Difference between revisions of "2001 AMC 12 Problems/Problem 11"

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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #11]] and [[2001 AMC 10 Problems|2001 AMC 10 #23]]}}
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== Problem ==
 
== Problem ==
  

Revision as of 22:37, 16 March 2011

The following problem is from both the 2001 AMC 12 #11 and 2001 AMC 10 #23, so both problems redirect to this page.

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 $\frac 6{10} = \boxed{\frac {3}{5}}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions