Question

Bag A contains 4 white balls and 3 black balls. While Bag B contains 3 white balls and 5 black balls. Two balls are drawn from Bag A and placed in Bag B. Then, what is the probability of drawing a white ball from Bag B?

Answer 1

Case I : Let Both white balls are transfered from bag I to bag II and then a white ball is drawn from bag II.

Required probability = ${\displaystyle \frac{{}^4{C}_2}{{}^7{C}_2}}$ × ${\displaystyle \frac{{}^5{C}_1}{{}^{10}{C}_1}}$

${\displaystyle =\frac{4\times 3}{7\times 6}\times \frac{5}{10}}$

${\displaystyle =\frac{2}{7}\times \frac{1}{2}=\frac{1}{7}}$

Case II: Let both black balls are transfered from bag I to bag II and then a white ball is drawn from bag II. 

Required probability = ${\displaystyle \frac{{}^3{C}_2}{{}^7{C}_2}}$ × ${\displaystyle \frac{{}^3{C}_1}{{}^{10}{C}_1}}$

${\displaystyle =\frac{3\times 2}{7\times 6}\times \frac{3}{10}}$

${\displaystyle =\frac{1}{7}\times \frac{3}{10}=\frac{3}{70}}$

Case III: 1 white and 1 black ball is transfered and then a white ball is drawn from bag II.

Required probability = ${\displaystyle \frac{{}^4{C}_1\times {}^3{C}_1}{{}^7{C}_2}\times \frac{{}^4{C}_1}{{}^{10}{C}_1}}$

${\displaystyle =\frac{4\times 3}{7\times 3}\times \frac{4}{10}}$

${\displaystyle =\frac{16}{70}}$
Total probability = Case I + Case II + Case III
                           ${\displaystyle =\frac{1}{7}+\frac{3}{70}+\frac{16}{70}}$

                            ${\displaystyle =\frac{10}{70}+\frac{19}{70}}$

                             ${\displaystyle =\frac{29}{70}}$