Let A and B denote respectively the events the first and second balls both drawn are white.
Then, we have to find `P(A nn B)`.
Now, `P(A)=P` (white ball in the first drawn) `=8/12`
After the occurrence of events A, we are left with 7 white and 4 red balls.
The probability of frawing second white ball, given that the first ball drawn is white, is clearly the conditional probability of occurrence of B, given that A has occurred.
`:. P(B//A)=7/11`.
By multiplication rule of probability, we have
`P(A nn B)=P(A).P(B//A)=(8/12xx7/11)=14/33`.
Hence, the required probability is `14/33`.
