2001 AMC 12 Problems/Problem 11

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?

$\text{(A) }\frac {3}{10} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {1}{2} \qquad \text{(D) }\frac {3}{5} \qquad \text{(E) }\frac {7}{10}$

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 $1 - \frac 6{10} = \frac 4{10} = \boxed{\frac {2}{5}}$ .

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2001 AMC 12 Problems/Problem 11

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