Question

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball is drawn is black.

Answer 1

Let Bi = ith ball drawn is black. 
Wi = ith ball drawn is white; i =1,2 and A = Third ball drawn is black
 We observe that the black ball can be drawn in the third draw in one of the following mutually exclusive ways. 
(1) Both first and second balls drawn are white and third ball drawn is black,
ie, (W1 ∩ W2) ∩ A. 
(2) Both first and second balls are black and third ball drawn is black ie, (B1 ∩ B2) ∩ A.
(3) The first ball drawn is white, the second ball drawn is black and the third ball drawn is black ie, (W1 ∩ B2) ∩ A.
(4) The first ball drawn is black, the second ball drawn is white and the third ball drawn is black ie, (B1∩ W2) ∩ A. 

`therefore P(A) = P[{(W_1∩W_2)∩ A}∪{(B_1∩ B_2)..A}`
                  `∪{(W1∩ B_2)∩ A}∪{(B_1∩ W_2)∩A}]` 
`= P{(W_1∩ W_2)∩ A} + P{(B_1∩ B_2)∩ A}`
        `+ P{(W_1∩ B_2)∩ A} + P {(B_1∩ W_2)∩ A}`
`= P(W_1∩ W_2). P(A//(W_1∩ W_2)) + P(B_1∩ B_2)`

`P(A//(B_1∩ B_2)) + P(W_1∩ B_2).P(A//(W_1∩ B_2))`
                              `+P(B_1∩ W_2).P(A//(B_1∩ W_2))`

`= (2/4 xx 1/3) xx 1+(2/4xx3/5)xx4/6`

                          `+(2/4xx2/3)xx3/4+(2/4xx2/5)xx3/4`

`= 1/6+1/5+1/4+3/20 = 23/30`