SOLUTION: Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. suppose that we win #1 for each white ball selected and lose #1 for each red ball selected

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Question 1119461: Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. suppose that we win #1 for each white ball selected and lose #1 for each red ball selected. what is the probability that we win the money?
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. suppose that we win #1 for each white ball selected and lose #1 for each red ball selected. what is the probability that we win the money?
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Determine # of ways to "win"::
3white:: 1 way
2white and 1red:: 3C2*3C1 = 3*3 = 9 ways
2white and 1black:: 3C2*5C1 = 3*5 = 15 ways
1white and 2black:: 3C1*5C2 = 3*10 = 30 ways
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Determine # of possible triples:: 11C3 = (11*10*9)/(1*2*3) = 165
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Ans:: (1+9+15+30)/165 = 55/165
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Cheers,
Stan H.
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Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
Since the number of white balls is the same as the number of red balls, 


the game is totally symmetric. In other words, for any combination  (mW,nR) of having m White balls and n Red balls,


there is the same probability to have symmetric combination (nW,mR).


It means that for every chance to win (m-n) dollars, there is THE SAME chance to loose (m-n) dollars.


If the game is repeating infinitely long, the probability to gain money is ZERO - same as the probability to loose money.


The answer and the solution are clear without any calculations.


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