Difference between revisions of "2006 AMC 10A Problems/Problem 13"
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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> | <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. | ||
| + | |||
| + | The winning combinations are <math> (2,2) ; (4,4) ; (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> | ||
| + | |||
| + | Let <math>x</math> be the ammount won in a fair game. | ||
| + | |||
| + | By the definition of a fair game: | ||
| + | |||
| + | <math> \frac{1}{12} \cdot x = 5 </math> | ||
| + | |||
| + | Therefore: <math> x = 60 \Rightarrow D </math> | ||
| + | |||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 18:36, 15 July 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. Unknown error_msg)
Solution
There are 
possible combinations of 2 dice rolls.
The winning combinations are 
Since there are 
winning combinations and possible combinations of dice rolls, the probability of winning is 
Let 
be the ammount won in a fair game.
By the definition of a fair game:
Therefore: 
