Question

There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.

 

Answer 1

\[\text{ It is given that bag A contains 3 red and 5 black balls } \left( 3R, 5B \right) \text{ and bag B contains 2 red and 3 black balls } \left( 2R, 3B \right).\]
\[\text{ Now } , \]
\[P\left( \text{ one red and 2 black }  \right) = P\left(\text{  one red from bag A and two black from bag B }\right) + P\left( \text{ black ball from bag A and remaining balls from bag B } \right)\]
\[ = \frac{3}{8} \times \frac{3}{5} \times \frac{2}{4} + \frac{5}{8} \times \frac{2}{5} \times \frac{3}{4} \times 2\]
\[ = \frac{9}{80} + \frac{30}{80}\]
\[ = \frac{39}{80}\]
\[ \text{ Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B } .\]
\[\text{ While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black } .\]