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Question
A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.
Solution
Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw
\[\text{ Now } , \]
\[P\left( A \right) = \frac{4}{16} = \frac{1}{4}\]
\[P\left( B/A \right) = \frac{7}{15}\]
\[P\left( C/A \cap B \right) = \frac{5}{14}\]
\[ \therefore \text{ Required probability } = P\left( A \cap B \cap C \right) = P\left( A \right) \times P\left( B/A \right) \times P\left( C/A \cap B \right)\]
\[ = \frac{1}{4} \times \frac{7}{15} \times \frac{5}{14}\]
\[ = \frac{1}{24}\]
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