Question

A box contains \[15\] green balls 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. $ \dfrac{{12}}{5} $
B. $ 6 $
C. $ 4 $
D. $ \dfrac{6}{{25}} $

Answer 1

Hint: First calculate the probability of drawing a green ball out of the box and then the probability of drawing a yellow ball out of the box. Then use the formula of the variance of the number of green balls drawn.

Complete step-by-step answer:
The total number of balls in the box are $ 25 $ . The probability is always the ratio of the number of possible outcomes to the number of total outcomes.
The number of possible outcomes for drawing a green ball are $ 15 $ . So, the probability of drawing a green ball out of box is $ \dfrac{{15}}{{25}} = \dfrac{3}{5} $ .
The number of possible outcomes for drawing a yellow ball are $ 10 $ . So, the probability of drawing a yellow ball out of box is $ \dfrac{{10}}{{25}} = \dfrac{2}{5} $ .
It is given that the balls are drawn with replacement. So, there is no change in drawing any type of ball.
So, the probability of first outcome is $ p = \dfrac{3}{5} $ and the probability of second outcome is $ q = \dfrac{2}{5} $ .
As given only $ 10 $ balls are drawn from the box. So, the value of $ n $ is equal to $ 10 $ .
The variance of number of green balls drawn is:
 $
  npq = 10 \times \dfrac{3}{5} \times \dfrac{2}{5} \\
   = \dfrac{{12}}{5} \\
  $
So, the variance of the number of green balls drawn is equal to $ \dfrac{{12}}{5} $ .
So, the correct answer is “Option ”.A

Note: The formula for the probability is the number of favorable outcomes to the number of total outcomes. As given in the question the balls are drawn with replacement. So, the probability of drawing any type of ball remains the same all the time.