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...) |
(→Solution) |
||
| Line 17: | Line 17: | ||
== 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 { | + | 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 22:31, 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?
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 
possible outcomes, and each of them is equally likely. All we now have to do is to count in how many of these 
will the white chips run out first. These are precisely those sequences that end with a red chip, and there are 
of them. Hence the probability that in the original experiment the last drawn chip is white is 
.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| 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 | |
