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

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