1. Key Concepts and Formulas

• Discrete Random Variable: A variable whose value can only take a finite or countable number of distinct values. In this problem, $x$ (number of defective oranges) is a discrete random variable.
• Probability Distribution: For a discrete random variable $x$, its probability distribution lists all possible values $x_i$ and their corresponding probabilities $P(x=x_i)$. The sum of all probabilities must equal 1.
• Expected Value (Mean): Denoted as $E(x)$ or $\mu$, it represents the average value of the random variable. For a discrete random variable, it's calculated as: $E(x) = \sum_{i} x_i P(x=x_i)$
• Expected Value of $x^2$: Denoted as $E(x^2)$, it is used in the variance calculation: $E(x^2) = \sum_{i} x_i^2 P(x=x_i)$
• Variance: Denoted as $Var(x)$ or $\sigma^2$, it measures the spread or dispersion of the values of a random variable around its mean. For a discrete random variable, it is calculated using the formula: $Var(x) = E(x^2) - [E(x)]^2$
• Combinations: Used to calculate the number of ways to choose $k$ items from a set of $n$ items without regard to the order, given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. This is crucial for problems involving selections without replacement, like drawing items from a lot.

2. Step-by-Step Solution

Step 1: Identify the Problem Parameters and Possible Values of $x$

We are given:

• Number of defective oranges: $D = 3$
• Number of good oranges: $G = 7$
• Total number of oranges: $N = D + G = 3 + 7 = 10$
• Number of oranges drawn: $n = 2$

The random variable $x$ denotes the number of defective oranges drawn. Since we draw 2 oranges from a lot containing 3 defective ones, the possible values for $x$ are:

• $x=0$: Both drawn oranges are good.
• $x=1$: One drawn orange is defective, and one is good.
• $x=2$: Both drawn oranges are defective. The maximum number of defective oranges we can draw is limited by the number of oranges drawn (2) and the total number of defective oranges available (3). Thus, $x \in \{0, 1, 2\}$.

Step 2: Calculate the Total Number of Outcomes

The total number of ways to draw 2 oranges from 10 oranges is given by the combination formula $\binom{N}{n}$: $\text{Total outcomes} = \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$ This value will be the denominator for calculating the probabilities $P(x=x_i)$.

Step 3: Determine the Probability Distribution of $x$

We calculate $P(x=x_i)$ for each possible value of $x$ using combinations:

• For $x=0$ (0 defective, 2 good): We choose 0 defective oranges from 3, AND 2 good oranges from 7. $\text{Number of ways for } x=0 = \binom{3}{0} \times \binom{7}{2} = 1 \times \frac{7 \times 6}{2 \times 1} = 1 \times 21 = 21$ $P(x=0) = \frac{21}{45}$

• For $x=1$ (1 defective, 1 good): We choose 1 defective orange from 3, AND 1 good orange from 7. $\text{Number of ways for } x=1 = \binom{3}{1} \times \binom{7}{1} = 3 \times 7 = 21$ $P(x=1) = \frac{21}{45}$

• For $x=2$ (2 defective, 0 good): We choose 2 defective oranges from 3, AND 0 good oranges from 7. $\text{Number of ways for } x=2 = \binom{3}{2} \times \binom{7}{0} = \frac{3 \times 2}{2 \times 1} \times 1 = 3 \times 1 = 3$ $P(x=2) = \frac{3}{45}$

Self-Check: Sum of probabilities: $P(x=0) + P(x=1) + P(x=2) = \frac{21}{45} + \frac{21}{45} + \frac{3}{45} = \frac{45}{45} = 1$. This confirms our probability calculations are correct.

