in the population, and $n$ is the number of draws.

2. Step-by-Step Solution

Step 1: Identify Given Information and Define the Random Variable First, let's list the known quantities from the problem statement:

• Number of rotten apples ($K$) = 3
• Number of good apples = 15
• Total number of apples ($N$) = $3 + 15 = 18$
• Number of apples drawn ($n$) = 2

Let $X$ be the random variable representing the number of rotten apples in a draw of two apples. Since we are drawing 2 apples and there are 3 rotten apples available, the possible values for $X$ are:

• $X=0$: We draw 0 rotten apples (and 2 good apples).
• $X=1$: We draw 1 rotten apple (and 1 good apple).
• $X=2$: We draw 2 rotten apples (and 0 good apples).

Step 2: Calculate the Total Number of Possible Outcomes The total number of ways to draw 2 apples from the 18 available apples is calculated using combinations, as the order of selection does not matter. This total will be the denominator for our probability calculations. $\text{Total number of ways to draw 2 apples} = ^{18}C_2$ $^{18}C_2 = \frac{18!}{2!(18-2)!} = \frac{18 \times 17}{2 \times 1} = 9 \times 17 = 153$ So, there are 153 possible combinations when drawing two apples.

Step 3: Construct the Probability Distribution of $X$ Next, we calculate the probability for each possible value of $X$ by determining the number of favorable outcomes for each case and dividing by the total number of outcomes.

• Case 1: $P(X=0)$ (Probability of drawing 0 rotten apples) To get 0 rotten apples, we must choose 0 rotten apples from the 3 available AND 2 good apples from the 15 available. $\text{Number of ways for } X=0 = ^3C_0 \times ^{15}C_2$ $^3C_0 = 1$ $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$ So, the number of ways for $X=0$ is $1 \times 105 = 105$. $P(X=0) = \frac{105}{153}$

• Case 2: $P(X=1)$ (Probability of drawing 1 rotten apple) To get 1 rotten apple, we must choose 1 rotten apple from the 3 available AND 1 good apple from the 15 available. $\text{Number of ways for } X=1 = ^3C_1 \times ^{15}C_1$ $^3C_1 = 3$ $^{15}C_1 = 15$ So, the number of ways for $X=1$ is $3 \times 15 = 45$. $P(X=1) = \frac{45}{153}$

• Case 3: $P(X=2)$ (Probability of drawing 2 rotten apples) To get 2 rotten apples, we must choose 2 rotten apples from the 3 available AND 0 good apples from the 15 available. $\text{Number of ways for } X=2 = ^3C_2 \times ^{15}C_0$ $^3C_2 = \frac{3 \times 2}{2 \times 1} = 3$ $^{15}C_0 = 1$ So, the number of ways for $X=2$ is $3 \times 1 = 3$. $P(X=2) = \frac{3}{153}$

Self-Check: Sum of Probabilities It's essential to verify that the sum of all probabilities equals 1. $P(X=0) + P(X=1) + P(X=2) = \frac{105}{153} + \frac{45}{153} + \frac{3}{153} = \frac{105 + 45 + 3}{153} = \frac{153}{153} = 1$ The probabilities are consistent.

Step 4: Calculate the Expected Value $E(X)$ The expected value $E(X)$ is the mean number of rotten apples we expect to draw. Using the formula $E(X) = \sum x_i P(X=x_i)$: $E(X) = \left(0 \times P(X=0)\right) + \left(1 \times P(X=1)\right) + \left(2 \times P(X=2)\right)$ Substituting the calculated probabilities: $E(X) = \left(0 \times \frac{105}{153}\right) + \left(1 \times \frac{45}{153}\right) + \left(2 \times \frac{3}{153}\right)$ $E(X) = 0 + \frac{45}{153} + \frac{6}{153} = \frac{45 + 6}{153} = \frac{51}{153}$ This fraction can be simplified by dividing both numerator and denominator by 51: $E(X) = \frac{51 \div 51}{153 \div 51} = \frac{1}{3}$ So, on average, we expect to draw $1/3$ of a rotten apple.

