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 problem asks for the variance of a random variable \(X\), which represents the number of defective items in a random sample of 5 items drawn without replacement from a lot of 12 items containing 3 defectives. After finding the variance in the form of a simplified fraction \(\frac{m}{n}\), we need to compute the value of \(n - m\).

Concept Used:

The random variable \(X\) follows a hypergeometric distribution. This distribution describes the probability of \(k\) successes (defective items) in a sample of size \(n\) drawn without replacement from a population of size \(N\) containing \(K\) successes.

The key parameters are:

• \(N\): Total population size.
• \(K\): Total number of items with the desired feature (successes) in the population.
• \(n\): Sample size.

The variance of a random variable \(X\) following a hypergeometric distribution is given by 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 term \( \left(\frac{N-n}{N-1}\right) \) is the finite population correction factor.

Step-by-Step Solution:

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

• The total number of items in the lot is \(N = 12\).
• The number of defective items in the lot is \(K = 3\).
• The sample size is \(n = 5\).

Step 2: Apply the formula for the variance of the hypergeometric distribution.

The formula for the variance 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.

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

Now, we simplify each term:

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

Substitute these simplified fractions back into the variance formula:

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

Step 4: Calculate the final value of the variance.

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

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

The variance is given as \(\frac{m}{n} = \frac{105}{176}\). We need to check if this fraction is in its simplest form, i.e., \(\gcd(m, n) = 1\).

The prime factorization of the numerator is \(m = 105 = 3 \times 5 \times 7\).

The prime factorization of the denominator is \(n = 176 = 16 \times 11 = 2^4 \times 11\).

Since the numerator and denominator share no common prime factors, \(\gcd(105, 176) = 1\). Thus, we have \(m = 105\) and \(n = 176\).

The problem asks for the value of \(n - m\).

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

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

Answer 2

\[a = 1 - \frac{\binom{3}{5}}{\binom{12}{5}}\]
\[b = 3 \cdot \frac{\binom{9}{4}}{\binom{12}{5}}\]
\[c = 3 \cdot \frac{\binom{9}{3}}{\binom{12}{5}}\]
\[d = 1 \cdot \frac{\binom{9}{2}}{\binom{12}{5}}\]
\[u = 0 \cdot a + 1 \cdot b + 2 \cdot c + 3 \cdot d = 1.25\]
\[\sigma^2 = 0 \cdot a + 1 \cdot b + 4 \cdot c + 9 \cdot d - u^2\]
\[\sigma^2 = \frac{105}{176}\]
\[\text{Ans.} \quad 176 - 105 = 71\]