Question 1151171: A box contains 6 red, 4 white and 5 black balls. A man draws 4 balls from the box at random. Find the probability that among the balls drawn there is at least one ball of each color
Found 2 solutions by math_helper, Edwin McCravy:
Answer by math_helper(2461)   (Show Source):
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Assuming there is no replacement of balls...
One color will be chosen twice, the other two colors will each be chosen exactly once, and we consider each color being drawn twice separately:
P(one of each color chosen on 4 draws) =
(C(6,2)*C(4,1)*C(5,1) + C(6,1)*C(4,2)*C(5,1) + C(6,1)*C(4,1)*C(5,2)) / C(15,4)
where the 15 comes from the total number of balls, and C(n,r) = n!/((n-r)!r!)
This works out to about 52.7%
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website!
To be successful he must either draw {R,R,W,B}, {R,W,W,B} or {R,W,B,B}.
[It doesn't matter what order he draws the four balls in, or whether he just
reaches in and draws them all at once, or two at a time, so we use
combinations.]
Case 1: How many ways he can draw {R,R,W,B}
(6 reds CHOOSE 2)∙(4 whites CHOOSE 1)∙(5 blacks CHOOSE 1) =
(6C2)∙(4C1)∙(5C1) = 15∙4∙5 = 300 ways
Case 2: How many ways he can draw {R,W,W,B}
(6 reds CHOOSE 1)∙(4 whites CHOOSE 2)∙(5 blacks CHOOSE 1) =
(6C1)∙(4C2)∙(5C1) = 6∙6∙5 = 180 ways
Case 3: How many ways can he draw {R,W,B,B}
(6 reds CHOOSE 1)∙(4 whites CHOOSE 1)∙(5 blacks CHOOSE 2) =
(6C1)∙(4C1)∙(5C2) = 6∙4∙10 = 240
Total number of successful draws = 300+180+240 = 720
Total number of possible draws, successful or unsuccessful =
15 balls CHOOSE 4 = 15C4 = 1365
Desired probability. 720/1365 which reduces to 48/91 about 52.7%
of the time he'll get at least one of each color.
Edwin
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