Difference between revisions of "2006 AMC 10A Problems/Problem 13"

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== Problem ==
 
== Problem ==
A player pays $5 to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)  
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A player pays <math>5 to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)  
  
<math>\mathrm{(A) \ } $12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ } $50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ } $100\qquad</math>
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<math>\mathrm{(A) \ } </math>12\qquad\mathrm{(B) \ } <math>30\qquad\mathrm{(C) \ } </math>50\qquad\mathrm{(D) \ } <math>60\qquad\mathrm{(E) \ } </math>100\qquad</math>
 
== Solution ==
 
== Solution ==
 
There are <math>36</math> possible combinations of 2 dice rolls.  
 
There are <math>36</math> possible combinations of 2 dice rolls.  
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== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]
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*[[2006 AMC 10A Problems/Problem 12|Previous Problem]]
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*[[2006 AMC 10A Problems/Problem 14|Next Problem]]
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[[Category:Introductory Combinatorics Problems]]

Revision as of 14:50, 4 August 2006

Problem

A player pays $5 to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

<math>\mathrm{(A) \ }$ (Error compiling LaTeX. ! Missing $ inserted.)12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ }$50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ }$100\qquad</math>

Solution

There are $36$ possible combinations of 2 dice rolls.

The winning combinations are $(2,2) ; (4,4) ; (6,6)$.

Since there are $3$ winning combinations and $36$ possible combinations of dice rolls, the probability of winning is $\frac{3}{36}=\frac{1}{12}$.

Let $x$ be the amount won in a fair game.

By the definition of a fair game:

$\frac{1}{12} \cdot x = 5$.

Therefore: $x = 60 \Rightarrow D$.

See Also

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