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

Algebra ->  Probability-and-statistics -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
The probability the first ball picked is white is: P(ball 1 is white) = +n%2F%282n%29+=+1%2F2+
The probability the 2nd is also white: P(ball 2 is also white) = +%28n-1%29%2F%282n-1%29++
The probability the 3rd is also white: P(ball 3 is also white) = +%28n-2%29%2F%282n-2%29++
P(all 3 picks are white) = +%281%2F2%29%2A%28%28n-1%29%2F%282n-1%29%2A%28n-2%29%2F%282n-2%29%29+=++%281%2F12%29+
Cross multiplying and then simplifying with all terms moved over to the left:
+%28n%5E2+-+6n+%2B+5%29+=+0+
+%28n-5%29%28n-1%29+=+0+
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.
—
Ans: +highlight%28n+=+5%29+
—
Check: (1/2)*(4/9)*(3/8) = (12/144) = 1/12 (ok)
—
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.