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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>
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<math>
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\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}
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</math>
  
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==Solution==
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==Solution 1==
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There are <math>\frac{5!}{2!3!}=10</math> distinct arrangments of three <math>X</math>'s and two <math>O</math>'s.  
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There are <math>\frac{5!}{2!3!} = 10</math> distinct arrangements of <math>3</math> <math>X</math>s and <math>2</math> <math>O</math>s, and only <math>1</math> distinct arrangement that reads <math>XOXOX</math>.
  
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There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math>
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Therefore the desired probability is simply <math>\boxed{\textbf{(B) }\frac{1}{10}}</math>.
  
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Therfore the desired probability is <math>\frac{1}{10} \Rightarrow B</math>
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==Solution 2==
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Arranging the <math>3</math> <math>X</math>s and <math>2</math> <math>O</math>s in a row is equivalent to fitting the <math>O</math>s into the gaps between the <math>X</math>s.
  
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==See Also==
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There are a total of <math>3+1 = 4</math> gaps between the <math>X</math>s, so fitting <math>2</math> <math>O</math>s into these gaps gives <math>2</math> possible outcomes:
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem 8|Previous Problem]]
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1. The <math>2</math> <math>O</math>s are put into different gaps. In this case, the number of possible arrangements is <math>\binom{4}{2} = 6</math>.
  
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*[[2005 AMC 10A Problems/Problem 10|Next Problem]]
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2. The <math>2</math> <math>O</math>s are put into the same gap. In this case, as there are <math>4</math> gaps, we simply obtain <math>4</math> possible arrangements.
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Therefore the probability of an arrangement that reads <math>XOXOX</math> is <math>\frac{1}{4 + 6} = \boxed{\textbf{(B) }\frac{1}{10}}</math>.
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~Dew grass meadow
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==See also==
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{{AMC10 box|year=2005|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 17:14, 1 July 2025

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 1

There are $\frac{5!}{2!3!} = 10$ distinct arrangements of $3$ $X$s and $2$ $O$s, and only $1$ distinct arrangement that reads $XOXOX$.

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

Solution 2

Arranging the $3$ $X$s and $2$ $O$s in a row is equivalent to fitting the $O$s into the gaps between the $X$s.

There are a total of $3+1 = 4$ gaps between the $X$s, so fitting $2$ $O$s into these gaps gives $2$ possible outcomes:

1. The $2$ $O$s are put into different gaps. In this case, the number of possible arrangements is $\binom{4}{2} = 6$.

2. The $2$ $O$s are put into the same gap. In this case, as there are $4$ gaps, we simply obtain $4$ possible arrangements.

Therefore the probability of an arrangement that reads $XOXOX$ is $\frac{1}{4 + 6} = \boxed{\textbf{(B) }\frac{1}{10}}$.

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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