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Difference between revisions of "2005 AMC 10A Problems/Problem 9"

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Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?
 
Three tiles are marked <math>X</math> and two other tiles are marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>?
  
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<math> \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(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>
  
 
==Solution==
 
==Solution==
 
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.  
  
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There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math>
+
There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math>.
  
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Therefore the desired [[probability]] is <math>\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}</math>
+
Therefore the desired [[probability]] is <math>\boxed{\textbf{(B) }\frac{1}{10}}</math>
  
 
==See also==
 
==See also==

Revision as of 11:01, 13 December 2021

Problem

Three tiles are marked $X$ and two other tiles are marked $O$. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads $XOXOX$?

$\textbf{(A) } \frac{1}{12}\qquad \textbf{(B) } \frac{1}{10}\qquad \textbf{(C) } \frac{1}{6}\qquad \textbf{(D) } \frac{1}{4}\qquad \textbf{(E) } \frac{1}{3}$

Solution

There are $\frac{5!}{2!3!}=10$ distinct arrangements of three $X$'s and two $O$'s.

There is only $1$ distinct arrangement that reads $XOXOX$.

Therefore the desired probability is $\boxed{\textbf{(B) }\frac{1}{10}}$

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

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