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Question
A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.
Solution
\[ \text{ Given } :\]
\[ \text{ Bag } 1=\left( 3R+5B \right)\text{ balls} \]
\[\text{ Bag } 2=\left( 6R+4B \right)\text{ balls } \]
\[P\left( \text{ one is red and one is black } \right) = P\left( \text{ red from bag 1 and black from bag 2 } \right) + P\left( \text{ red from bag 2 and black from bag 1 } \right)\]
\[ = \frac{3}{8} \times \frac{4}{10} + \frac{5}{8} \times \frac{6}{10}\]
\[ = \frac{12}{80} + \frac{30}{80}\]
\[ = \frac{42}{80}\]
\[ = \frac{21}{40}\]
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