CUET · MATHS · PYQ PAPER 2025
A bag contains 6 red balls, 4 green balls and 10 blue balls. Three balls are drawn with replacement. Find the probability of getting at least one green ball
- A \(\frac{1}{4}\)
- B \(\frac{61}{125}\)
- C \(\frac{1}{2}\)
- D \(\frac{29}{57}\)
Answer & Solution
Correct Answer
(B) \(\frac{61}{125}\)
Step-by-step Solution
Detailed explanation
\(P(\text{no green}) = \frac{6+10}{6+4+10} = \frac{16}{20} = \frac{4}{5}\) \(P(\text{no green in 3 draws}) = \left(\frac{4}{5}\right)^3 = \frac{64}{125}\) \(P(\text{at least one green}) = 1 - P(\text{no green in 3 draws}) = 1 - \frac{64}{125} = \frac{61}{125}\)
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