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2001 AMC 12 Problems/Problem 11

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Problem

A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

\text{(A) }\frac {3}{10}\qquad\text{(B) }\frac {2}{5}\qquad\text{(C) }\frac {1}{2}\qquad\text{(D) }\frac {3}{5}\qquad\text{(E...

Solution

Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. There are {5\choose 2}=10 possible outcomes, and each of them is equally likely. All we now have to do is to count in how many of these 10 will the white chips run out first. These are precisely those sequences that end with a red chip, and there are {4\choose 2} = 6 of them. Hence the probability that in the original experiment the last drawn chip is white is \frac 6{10} = \boxed{\frac {3}{5}}.

See Also

2001 AMC 12 (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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