1. Key Concepts and Formulas

• Discrete Random Variable: A variable whose value can only take a countable number of distinct values. Here, $X$ represents the number of defective items.
• Probability Distribution: For a discrete random variable $X$, its probability distribution lists all possible values $x_i$ and their corresponding probabilities $P(X=x_i)$.
• Expected Value (Mean): For a discrete random variable $X$, the expected value $E[X]$ is given by: $E[X] = \sum x_i P(X=x_i)$
• Expected Value of $X^2$: $E[X^2] = \sum x_i^2 P(X=x_i)$
• Variance: The variance $\sigma^2$ quantifies the spread of a distribution around its mean. For a discrete random variable $X$, it is calculated as: $\sigma^2 = E[X^2] - (E[X])^2$
• Combinations: Since items are drawn without replacement and the order of selection does not matter, we use combinations. The number of ways to choose $r$ items from $n$ is: $^n C_r = \frac{n!}{r!(n-r)!}$
• Hypergeometric Distribution: This distribution applies when sampling without replacement from a finite population containing two types of items (e.g., defective/non-defective). The problem fits this description. For a Hypergeometric distribution with population size $N$, number of successes $D$, and sample size $n$:
  • Expected Value: $E[X] = n \frac{D}{N}$
  • Variance: $Var(X) = n \frac{D}{N} \left(1 - \frac{D}{N}\right) \left(\frac{N-n}{N-1}\right)$ These formulas can be used for verification.

2. Step-by-Step Solution

Step 1: Identify Parameters and Possible Values of $X$ We are given:

• Total number of items in the lot ($N$): 10
• Number of defective items in the lot ($D$): 3
• Number of non-defective items in the lot ($N-D$): $10 - 3 = 7$
• Sample size ($n$): 5 items are drawn.
• Random Variable ($X$): Number of defective items in the sample.

To determine the possible values of $X$, we consider the constraints:

1. $X$ must be non-negative: $X \ge 0$.
2. $X$ cannot exceed the total number of defective items in the lot: $X \le D = 3$.
3. The number of non-defective items in the sample, $(n-X)$, cannot exceed the total non-defective items in the lot: $n-X \le N-D \implies 5-X \le 7 \implies X \ge -2$.
4. $X$ cannot exceed the sample size: $X \le n = 5$. Combining these, the possible integer values for $X$ are $max(0, -2) \le X \le min(3, 5)$, which simplifies to $0 \le X \le 3$. So, $X \in \{0, 1, 2, 3\}$.

Step 2: Calculate the Total Number of Ways to Draw a Sample The total number of ways to choose 5 items from 10 is given by: $\text{Total samples} = ^{10}C_5 = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

Step 3: Calculate Probabilities $P(X=x)$ for each value of $X$ For each possible value of $X=x$, the number of favorable ways to draw $x$ defective items and $(5-x)$ non-defective items is given by $^D C_x \times ^{N-D} C_{n-x} = ^3 C_x \times ^7 C_{5-x}$. Then, $P(X=x) = \frac{^3 C_x \times ^7 C_{5-x}}{^{10}C_5}$.

• For $X=0$ (0 defective items): $\text{Favorable ways} = ^3C_0 \times ^7C_5 = 1 \times 21 = 21$ $P(X=0) = \frac{21}{252} = \frac{1}{12}$
• For $X=1$ (1 defective item): $\text{Favorable ways} = ^3C_1 \times ^7C_4 = 3 \times 35 = 105$ $P(X=1) = \frac{105}{252} = \frac{5}{12}$
• For $X=2$ (2 defective items): $\text{Favorable ways} = ^3C_2 \times ^7C_3 = 3 \times 35 = 105$ $P(X=2) = \frac{105}{252} = \frac{5}{12}$
• For $X=3$ (3 defective items): $\text{Favorable ways} = ^3C_3 \times ^7C_2 = 1 \times 21 = 21$ $P(X=3) = \frac{21}{252} = \frac{1}{12}$ Verification: Sum of probabilities: $\frac{1}{12} + \frac{5}{12} + \frac{5}{12} + \frac{1}{12} = \frac{12}{12} = 1$. The probabilities are correct.

