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Question
There are three urns A, B, and C. Urn A contains 4 red balls and 3 black balls. urn B contains 5 red balls and 4 black balls. Urn C contains 4 red and 4 black balls. One ball is drawn from each of these urns. What is the probability that 3 balls drawn consists of 2 red balls and a black ball?
Solution
\[\text{ Given} :\]
\[\text{ Urn } A\left( 4R + 3B \right)\]
\[\text{ Urn } B\left( 5R + 4B \right)\]
\[\text{ Urn } C\left( 4R + 4B \right)\]
P (two red and one black ) = P ( Red from urn A ) × ( Black from urn B ) × P ( Red from urn C ) + P ( Red from urn A ) × ( Red from urn B ) × P ( Black from urn C ) + P ( Black from urn A ) × ( Red from urn B ) × P ( Red from urn C )
\[ = \frac{3}{7} \times \frac{5}{9} \times \frac{4}{8} + \frac{4}{7} \times \frac{4}{9} \times \frac{4}{8} + \frac{4}{7} \times \frac{5}{9} \times \frac{4}{8}\]
\[ = \frac{5}{42} \times \frac{16}{126} \times \frac{20}{126}\]
\[ = \frac{15 + 16 + 20}{126}\]
\[ = \frac{51}{126} = \frac{17}{42}\]
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