Question

An urn has 12 balls. 8 are white balls and 4 are black balls.

If we draw a sample of 3 balls (i.e., picking without replacement) and given that the first two balls selected were a black ball and a white ball, what is the conditional probability of the third ball drawn being white?

Answer 1

let drawing of white and black ball is represented by W and B

P(first two select were black and white)=P(WBW)+P(WBB)+P(BWW)+P(BWB)

=(8/12)*(4/11)*(7/10)+(8/12)*(4/11)*(3/10)+(4/12)*(8/11)*(7/10)+(4/12)*(8/11)*(3/10)

=16/33

P(first two are black and white and third is white)=P(WBW)+P(BWW)

=(8/12)*(4/11)*(7/10)+(4/12)*(8/11)*(7/10)

=56/165

therefore P(third is white given first two are black and white)

=(56/165)/(16/33)=7/10

(Note: this can easily be solved by the fact that after removing 1st two balls one of which is white and other black, there remains 10 balls in which there are 7 whites, therefore probability of white =7/10)