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Question
A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
(ii) both are black
(iii) one is white and one is black
Solution
\[\text{ Given } :\]
\[\text{ Bag } 1=\left( 4W+2B \right)\text{ balls} \]
\[\text{ Bag }2=\left( 3W+5B \right)\text{ balls } \]
\[\left( i \right) P\left( \text{ both are white } \right) = \frac{4}{6} \times \frac{3}{8}\]
\[ = \frac{1}{4}\]
\[\left( ii \right) P\left( \text{ both are black } \right) = \frac{2}{6} \times \frac{5}{8}\]
\[ = \frac{5}{24}\]
\[\left( iii \right) P\left( \text{ one is white and one is black } \right) = P\left( \text{ white from bag 1 and black from bag } 2 \right) + P\left( \text{ white from bag 2 and black from bag } 1 \right)\]
\[ = \frac{4}{6} \times \frac{5}{8} + \frac{3}{8} \times \frac{2}{6}\]
\[ = \frac{20}{48} + \frac{6}{48}\]
\[ = \frac{26}{48}\]
\[ = \frac{13}{24}\]
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