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Question
An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?
Solution
Consider the given events.
A = A white or red ball in the first draw
B = A white or red ball in the second draw
\[\text{ Now } , \]
\[P\left( A \right) = \frac{7}{12}\]
\[P\left( B/A \right) = \frac{6}{11}\]
\[ \therefore P\left( A \cap B \right) = P\left( A \right) \times P\left( B/A \right)\]
\[ = \frac{7}{12} \times \frac{6}{11}\]
\[ = \frac{7}{22}\]
\[ \therefore \text{ Required probability } = 1 - P\left( A \cap B \right)\]
\[ = 1 - \frac{7}{22}\]
\[ = \frac{15}{22}\]
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