A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is
Show Hint
- $\frac{11}{15}$
- $\frac{28}{75}$
- $\frac{2}{15}$
- $\frac{3}{5}$
The Correct Option is B
Approach Solution - 1
2. Calculate the expected value $E(X)$: \[ E(X) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{3}{5} \]
3. Calculate the variance $Var(X)$: \[ Var(X) = \left(0 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(1 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(2 - \frac{3}{5}\right)^2 \cdot \frac{1}{15} \] \[ = \frac{9}{25} \cdot \frac{7}{15} + \frac{4}{25} \cdot \frac{7}{15} + \frac{1}{25} \cdot \frac{1}{15} \] \[ = \frac{63}{375} + \frac{28}{375} + \frac{1}{375} = \frac{92}{375} = \frac{28}{75} \] Therefore, the correct answer is (2) $\frac{28}{75}$.
Approach Solution -2
We have 10 pens, out of which 3 are defective. A sample of 2 pens is drawn at random without replacement. Let \(X\) denote the number of defective pens in the sample. We need to find the variance of \(X\).
Concept Used:
When sampling without replacement from a finite population, \(X\) (the number of defective items) follows the hypergeometric distribution:
\[ P(X = x) = \frac{\binom{3}{x}\binom{7}{2-x}}{\binom{10}{2}}, \quad x=0,1,2. \] \[ E(X) = n\frac{K}{N}, \qquad Var(X) = n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}. \] where \(N=10,\ K=3,\ n=2.\)
Step-by-Step Solution:
Step 1: Compute the mean \(E(X)\):
\[ E(X) = n\frac{K}{N} = 2 \times \frac{3}{10} = \frac{3}{5}. \]
Step 2: Compute the variance \(Var(X)\):
\[ Var(X) = n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}. \] Substitute the values: \[ Var(X) = 2 \times \frac{3}{10} \times \frac{7}{10} \times \frac{8}{9}. \]
Step 3: Simplify:
\[ Var(X) = \frac{2 \times 3 \times 7 \times 8}{10 \times 10 \times 9} = \frac{336}{900} = \frac{28}{75}. \]
Final Computation & Result:
\[ \boxed{Var(X) = \frac{28}{75}}. \]
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