Finite 3.3 Probability

Question

 Assume that you have 6 red balls and 9 blue balls from which randomly to select two without replacement.

(1) What is the probability that exactly one red ball is chosen? 

 

(2) What is the probability that the second ball is blue given that at least one of the balls is blue?

Answer 1

1) the probability that one red and one blue ball is selected is C(9,1)C(6,1)/C(15,2) = 9*6/105 = 18/35

2) The probability of selecting two blue balls is C(9,2)/C(15,2) = 36/105 = 12/35

Thus, the probability of selecting at least one blue is 12/35 + 18/35 = 30/35

The probability that the second ball is blue given that there is at least one blue is

(1/2*18/35 + 12/35)/(30/35) = 21/35/(30/35) = 21/30=7/10

You multiply 18/35 by 1/2 because, if you have a red and blue, having a blue first or second is equally likely. You have also used the fact that, if the second ball is blue, at least one is blue.