There are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from the first bag to the second bag and then a white ball be drawn from the second bag.
(i) Two white balls are transferred from the first bag to the second bag
(ii) Two black balls are transferred from the first bag to the second bag
(iii) One white and one black ball are transferred from the first bag to the second bag
Let the events be described as under
A: 2 white balls drawn from 1st bag, transferred to 2nd bag
B: 2 black balls drawn from 1st bag, transferred to 2nd bag
C: 1 white and 1 black ball drawn from 1st bag, transferred to 2nd bag
D: 1 white ball drawn from the second bag.
∴ P(A) = \(\frac{^4C_2}{^6C_2} = \frac{4\times 3}{6 \times 5} = \frac{6}{15}\)
P(B) = \(\frac{^2C_2}{^6C_2} =\frac{1}{15}\)
P(C) = \(\frac{^4C_1 \times ^2C_1}{^6C_2} = \frac{4\times 2}{\frac{6\times 5}{2}} = \frac{8}{15}\)
Now when A has occurred, we have 7 white and 4 black balls in 2nd bag.
∴ P(Getting a white ball from 2nd bag) = \(P(D/A) = \frac{7}{11}\)
Similarly when B has occurred, we have 5 white and 6 black balls in 2nd bag
∴ P(Getting a white ball from 2nd bag) = \(P(D/B)=\frac{5}{11}\)
when C has occurred, we have 6 white and 5 black balls in 2nd bag,
∴ P(Getting a white ball from 2nd bag) = \(P(D/C) =\frac{6}{11}\)
∴ By law of total probability, P(Ball drawn from 2nd bag is white)
P(D) = P(A) × P(D/A) + P(B) × P(D/B) + P(C) × P(D/C)
= \(\frac{6}{15}\times\frac{7}{11}+\frac{1}{15}\times \frac{5}{11} +\frac{8}{15} \times \frac{6}{11}\)
\(=\frac{42}{165}+\frac{5}{165}+\frac{48}{165}= \frac{95}{165} =\frac{19}{33}\)
