1. Key Concepts and Formulas

• Hypergeometric Distribution: This distribution models the number of successes (defective items) in a sample drawn without replacement from a finite population. It's characterized by three parameters:
  • $N$: Total population size.
  • $K$: Total number of successes in the population.
  • $n$: Sample size.
• Variance of a Hypergeometric Random Variable ($X$): The formula for the variance is crucial for this problem. If $X \sim \text{Hypergeometric}(N, K, n)$, then its variance is given by: $Var(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ The term $\frac{N-n}{N-1}$ is known as the finite population correction factor (FPCF), which accounts for sampling without replacement.

2. Step-by-Step Solution

Step 1: Identify the parameters of the Hypergeometric Distribution. The problem provides all necessary values to define the Hypergeometric distribution:

• Total number of items in the lot ($N$): 12
• Number of defective items in the lot ($K$): 3 (These are considered 'successes' for our random variable $X$.)
• Number of non-defective items in the lot ($N-K$): $12 - 3 = 9$
• Size of the sample drawn ($n$): 5

We also need the values for the finite population correction factor:

• $N-n$: $12 - 5 = 7$
• $N-1$: $12 - 1 = 11$

Step 2: Apply the variance formula for the Hypergeometric Distribution. Substitute the identified parameters into the variance formula: $Var(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}$ $Var(X) = 5 \times \frac{3}{12} \times \frac{9}{12} \times \frac{7}{11}$

Step 3: Simplify the terms and calculate the variance. Before multiplying, simplify the fractions:

• $\frac{3}{12} = \frac{1}{4}$
• $\frac{9}{12} = \frac{3}{4}$

Now substitute these simplified fractions back into the expression for $Var(X)$: $Var(X) = 5 \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{11}$ Multiply the numerators and the denominators: $Var(X) = \frac{5 \times 1 \times 3 \times 7}{4 \times 4 \times 11}$ $Var(X) = \frac{105}{176}$

Step 4: Determine $m$ and $n$ and ensure they are in simplest form. The problem states that the variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$. From our calculation, $Var(X) = \frac{105}{176}$. To check if this fraction is in its simplest form, we find the prime factorization of the numerator and the denominator:

• Prime factorization of $105 = 3 \times 5 \times 7$
• Prime factorization of $176 = 2^4 \times 11$ Since there are no common prime factors, the fraction $\frac{105}{176}$ is already in its simplest form. Therefore, we have $m = 105$ and $n = 176$.

Step 5: Calculate $n-m$. Finally, we compute the required value $n-m$: $n-m = 176 - 105 = 71$

3. Common Mistakes & Tips

• Binomial vs. Hypergeometric: A frequent error is to use the Binomial distribution formula ($np(1-p)$) for variance. Remember, Binomial applies to sampling with replacement or from an infinite population, while Hypergeometric is for sampling without replacement from a finite population. The problem explicitly states "without replacement," making Hypergeometric the correct choice.
• Forgetting the FPCF: The finite population correction factor ($\frac{N-n}{N-1}$) is crucial for Hypergeometric variance. Omitting it would lead to an incorrect result (in this case, $\frac{15}{16}$ instead of $\frac{105}{176}$).
• Simplification: Always simplify the fraction $\frac{m}{n}$ to its lowest terms before identifying $m$ and $n$ to satisfy the $\operatorname{gcd}(m, n)=1$ condition. This ensures you have the correct $m$ and $n$ values for the final calculation.

4. Summary

This problem required us to calculate the variance of a random variable representing the number of defective items in a sample drawn without replacement. This scenario perfectly fits the Hypergeometric distribution. By correctly identifying the parameters ($N=12, K=3, n=5$) and applying the standard variance formula for a Hypergeometric distribution, we calculated $Var(X) = \frac{105}{176}$. Since this fraction is in its simplest form, we found $m=105$ and $n=176$. The final step was to compute $n-m$, which resulted in $71$.

5. Final Answer The final answer is $\boxed{71}$.