Question

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If \( x \) denotes the number of defective oranges, then the variance of \( x \) is:

• A. \(\frac{26}{75}\)
• B. \(\frac{14}{25}\)
• C. \(\frac{28}{75}\)
• D. \(\frac{18}{25}\)

Hint:

Understanding the concept of hypergeometric distribution is crucial for solving problems involving random drawing without replacement.

Answer 1

The Correct Option is C

Step 1: Define the random variable \( x \). \( x \) denotes the quantity of defective oranges selected.

Step 2: Ascertain probabilities for \( x \). \[ P(x=0) = \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15} \] \[ P(x=1) = \frac{\binom{3}{1} \binom{7}{1}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15} \] \[ P(x=2) = \frac{\binom{3}{2} \binom{7}{0}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15} \] 

Step 3: Compute the expected value \( E(x) \). \[ E(x) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5} \] 

Step 4: Compute the expected value \( E(x^2) \). \[ E(x^2) = 0^2 \cdot \frac{7}{15} + 1^2 \cdot \frac{7}{15} + 2^2 \cdot \frac{1}{15} = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15} \] 

Step 5: Calculate the variance \({Var}(x)\). \[ {Var}(x) = E(x^2) - [E(x)]^2 = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{55}{75} - \frac{27}{75} = \frac{28}{75} \]