Question

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

• A. $\frac{11}{15}$
• B. $\frac{28}{75}$
• C. $\frac{2}{15}$
• D. $\frac{3}{5}$

Hint:

Calculate the probability distribution, expected value, and variance to find the variance of a random variable.

Answer 1

The Correct Option is B

1. Calculate the probability distribution of $X$: - $P(X = 0) = \frac{^7C_2}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}$ - $P(X = 1) = \frac{^7C_1 \cdot ^3C_1}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}$ - $P(X = 2) = \frac{^3C_2}{^{10}C_2} = \frac{3}{45} = \frac{1}{15}$
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) $\frac{28}{75}$.