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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{61}{125}\)
The bag has 6 + 4 + 10 = 20 balls. Since the balls are drawn with replacement, the three draws are independent and each has
\( \displaystyle P(\text{green}) = \frac{4}{20} = \frac{1}{5},\quad P(\text{not green}) = \frac{4}{5} \)
“At least 1 green” is the complement of “no green in all three draws”:
\( \displaystyle P(\text{no green}) = \left(\frac{4}{5}\right)^3 = \frac{64}{125} \)
\( \displaystyle P(\text{at least 1 green}) = 1 - \frac{64}{125} = \frac{61}{125} \)