A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of number of yellow balls drawn is :

Options

A. 4
B. `(6)/(25)`
C. `(12)/(5)`
D. 6

Answer

To find the variance of the number of yellow balls drawn from a box containing 15 green and 10 yellow balls when 10 balls are drawn with replacement, we can use the properties of the binomial distribution.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls:**
- The box contains 15 green balls and 10 yellow balls.
- Total number of balls = 15 + 10 = 25.

2. **Define the Number of Trials (N):**
- We are drawing 10 balls, so the number of trials \( N = 10 \).

3. **Determine the Probability of Success (P):**
- The probability of drawing a yellow ball (success) is:
\[ P = \frac{\text{Number of yellow balls}}{\text{Total number of balls}} = \frac{10}{25} = \frac{2}{5}. \]

4. **Calculate the Probability of Failure (Q):**
- The probability of drawing a green ball (failure) is:
\[ Q = 1 - P = 1 - \frac{2}{5} = \frac{3}{5}. \]

5. **Use the Variance Formula for Binomial Distribution:**
- The variance \( \sigma^2 \) for a binomial distribution is given by:
\[ \sigma^2 = N \cdot P \cdot Q. \]
- Substituting the values we have:
\[ \sigma^2 = 10 \cdot \frac{2}{5} \cdot \frac{3}{5}. \]

6. **Perform the Calculation:**
- First, calculate \( P \cdot Q \):
\[ P \cdot Q = \frac{2}{5} \cdot \frac{3}{5} = \frac{6}{25}. \]
- Now, substitute this back into the variance formula:
\[ \sigma^2 = 10 \cdot \frac{6}{25} = \frac{60}{25} = \frac{12}{5}. \]

7. **Final Result:**
- The variance of the number of yellow balls drawn is:
\[ \sigma^2 = \frac{12}{5}. \]