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Question
A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.
Solution
\[P\left( \text{ atleast 2 balls are green } \right) = 1 - P\left( \text{ at most one ball is green } \right)\]
\[ = 1 - \left[ P\left( \text{ first green } \right) + P\left( \text{ second green } \right) + P\left( \text{ third green } \right) + P\left( \text{ no green } \right) \right]\]
\[ = 1 - \left[ \frac{6}{14} \times \frac{8}{13} \times \frac{7}{12} + \frac{8}{14} \times \frac{6}{13} \times \frac{7}{12} + \frac{8}{14} \times \frac{7}{13} \times \frac{6}{12} + \frac{8}{14} \times \frac{7}{13} \times \frac{6}{12} \right]\]
\[ = 1 - \left[ \frac{336}{2184} + \frac{336}{2184} + \frac{336}{2184} + \frac{336}{2184} \right]\]
\[ = 1 - \frac{1344}{2184}\]
\[ = \frac{840}{2184}\]
\[ = \frac{5}{13}\]
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