The probability distribution table is:

$$\begin{array}{|c|c|c|c|} \hline x_i & 0 & 1 & 2 \\ \hline P(x=x_i) & \frac{21}{45} & \frac{21}{45} & \frac{3}{45} \\ \hline \end{array}$$

Step 4: Calculate the Expected Value $E(x)$

Using the formula $E(x) = \sum x_i P(x=x_i)$: $E(x) = (0) \times \frac{21}{45} + (1) \times \frac{21}{45} + (2) \times \frac{3}{45}$ $E(x) = 0 + \frac{21}{45} + \frac{6}{45} = \frac{27}{45}$ Simplifying the fraction: $E(x) = \frac{3}{5}$

Step 5: Calculate the Expected Value $E(x^2)$

Using the formula $E(x^2) = \sum x_i^2 P(x=x_i)$: $E(x^2) = (0^2) \times \frac{21}{45} + (1^2) \times \frac{21}{45} + (2^2) \times \frac{3}{45}$ $E(x^2) = 0 \times \frac{21}{45} + 1 \times \frac{21}{45} + 4 \times \frac{3}{45}$ $E(x^2) = 0 + \frac{21}{45} + \frac{12}{45} = \frac{33}{45}$ Simplifying the fraction: $E(x^2) = \frac{11}{15}$

Step 6: Calculate the Variance $Var(x)$

Using the formula $Var(x) = E(x^2) - [E(x)]^2$: $Var(x) = \frac{11}{15} - \left(\frac{3}{5}\right)^2$ $Var(x) = \frac{11}{15} - \frac{9}{25}$ To subtract these fractions, find the Least Common Multiple (LCM) of 15 and 25. $15 = 3 \times 5$ $25 = 5 \times 5$ LCM$(15, 25) = 3 \times 5 \times 5 = 75$. $Var(x) = \frac{11 \times 5}{15 \times 5} - \frac{9 \times 3}{25 \times 3}$ $Var(x) = \frac{55}{75} - \frac{27}{75}$ $Var(x) = \frac{55 - 27}{75} = \frac{28}{75}$

3. Common Mistakes & Tips

• Forgetting to square $E(x)$: A common error is to calculate $E(x^2) - E(x)$ instead of $E(x^2) - [E(x)]^2$.
• Incorrectly calculating combinations: Ensure you correctly apply the combination formula $\binom{n}{k}$ for selecting items without replacement.
• Not checking the sum of probabilities: Always verify that $\sum P(x=x_i) = 1$. This helps catch errors in individual probability calculations early.
• Simplifying fractions too early: While simplifying fractions is good practice, sometimes keeping a common denominator throughout intermediate steps (especially for sums) can reduce calculation errors.
• Recognizing Hypergeometric Distribution: This problem is an example of a Hypergeometric Distribution. For such distributions, the mean and variance can be calculated directly using formulas:
  • $E(x) = n \frac{K}{N}$
  • $Var(x) = n \frac{K}{N} \left(1 - \frac{K}{N}\right) \left(\frac{N-n}{N-1}\right)$ Using $N=10, K=3, n=2$: $E(x) = 2 \times \frac{3}{10} = \frac{6}{10} = \frac{3}{5}$ (Matches our calculation) $Var(x) = 2 \times \frac{3}{10} \times \left(1 - \frac{3}{10}\right) \times \left(\frac{10-2}{10-1}\right) = \frac{3}{5} \times \frac{7}{10} \times \frac{8}{9} = \frac{168}{450} = \frac{28}{75}$ (Matches our calculation). While using these formulas can be a shortcut, understanding the first principles calculation is fundamental.

4. Summary

To find the variance of the number of defective oranges drawn, we first determined the possible values of the random variable $x$ and calculated the probability of each value using combinations. Then, we computed the expected value $E(x)$ and the expected value of $x^2$, $E(x^2)$, by summing $x_i P(x=x_i)$ and $x_i^2 P(x=x_i)$ respectively. Finally, we applied the variance formula $Var(x) = E(x^2) - [E(x)]^2$ to arrive at the result. The calculation yielded a variance of $\frac{28}{75}$.

5. Final Answer

The variance of $x$ is $\frac{28}{75}$, which corresponds to option (D).

The final answer is $\boxed{\text{28 / 75}}$.