A bag contains 6 red balls, 4 green balls and 10 blue balls. Three balls are drawn with replacement. The probability of getting at least 1 green ball is:

$\frac{61}{125}$

The correct answer is Option (2) → $\frac{61}{125}$

Total balls = 6 + 4 + 10 = 20

Probability of green ball = $P(G) = \frac{4}{20} = \frac{1}{5}$

Probability of not green = $P(\text{not G}) = 1 - \frac{1}{5} = \frac{4}{5}$

Three balls drawn with replacement → independent draws

Probability of no green balls in 3 draws = $(\frac{4}{5})^3 = \frac{64}{125}$

Probability of at least 1 green ball = $1 - \frac{64}{125} = \frac{61}{125}$