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Question
A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.
Solution
Consider the given events.
A = A white or black ball in the first draw
B = A white or black ball in the second draw
C = A white or black ball in the third draw
\[\text{ Now } , \]
\[P\left( A \right) = \frac{8}{15}\]
\[P\left( B/A \right) = \frac{7}{14} = \frac{1}{2}\]
\[P\left( C/A \cap B \right) = \frac{6}{13}\]
\[ \therefore \text{ Required probability } = P\left( A \cap B \cap C \right)\]
\[ = P\left( A \right) \times P\left( B/A \right) \times P\left( C/A \cap B \right)\]
\[ = \frac{8}{15} \times \frac{1}{2} \times \frac{6}{13}\]
\[ = \frac{8}{65}\]
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