SOLUTION: A urn contains n black balls and n white balls. Three balls are chosen from the urn at random and without replacement. What is the value of n if the probability is 1/12 that all th

Question 1113083: A urn contains n black balls and n white balls. Three balls are chosen from the urn at random and without replacement. What is the value of n if the probability is 1/12 that all three balls are white?
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
The probability the first ball picked is white is: P(ball 1 is white) =
The probability the 2nd is also white: P(ball 2 is also white) =
The probability the 3rd is also white: P(ball 3 is also white) =
P(all 3 picks are white) =
Cross multiplying and then simplifying with all terms moved over to the left:

Potential solutions are: 1 and 5 (1 and 5 are solutions to the above equation, but might not be solutions to the problem).
Discard n=1 because that solution is too small to draw 3 balls.
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Ans:
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Check: (1/2)*(4/9)*(3/8) = (12/144) = 1/12 (ok)
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Edited to remove extra factor of n, which was a carry over from my paper solution where I didn't reduce n/(2n) to 1/2.

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