Step 4: Calculate the Expected Value $E[X]$ Using the formula $E[X] = \sum x_i P(X=x_i)$: $E[X] = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$ $E[X] = \left(0 \times \frac{1}{12}\right) + \left(1 \times \frac{5}{12}\right) + \left(2 \times \frac{5}{12}\right) + \left(3 \times \frac{1}{12}\right)$ $E[X] = 0 + \frac{5}{12} + \frac{10}{12} + \frac{3}{12} = \frac{18}{12} = \frac{3}{2} = 1.5$ Verification (Hypergeometric Mean Formula): $E[X] = n \frac{D}{N} = 5 \times \frac{3}{10} = \frac{15}{10} = 1.5$. This matches.

Step 5: Calculate $E[X^2]$ Using the formula $E[X^2] = \sum x_i^2 P(X=x_i)$: $E[X^2] = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2)) + (3^2 \times P(X=3))$ $E[X^2] = \left(0 \times \frac{1}{12}\right) + \left(1 \times \frac{5}{12}\right) + \left(4 \times \frac{5}{12}\right) + \left(9 \times \frac{1}{12}\right)$ $E[X^2] = 0 + \frac{5}{12} + \frac{20}{12} + \frac{9}{12} = \frac{34}{12} = \frac{17}{6}$

Step 6: Calculate the Variance $\sigma^2$ Using the formula $\sigma^2 = E[X^2] - (E[X])^2$: $\sigma^2 = \frac{17}{6} - \left(\frac{3}{2}\right)^2 = \frac{17}{6} - \frac{9}{4}$ To subtract, find a common denominator (12): $\sigma^2 = \frac{17 \times 2}{12} - \frac{9 \times 3}{12} = \frac{34}{12} - \frac{27}{12} = \frac{7}{12}$ Verification (Hypergeometric Variance Formula): $Var(X) = n \frac{D}{N} \left(1 - \frac{D}{N}\right) \left(\frac{N-n}{N-1}\right)$ $Var(X) = 5 \times \frac{3}{10} \times \left(1 - \frac{3}{10}\right) \times \left(\frac{10-5}{10-1}\right)$ $Var(X) = 5 \times \frac{3}{10} \times \frac{7}{10} \times \frac{5}{9} = \frac{5 \times 3 \times 7 \times 5}{10 \times 10 \times 9} = \frac{525}{900} = \frac{21}{36} = \frac{7}{12}$ This matches our calculation.

Step 7: Calculate $96 \sigma^2$ The problem asks for the value of $96 \sigma^2$. $96 \sigma^2 = 96 \times \frac{7}{12} = (8 \times 12) \times \frac{7}{12} = 8 \times 7 = 56$

3. Common Mistakes & Tips

• Incorrect Distribution: A common mistake is to assume a Binomial distribution instead of a Hypergeometric distribution. Remember, sampling without replacement from a finite population implies Hypergeometric.
• Calculation Errors: Be careful with arithmetic and simplification of fractions, especially when calculating combinations and summing terms for $E[X]$ and $E[X^2]$.
• Range of X: Ensure the possible values of $X$ are correctly identified based on the constraints of both the sample size and the number of available items in the population.
• Verification: Utilize the direct formulas for the mean and variance of a Hypergeometric distribution to quickly verify your step-by-step calculations, especially in a time-sensitive exam.

4. Summary

This problem required us to calculate the variance of a discrete random variable following a Hypergeometric distribution. We began by identifying the problem parameters and the possible values the random variable $X$ could take. Next, we meticulously calculated the probability of each possible value of $X$ using combinations. These probabilities were then used to compute the expected value $E[X]$ and $E[X^2]$. Finally, we applied the variance formula $\sigma^2 = E[X^2] - (E[X])^2$ to find $\sigma^2$ and then determined $96 \sigma^2$. The results were consistent with the direct formulas for the Hypergeometric distribution.

5. Final Answer

The final answer is $\boxed{56}$.