SOLUTION: There are 9 black balls and 6 red balls in an urn. if 3 balls are drawn without replacement, what is the probability that no more than 1 black ball is drawn?

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Question 857412: There are 9 black balls and 6 red balls in an urn. if 3 balls are drawn without replacement, what is the probability that no more than 1 black ball is drawn?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
To get the numerator of the probability:

Two cases: 

Case 1: choose 0 black balls and all 3 red balls: 

Choose all three red balls C(6,3) = 20 ways

Case 2: choose 1 black ball and 2 red balls:

Choose the black ball C(9,1) = 9 ways
Choose the 2 red balls C(6,2) = 15 ways
That's 9×15 = 135 ways


That's 20 + 135 = 155 ways.

To get the denominator:
Choose any 3 balls from the 15 in C(15,3) = 455 ways.

Probability that no more than 1 black ball is drawn =  = ,
approximately 34% of the time.

Edwin
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