Question

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If the variance of $X$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $n - m$ is equal to ________.

Answer 1

Correct Answer: 71

The objective is to determine the variance of a random variable \(X\), representing the count of defective items in a sample of 5 drawn without replacement from a lot of 12 items containing 3 defectives. Following the calculation of the variance as a simplified fraction \(\frac{m}{n}\), the value of \(n - m\) is required.

Methodology:

The random variable \(X\) adheres to a hypergeometric distribution. This distribution models the probability of obtaining \(k\) successes (defective items) in a sample of size \(n\) selected without replacement from a population of size \(N\) with \(K\) successes.

The defining parameters are:

• \(N\): Total population size.
• \(K\): Total count of items with the specified characteristic (successes) within the population.
• \(n\): Sample size.

The variance for a hypergeometric distribution is computed using the formula:

\[ \text{Var}(X) = n \left(\frac{K}{N}\right) \left(1 - \frac{K}{N}\right) \left(\frac{N-n}{N-1}\right) \]

The factor \( \left(\frac{N-n}{N-1}\right) \) is recognized as the finite population correction factor.

Procedural Breakdown:

Step 1: Extract the parameters of the hypergeometric distribution from the problem statement.

• Total items in the lot: \(N = 12\).
• Number of defective items: \(K = 3\).
• Sample size: \(n = 5\).

Step 2: Employ the variance formula for the hypergeometric distribution.

The variance formula is:

\[ \text{Var}(X) = n \left(\frac{K}{N}\right) \left(\frac{N-K}{N}\right) \left(\frac{N-n}{N-1}\right) \]

Step 3: Substitute the identified parameter values into the formula.

\[ \text{Var}(X) = 5 \left(\frac{3}{12}\right) \left(\frac{12-3}{12}\right) \left(\frac{12-5}{12-1}\right) \]

Simplify each fractional component:

\[ \frac{3}{12} = \frac{1}{4} \] \[ \frac{12-3}{12} = \frac{9}{12} = \frac{3}{4} \] \[ \frac{12-5}{12-1} = \frac{7}{11} \]

Integrate these simplified fractions back into the variance computation:

\[ \text{Var}(X) = 5 \times \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{7}{11}\right) \]

Step 4: Compute the final variance value.

\[ \text{Var}(X) = \frac{5 \times 1 \times 3 \times 7}{4 \times 4 \times 11} = \frac{105}{176} \]

Step 5: Identify \(m\) and \(n\) and compute \(n-m\).

Given the variance as \(\frac{m}{n} = \frac{105}{176}\), verify if the fraction is in its simplest form, i.e., \(\gcd(m, n) = 1\).

Prime factorization of the numerator: \(m = 105 = 3 \times 5 \times 7\).

Prime factorization of the denominator: \(n = 176 = 16 \times 11 = 2^4 \times 11\).

As there are no common prime factors between the numerator and denominator, \(\gcd(105, 176) = 1\). Therefore, \(m = 105\) and \(n = 176\).

The problem requires the calculation of \(n - m\).

\[ n - m = 176 - 105 \] \[ n - m = 71 \]

The resultant value of \(n - m\) is 71.