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

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== Problem ==
 
== Problem ==
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.)  
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A player pays $5 to play a game. A die is rolled. If the number on the die is [[odd integer | odd]], the game is lost. If the number on the die is [[even integer | 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) \ } </math>12\qquad\mathrm{(B) \ } <math>30\qquad\mathrm{(C) \ } </math>50\qquad\mathrm{(D) \ } <math>60\qquad\mathrm{(E) \ } </math>100\qquad</math>
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<math>\mathrm{(A) \ } $12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ } $50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ } $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.  
  
The winning combinations are <math> (2,2) ; (4,4) ; (6,6) </math>.
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The winning combinations are <math> (2,2) </math>, <math>(4,4)</math> and  <math> (6,6) </math>.
  
 
Since there are <math>3</math> winning combinations and <math>36</math> possible combinations of dice rolls, the probability of winning is <math>\frac{3}{36}=\frac{1}{12}</math>.
 
Since there are <math>3</math> winning combinations and <math>36</math> possible combinations of dice rolls, the probability of winning is <math>\frac{3}{36}=\frac{1}{12}</math>.
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Let <math>x</math> be the amount won in a fair game.  
 
Let <math>x</math> be the amount won in a fair game.  
  
By the definition of a fair game:
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By the definition of a fair game,
  
 
<math>  \frac{1}{12} \cdot x = 5 </math>.
 
<math>  \frac{1}{12} \cdot x = 5 </math>.
  
Therefore: <math> x = 60 \Rightarrow D </math>.
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Therefore <math> x = \$60 \Longrightarrow D </math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 15:56, 12 October 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.)

$\mathrm{(A) \ } $12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ } $50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ } $100\qquad$ (Error compiling LaTeX. ! LaTeX Error: \mathrm allowed only in math mode.)

Solution

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

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

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 = $60 \Longrightarrow D$.

See Also

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