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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?
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Solution

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 = `(""^4C_2) / (""^7C_2)` × `(""^5C_1) / (""^10C_1)`
`= (4xx3)/(7xx6)xx5/10`
`=2/7xx1/2=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 = `(""^3C_2) / (""^7C_2)` × `(""^3C_1) / (""^10C_1)`
`= (3xx2)/(7xx6) xx 3/10`
`= 1/7xx3/10=3/70`
Case III: 1 white and 1 black ball is transfered and then a white ball is drawn from bag II.
Required probability = `(""^4C_1xx""^3C_1)/(""^7C_2) xx (""^4C_1)/(""^10C_1)`
`=(4xx3)/(7xx3)xx4/10`
`= 16/70`
Total probability = Case I + Case II + Case III
`= 1/7 + 3/70 + 16/70`
`= 10/70+ 19/70`
`= 29/70`
