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

Explanation

For the first two draws, following events may occur.
`E_(1):` Both are balls are white
`E_(2):` First is white and second is black
`E_(3):` First is black and second is white
`E_(4):` Both the balls are black
Let E represent the event that the third ball is black. Then,
`P(E_(1))=2/4xx1/3=1/6`
`P(E_(2))=2/4xx2/3=1/3`
`P(E_(3))=2/4xx2/5=1/5`
`P(E_(4))=2/4xx3/5=3/10`
The four events `E_(1),E_(2),E_(3)and E_(4)` are mutually exclusive and exhaustive. Using the theorem of total probability,
`P(E)=P(E_(1))P(E//E_(1))+P(E_(2))P(E//E_(2))+P(E_(3))P(E//E_(3))`
`" "+P(E_(4))P(E//E_(4))`
`=1/6xx2/2+1/3xx3/4+1/5xx3/4+3/10xx4/6=23/30`