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) **

= \(\frac{6}{15}\times\frac{7}{11}+\frac{1}{15}\times \frac{5}{11} +\frac{8}{15} \times \frac{6}{11}\)

**\(=\frac{42}{165}+\frac{5}{165}+\frac{48}{165}= \frac{95}{165} =\frac{19}{33}\)**