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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:
Options
\[\frac{1}{2}\]
\[\frac{2}{5}\]
\[\frac{1}{5}\]
1.8
MCQ
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Solution
\[\boldsymbol{\frac{1}{5}}\]
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) \[{} = \frac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \frac{8}{40} = \frac{1}{5}.\]
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