SOLUTION: Hi, I got stuck in a question:There are 5 Black balls, 4 red balls, and 1 blue ball in a bag. If 3 balls are drawn without replacement, find probability of: a) Exactly one black b

Question 1153646: Hi, I got stuck in a question:There are 5 Black balls, 4 red balls, and 1 blue ball in a bag. If 3 balls are drawn without replacement, find probability of:
a) Exactly one black ball is drawn given that at least one black ball is drawn.
b) The third ball drawn is yellow given that balls drawn are of different colours.
For a), I have tried to use conditional probability, but I can't reach the value smaller than 1,and so here I am seeking for help, looking forward to your reply and thanks for your effort.
Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
a) Exactly one black ball is drawn given that at least one black ball is drawn.

P(A) = Probability exactly one black ball is drawn:

= = = = 0.4167

P(B) = Probability at least black ball is drawn:

= = = = = 0.9167

Probability exactly one black ball is drawn given that at least one black ball is drawn:

P(A|B) =

P(A and B) means the probability that "exactly one black ball is drawn AND at least one black ball is drawn." Both conditions must be met. Logic would tell you that this is the same as the probability that "exactly one black ball is drawn." We figured this out above, and that probability is 0.4167.

So, we know that P(A and B) is 0.4167. We also know that P(B) is 0.9167.

P(A|B) = = = 0.4546
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