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A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.
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Solution
Bag X = 4 white, 2 black.
Bag Y = 3 white, 3 black
Let A be the event of selecting one white and one black ball.
E1 first bag selected
E2 second bag selected
`P(E_1)=1/2 P(E_2)=1/2`
`P(A/E_1)=4/6xx2/5+2/6xx4/5=16/30`
`P(AE_2)=3/6xx3/5+3/6xx3/6xx3/5=18/30`
`P(E_2"/"A)=(P(E_2)P(A"/"E_2))/(P(E_1)P(A"/"E_1)+P(E_2)P(A"/"E_2)`
`P(E_2"/"A)= (1/2xx18/30)/(1/2xx16/30+1/2xx18/30)`
`=18/(16+18)`
=`18/34`
=`9/17`
Concept: Conditional Probability
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