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. $\frac{6}{25}$
• C. $6$
• D. 4

Answer 1

The correct option is A

$\frac{12}{5}$

Explanation for the correct option:

Step 1: Find out the probability of drawing a green ball

Given data

Green Ball $=15$

Yellow Ball $=10$

Total Ball $=10+15$

$=25$

let $P\left(G\right)=$Probability of drawing a green ball

Probability $=\frac{{}^{15}{C}_1}{{}^{25}{C}_1}$

$=\frac{15}{25}$ $\left[\mathbf{\because }{\mathbf{}}^{\mathbf{n}}{\mathit{C}}_{\mathbf{1}\mathbf{}}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{*}\left(n-1\right)\mathbf{!}}{\left(n-1\right)\mathbf{!}\mathbf{}\mathbf{*}\mathbf{1}\mathbf{!}}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\right]$

$=\frac{3}{5}$

As we know,

$P\left(G\right)+P\left(G\right)\text{'}$$=1$

$\frac{3}{5}+P\left(G\right)\text{'}=1$

$P\left(G\right)\text{'}$$=\frac{2}{5}$

Step 2: Find Out the variance

Variance (σ² ) $=10*\frac{3}{5}*\frac{2}{5}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{v}\mathit{a}\mathit{r}\mathit{i}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathbf{=}\mathit{n}\mathit{p}\mathit{q}$

Variance (σ²) $=\frac{12}{5}$

Hence, Option $\left(A\right)$ is the correct answer