Asked by Anonymous
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?
Answers
Answered by
Reiny
So you want the sequence
XXB , that is, notblue-notblue-blue
prob = (6/8)(5/7)(2/6) = 5/28
XXB , that is, notblue-notblue-blue
prob = (6/8)(5/7)(2/6) = 5/28
Answered by
Anonymous
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.
Answered by
Reiny
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
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