Difference between revisions of "2006 AMC 10A Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | 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}{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 prize money, so the answer is | + | 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 |
| − | <math>\boxed{\ | + | <math>\boxed{\textbf{(D) } \$60}.</math> |
== Video Solution == | == Video Solution == | ||
Latest revision as of 08:39, 25 July 2023
Contents
Problem
A player pays 
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.)
Solution
The probability of rolling an even number on the first turn is 
and the probability of rolling the same number on the next turn is 
. The probability of winning is 
. If the game is to be fair, the amount paid, 
dollars, must be 
the amount of the prize money, so the answer is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
