Question 1174274: A box contains 3 orange balls 3 yellow balls and 2 white balls. Three balls are selected without replacement. Find the probability of 2 yellow balls and a white ball.
Found 2 solutions by math_helper, greenestamps:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! 
Here's how:
8 balls in total (3Orange(O) + 3Y + 2W)
Successful outcomes are: YYW, YWY, and WYY
It is the sum of the probabilities of these three that
gives the overall probability. First, find the
probability of each of the successful outcomes:
YYW: (3/8)(2/7)(2/6) = 12/336
YWY: (3/8)(2/7)(2/6) = 12/336
WYY: (2/8)(3/7)(2/6) = 12/336
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add up these outcomes: 36/336
36/336 reduces to 3/28
Could we have just computed P(YYW) and then noticed that there
are 3 ways to arrange YYW giving 3*12/336 = 36/336 ? For this
problem the answer is yes.
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
The total number of ways of choosing 3 of the 8 balls is "8 choose 3", C(8,3).
The number of ways of choosing 2 of the 3 yellow balls and 1 of the 2 white balls is "3 choose 2" times "2 choose 1", C(3,2)*C(2,1).
The probability of choosing 2 yellow and 1 white is the ratio of those numbers of ways.
ANSWER: P(2 yellow, 1 white) = 3/28
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