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

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== Problem ==
 
== Problem ==
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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.)  
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A player pays <math>\textdollar 5</math> 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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<math>\textbf{(A) } \textdollar12\qquad\textbf{(B) } \textdollar30\qquad\textbf{(C) } \textdollar50\qquad\textbf{(D) } \textdollar60\qquad\textbf{(E) } \textdollar 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 ==
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There are <math>36</math> possible combinations of 2 dice rolls.
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The probability of rolling an even number on the first turn is <math>\frac{1}{2}</math> and the probability of rolling the same number on the next turn is <math>\frac{1}{6}</math>. The probability of winning is <math>\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5</math> dollars, must be <math>\frac{1}{12}</math> the amount of the prize money, so the answer is
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The winning combinations are <math> (2,2) ; (4,4) ; (6,6) </math>.
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<math>\boxed{\textbf{(D) } \$60}.</math>
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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.
 
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By the definition of a fair game:
 
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<math> \frac{1}{12} \cdot x = 5 </math>.
 
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Therefore: <math> x = 60 \Rightarrow D </math>.
 
  
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== See Also ==
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== Video Solution ==
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*[[2006 AMC 10A Problems]]
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https://youtu.be/IRyWOZQMTV8?t=3410
  
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*[[2006 AMC 10A Problems/Problem 12|Previous Problem]]
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~ pi_is_3.14
  
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*[[2006 AMC 10A Problems/Problem 14|Next Problem]]
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== See also ==
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{{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 08:39, 25 July 2023

Problem

A player pays $\textdollar 5$ 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.)

$\textbf{(A) } \textdollar12\qquad\textbf{(B) } \textdollar30\qquad\textbf{(C) } \textdollar50\qquad\textbf{(D) } \textdollar60\qquad\textbf{(E) } \textdollar 100\qquad$

Solution

The probability of rolling an even number on the first turn is $\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\frac{1}{6}$. The probability of winning is $\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}$. If the game is to be fair, the amount paid, $5$ dollars, must be $\frac{1}{12}$ the amount of the prize money, so the answer is $\boxed{\textbf{(D) } $60}.$

Video Solution

https://youtu.be/IRyWOZQMTV8?t=3410

~ pi_is_3.14

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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