Question

A bag contains 20 red balls, 12 blue balls and 8 green balls. Three balls are drawn one by one with replacement, the probability that the third ball being green is:

पर्याय

• \[\frac{1}{2}\]

• \[\frac{2}{5}\]

• \[\frac{1}{5}\]

• 1.8

Answer 1

\[\boldsymbol{\frac{1}{5}}\]

Explanation: 

Total no. of balls = 20 + 12 + 8 = 40

No. of green balls = 8

Since, balls are drawn one by one with replacement.

∴ P(third ball drawn be green) = P(ball drawn to be green) \[{} = \frac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \frac{8}{40} = \frac{1}{5}.\]