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 2 dice rolls.  
+
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>.
  
The winning combinations are <math> (2,2) </math>, <math>(4,4)</math> and  <math> (6,6) </math>.
+
Let <math>x</math> be the amount won in a fair game. By the definition of a fair game,
  
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>.
+
:<math>\frac{1}{12} \cdot x = 5 </math>.
  
Let <math>x</math> be the amount won in a fair game.  
+
Therefore, <math> x = \$60 \Longrightarrow \mathrm{D} </math>.
  
By the definition of a fair game,
+
== See also ==
 
+
{{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}
<math>  \frac{1}{12} \cdot x = 5 </math>.
 
 
 
Therefore <math> x = \$60 \Longrightarrow D </math>.
 
 
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
 
 
*[[2006 AMC 10A Problems/Problem 12|Previous Problem]]
 
 
 
*[[2006 AMC 10A Problems/Problem 14|Next Problem]]
 
  
 
[[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. ! LaTeX Error: \mathrm allowed only in math mode.)

Solution

There are $6 \cdot 6 = 36$ possible combinations of 2 dice rolls. The only possible 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 = $60 \Longrightarrow \mathrm{D}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS