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

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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. To draw out all the whites chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip left is red is <math>\boxed{\frac {3}{5}}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 02:25, 5 January 2012

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. To draw out all the whites chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip left is red is $\boxed{\frac {3}{5}}$.

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
2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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 10 Problems and Solutions