Here, there are four possibilities,
First Ball | Second Ball | Event |
White | White | E1 |
White | Black | E2 |
Black | White | E3 |
Black | Black | E4 |
Let E denote the event of drawing a black ball in the third attempt
P(E|E1)=1; because there is no white ball to be selected.
P(E|E2)=34, because there are 3 black balls and 1 white ball
P(E|E3)=34 because again there are 3 black and 1 white ball
P(E|E4)=46=2/3, because there are 4 black and 2 white balls
The four events E1,E2,E3, and E4are mutually exclusive and exhaustive. Using the the theorem of total probability,
∴P(E)=P(E1)P(E|E1)+P(E2)P(E|E2)+P(E3)P(E|E3)+P(E4)P(E|E4)
=16×1+13×34+15×34+310×23
=16+14+320+15
Ans. =2330.