Question

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

• A. 6
• B. 4
• C. $\frac{6}{25}$
• D. $\frac{12}{5}$

Answer 1

The Correct Option is D

To find the variance of the number of green balls drawn from the box, we first need to analyze the situation described in the problem.

We have a total of \(15\) green balls and \(10\) yellow balls. The total number of balls is \(25\).

Since the drawing of balls is done with replacement, this is a case of a binomial distribution where each draw is independent, and the probability of drawing a green ball remains constant.

The probability \(p\) of drawing a green ball in a single trial is given by:

\(p = \frac{15}{25} = \frac{3}{5}\)

Thus, the probability \(q\) of drawing a yellow ball is:

\(q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5}\)

We are drawing a total of \(10\) balls. Therefore, this situation can be described by a binomial distribution with parameters \(n = 10\) (number of trials) and \(p = \frac{3}{5}\).

The variance of a binomial distribution is given by the formula:

\(\text{Variance} = n \cdot p \cdot q\)

Substituting the known values:

\(\text{Variance} = 10 \cdot \frac{3}{5} \cdot \frac{2}{5}\)

\(\text{Variance} = 10 \cdot \frac{6}{25}\)

\(\text{Variance} = \frac{60}{25} = \frac{12}{5}\)

Therefore, the variance of the number of green balls drawn is \(\frac{12}{5}\). This confirms that the correct option is:

\(\frac{12}{5}\)