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Difference between revisions of "2005 AMC 10A Problems/Problem 9"
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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 <math>4</math>.
+
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 <math>4</math>x<math>3</math>/<math>2</math>=<math>6</math>
+
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 <math>1</math>/<math>4</math>+<math>6</math>=<math>1</math>/<math>10</math>
+
Therefore the probability of arrangements that reads XOXOX is (1 / 4) + 6 = 1 / 10
  
 
==See also==
 
==See also==

Revision as of 21:03, 29 July 2024

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 $\boxed{\textbf{(B) }\frac{1}{10}}$

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

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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