You have a bag with 3 red, 3 white, and 2 blue balls in it. You draw three balls without replacement. What is the probability that the 3rd ball chosen is blue?

Answer 1

So you want the sequence
XXB , that is, notblue-notblue-blue
prob = (6/8)(5/7)(2/6) = 5/28

Answer 2

no, the first or 2nd ball may also be blue. In fact, they may both be blue, giving 0 probability of 3rd blue in that case.

Answer 3

ok, I misread the question.

All Cases , where X is not blue, 2 blue, 6 non-blue

BBB --- not possible, only 2 blue
BBX
BXB ---> (2/8)(6/7)(1/6) = 1/28
XBB ---> (6/8)(2/7)(1/6) = 1/28
BXX
XBX
XXB ---> (6/8)(5/7)(2/6) = 5/28
XXX

your prob = 7/28