Step 5: Calculate the Expected Value of $X^2$, $E(X^2)$ To calculate the variance, we need $E(X^2)$. Using the formula $E(X^2) = \sum x_i^2 P(X=x_i)$: $E(X^2) = \left(0^2 \times P(X=0)\right) + \left(1^2 \times P(X=1)\right) + \left(2^2 \times P(X=2)\right)$ Substituting the probabilities: $E(X^2) = \left(0^2 \times \frac{105}{153}\right) + \left(1^2 \times \frac{45}{153}\right) + \left(2^2 \times \frac{3}{153}\right)$ $E(X^2) = \left(0 \times \frac{105}{153}\right) + \left(1 \times \frac{45}{153}\right) + \left(4 \times \frac{3}{153}\right)$ $E(X^2) = 0 + \frac{45}{153} + \frac{12}{153} = \frac{45 + 12}{153} = \frac{57}{153}$ This fraction can be simplified to $\frac{19}{51}$ by dividing by 3. We will keep it as $\frac{57}{153}$ for ease of calculation with the common denominator in the next step.

Step 6: Calculate the Variance $Var(X)$ Finally, we use the formula $Var(X) = E(X^2) - (E(X))^2$. We have $E(X^2) = \frac{57}{153}$ and $E(X) = \frac{1}{3}$. $Var(X) = \frac{57}{153} - \left(\frac{1}{3}\right)^2$ $Var(X) = \frac{57}{153} - \frac{1}{9}$ To subtract these fractions, we need a common denominator. We know that $153 = 9 \times 17$. So, we can rewrite $\frac{1}{9}$ as $\frac{1 \times 17}{9 \times 17} = \frac{17}{153}$. $Var(X) = \frac{57}{153} - \frac{17}{153}$ $Var(X) = \frac{57 - 17}{153}$ $Var(X) = \frac{40}{153}$

Step 7: (Optional) Verify using Hypergeometric Distribution Variance Formula As this problem fits the Hypergeometric distribution model, we can use its specific variance formula for verification: $Var(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ Plugging in our values ($N=18$, $K=3$, $n=2$): $Var(X) = 2 \times \left(\frac{3}{18}\right) \times \left(\frac{18-3}{18}\right) \times \left(\frac{18-2}{18-1}\right)$ $Var(X) = 2 \times \left(\frac{1}{6}\right) \times \left(\frac{15}{18}\right) \times \left(\frac{16}{17}\right)$ $Var(X) = \frac{1}{3} \times \frac{5}{6} \times \frac{16}{17}$ $Var(X) = \frac{1 \times 5 \times 16}{3 \times 6 \times 17} = \frac{80}{306}$ Simplifying the fraction by dividing both numerator and denominator by 2: $Var(X) = \frac{40}{153}$ Both methods yield the same result, confirming the calculation.

3. Common Mistakes & Tips

• Confusing $E(X^2)$ with $Var(X)$: A frequent error is to incorrectly equate $E(X^2)$ with $Var(X)$. Always remember that $Var(X) = E(X^2) - (E(X))^2$. $E(X^2)$ is an intermediate step, not the variance itself, unless $E(X)=0$.
• Combinations vs. Permutations: For selection problems where the order of items drawn does not matter, always use combinations ($^n C_r$).
• Careful with Fraction Arithmetic: Ensure accuracy when performing operations with fractions, especially when finding common denominators for subtraction.
• Recognize Distribution Types: Identifying the underlying probability distribution (like Hypergeometric in this case) can provide a powerful shortcut for calculating mean and variance, and serves as an excellent method for verifying results obtained from first principles.

4. Summary

This problem required us to calculate the variance of a discrete random variable representing the number of rotten apples drawn. We systematically determined the total number of ways to draw two apples and then calculated the probability for each possible number of rotten apples (0, 1, or 2). Using these probabilities, we computed the expected value $E(X)$ and the expected value of $X^2$, $E(X^2)$. Finally, we applied the fundamental formula for variance, $Var(X) = E(X^2) - (E(X))^2$, to arrive at the solution. The result was further validated using the specific formula for the variance of a Hypergeometric distribution, confirming the consistency of our calculations.

5. Final Answer

The final answer is $\boxed{\frac{40}{153}}$, which corresponds to option (B).