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Question
A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Solution
\[\text{ Given: Bag } = \left( 7W + 5B + 4R \right) \text{ balls } \]
\[P\left( \text{ atleast 3 balls are black } \right) = P\left( \text{ exactly 3 black } \right) + P\left( \text{ all 4 black } \right)\]
\[ = \left( \frac{11}{16} \times \frac{5}{15} \times \frac{4}{14} \times \frac{3}{13} \times 4 \right) + \left( \frac{5}{16} \times \frac{4}{15} \times \frac{3}{14} \times \frac{2}{13} \right)\]
\[ = \frac{11}{14 \times 13} + \frac{1}{2 \times 14 \times 13}\]
\[ = \frac{22 + 1}{364}\]
\[ = \frac{23}{364}\]
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