Difference between revisions of "2005 AMC 10A Problems/Problem 9"
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<math> \textbf{(A) } \frac{1}{12}\qquad \textbf{(B) } \frac{1}{10}\qquad \textbf{(C) } \frac{1}{6}\qquad \textbf{(D) } \frac{1}{4}\qquad \textbf{(E) } \frac{1}{3} </math> | <math> \textbf{(A) } \frac{1}{12}\qquad \textbf{(B) } \frac{1}{10}\qquad \textbf{(C) } \frac{1}{6}\qquad \textbf{(D) } \frac{1}{4}\qquad \textbf{(E) } \frac{1}{3} </math> | ||
| − | == | + | ==Solution1== |
There are <math>\frac{5!}{2!3!}=10</math> distinct arrangements of three <math>X</math>'s and two <math>O</math>'s. | There are <math>\frac{5!}{2!3!}=10</math> distinct arrangements of three <math>X</math>'s and two <math>O</math>'s. | ||
| Line 14: | Line 14: | ||
Imagine you need to fit the two Os into the gaps between the three Xs. | Imagine you need to fit the two Os into the gaps between the three Xs. | ||
| − | The gaps between the Xs are: _X_X_X_, a total of | + | The gaps between the Xs are: _X_X_X_, a total of 4. |
You need to fit two Os in the gaps. There are two possible outcomes: | You need to fit two Os in the gaps. There are two possible outcomes: | ||
| − | 1. The two Os are put into different gaps, in this case the number of arrangements is | + | 1. The two Os are put into different gaps, in this case the number of arrangements is 4 x 3 / 2 = 6 |
2. The two Os are put into the same gap, in this case there will be an extra 4. | 2. The two Os are put into the same gap, in this case there will be an extra 4. | ||
| − | Therefore the probability of arrangements that reads XOXOX is | + | Therefore the probability of arrangements that reads XOXOX is 1 / (4 + 6) = 1 / 10 |
| + | |||
| + | ~Dew grass meadow | ||
==See also== | ==See also== | ||
Latest revision as of 09:46, 30 July 2024
Contents
Problem
Three tiles are marked 
and two other tiles are marked 
. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads 
?
Solution1
There are 
distinct arrangements of three 's and two 's.
There is only 
distinct arrangement that reads .
Therefore the desired probability is 
Solution2
Imagine you need to fit the two Os into the gaps between the three Xs.
The gaps between the Xs are: _X_X_X_, a total of 4.
You need to fit two Os in the gaps. There are two possible outcomes:
1. The two Os are put into different gaps, in this case the number of arrangements is 4 x 3 / 2 = 6
2. The two Os are put into the same gap, in this case there will be an extra 4.
Therefore the probability of arrangements that reads XOXOX is 1 / (4 + 6) = 1 / 10
~Dew grass meadow
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
