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

(New 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 c...)
 
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== Solution ==
 
== 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 <math>{5\choose 2}=10</math> possible outcomes, and each of them is equally likely. All we now have to do is to count in how many of these <math>10</math> will the white chips run out first. These are precisely those sequences that end with a red chip, and there are <math>{4\choose 2} = 6</math> of them. Hence the probability that in the original experiment the last drawn chip is white is <math>\frac 6{10} = \boxed{\frac {3}{5}}</math>.
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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 <math>{5\choose 2}=10</math> possible outcomes, and each of them is equally likely. All we now have to do is to count in how many of these <math>10</math> will the white chips run out first. These are precisely those sequences that end with a red chip, and there are <math>{4\choose 2} = 6</math> of them. Hence the probability that in the original experiment the last drawn chip is white is <math>1 - \frac 6{10} = \frac 4{10} = \boxed{\frac {2}{5}}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=10|num-a=12}}
 
{{AMC12 box|year=2001|num-b=10|num-a=12}}

Revision as of 23:31, 22 February 2009

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

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