1. Key Concepts and Formulas

• Hypergeometric Distribution: This distribution models the probability of $k$ successes in $n$ draws, without replacement, from a finite population of size $N$ that contains $K$ successes.
  • Parameters:
    • $N$: Total population size.
    • $K$: Number of successes in the population.
    • $n$: Sample size (number of draws).
  • Mean ($\mu$ or $E[X]$): The expected number of successes is given by the formula: $\mu = n \frac{K}{N}$
  • Variance ($\sigma^2$ or $Var[X]$): The variability in the number of successes is given by: $\sigma^2 = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ The term $\frac{N-n}{N-1}$ is known as the Finite Population Correction Factor, which accounts for sampling without replacement from a finite population.

2. Step-by-Step Solution

Step 1: Identify the Distribution and its Parameters

The problem describes a scenario where items (apples) are drawn one by one without replacement from a finite population, and we are interested in the number of "rotten apples" (successes) in the sample. This perfectly fits the definition of a Hypergeometric Distribution.

Let's identify the parameters for this specific problem:

• Total population size ($N$): Total number of apples = $3 \text{ (rotten)} + 7 \text{ (good)} = 10$.
• Number of successes in the population ($K$): Number of rotten apples = $3$.
• Sample size ($n$): Number of apples drawn = $4$.
• The random variable $X$ denotes the number of rotten apples drawn. Based on the parameters, $X$ can take integer values from $\max(0, n - (N-K))$ to $\min(n, K)$.
  • $\max(0, 4 - (10-3)) = \max(0, 4-7) = \max(0, -3) = 0$.
  • $\min(4, 3) = 3$.
  • So, $X$ can take values $\{0, 1, 2, 3\}$.

Step 2: Calculate the Mean ($\mu$) of X

We use the formula for the mean of a Hypergeometric distribution: $\mu = n \frac{K}{N}$ Substitute the identified parameters: $\mu = 4 \times \frac{3}{10}$ $\mu = \frac{12}{10}$ $\mu = \frac{6}{5} = 1.2$

Step 3: Calculate the Variance ($\sigma^2$) of X

We use the formula for the variance of a Hypergeometric distribution: $\sigma^2 = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ Substitute the identified parameters:

• $n = 4$
• $K = 3$
• $N = 10$
• $N-K = 10-3 = 7$
• $N-n = 10-4 = 6$
• $N-1 = 10-1 = 9$

Now, plug these values into the variance formula: $\sigma^2 = 4 \times \frac{3}{10} \times \frac{7}{10} \times \frac{6}{9}$ Let's simplify the terms before multiplying: $\sigma^2 = \frac{12}{10} \times \frac{7}{10} \times \frac{6}{9}$ $\sigma^2 = \frac{6}{5} \times \frac{7}{10} \times \frac{2}{3} \quad \text{(simplifying } \frac{12}{10} \text{ to } \frac{6}{5} \text{ and } \frac{6}{9} \text{ to } \frac{2}{3})$ Now, multiply the numerators and denominators: $\sigma^2 = \frac{6 \times 7 \times 2}{5 \times 10 \times 3}$ $\sigma^2 = \frac{84}{150}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 6: $\sigma^2 = \frac{84 \div 6}{150 \div 6}$ $\sigma^2 = \frac{14}{25} = 0.56$

Step 4: Calculate the required expression $10(\mu^2 + \sigma^2)$

First, calculate $\mu^2$: $\mu^2 = \left(\frac{6}{5}\right)^2 = \frac{36}{25}$ Now, calculate $\mu^2 + \sigma^2$: $\mu^2 + \sigma^2 = \frac{36}{25} + \frac{14}{25}$ $\mu^2 + \sigma^2 = \frac{36+14}{25}$ $\mu^2 + \sigma^2 = \frac{50}{25}$ $\mu^2 + \sigma^2 = 2$ Finally, calculate $10(\mu^2 + \sigma^2)$: $10(\mu^2 + \sigma^2) = 10 \times 2$ $10(\mu^2 + \sigma^2) = 20$

3. Common Mistakes & Tips

• Misidentifying the Distribution: A common error is to confuse the Hypergeometric distribution with the Binomial distribution. Remember, the Hypergeometric distribution is for sampling without replacement from a finite population, while the Binomial distribution is for sampling with replacement or from an infinite population.
• Forgetting the Finite Population Correction Factor: The term $\frac{N-n}{N-1}$ in the variance formula is critical for Hypergeometric distributions. Omitting it would lead to an incorrect variance, often the variance of a Binomial distribution.
• Calculation Errors: Be careful with fractions and simplification. It's often easier to simplify fractions at intermediate steps to avoid large numbers.
• Parameter Identification: Double-check that $N$, $K$, and $n$ are correctly assigned from the problem statement.

4. Summary

This problem required us to calculate the mean and variance of the number of rotten apples drawn without replacement, which is a classic application of the Hypergeometric distribution. We first identified the distribution parameters: total population $N=10$, number of rotten apples (successes) $K=3$, and sample size $n=4$. Using the standard formulas for the mean ($\mu = n \frac{K}{N}$) and variance ($\sigma^2 = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$), we calculated $\mu = \frac{6}{5}$ and $\sigma^2 = \frac{14}{25}$. Finally, we substituted these values into the expression $10(\mu^2 + \sigma^2)$, which yielded $10\left(\left(\frac{6}{5}\right)^2 + \frac{14}{25}\right) = 10\left(\frac{36}{25} + \frac{14}{25}\right) = 10\left(\frac{50}{25}\right) = 10 \times 2 = 20$.

5. Final Answer

The final answer is $\boxed{20}$, which corresponds to option (A).