Difference between revisions of "2006 AMC 10A Problems/Problem 13"
m |
m (+ box, condense solution) |
||
| Line 4: | Line 4: | ||
<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 | + | There are <math>6 \cdot 6 = 36</math> possible [[combination]]s of 2 dice rolls. The only possible 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>. |
| − | + | Let <math>x</math> be the amount won in a fair game. By the definition of a fair game, | |
| − | + | :<math>\frac{1}{12} \cdot x = 5 </math>. | |
| − | + | Therefore, <math> x = \$60 \Longrightarrow \mathrm{D} </math>. | |
| − | + | == See also == | |
| − | + | {{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}} | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | == See | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 15:47, 26 February 2007
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 only possible winning combinations are 
, 
and 
. Since there are 
winning combinations and 
possible combinations of dice rolls, the probability of winning is 
.
Let 
be the amount won in a fair game. By the definition of a fair game,
.
.Therefore, 
.
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 | ||
