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

Answer 1

The correct option is A $\frac{12}{5}$

We can see that the Probability of getting a green ball is =

$\frac{\text{Favourable Events}}{\text{Total Number of balls}}=\frac{15}{25}$

Similarly, probability of getting a Yellow ball = $\frac{10}{25}$

Let $p=\frac{15}{25}$ and $q=\frac{10}{25}$

As it follows binomial distribution, $n=10$

$Variance=npq=10\times \frac{15}{25}\times \frac{10}{25}$

$=\frac{12}{5}$