Difference between revisions of "2007 AMC 10B Problems/Problem 22"
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There are <math>2 \cdot 3 \cdot 1 = 6</math> ways for your number to show up once, <math>1 \cdot 1 = 1</math> way for your number to show up twice, and <math>3 \cdot 3 = 9</math> ways for your number to not show up at all. Think of this as a set of <math>16</math> numbers with six <math>1s,</math> one <math>2,</math> and nine <math>-1s.</math> The average of this set is the expected return to the player. | There are <math>2 \cdot 3 \cdot 1 = 6</math> ways for your number to show up once, <math>1 \cdot 1 = 1</math> way for your number to show up twice, and <math>3 \cdot 3 = 9</math> ways for your number to not show up at all. Think of this as a set of <math>16</math> numbers with six <math>1s,</math> one <math>2,</math> and nine <math>-1s.</math> The average of this set is the expected return to the player. | ||
<cmath>\frac{6(1)+1(2)-9(1)}{16} = \frac{6+2-9}{16} = \boxed{\mathrm{(B) \ } -\frac{1}{16}}</cmath> | <cmath>\frac{6(1)+1(2)-9(1)}{16} = \frac{6+2-9}{16} = \boxed{\mathrm{(B) \ } -\frac{1}{16}}</cmath> | ||
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| + | ==Solution 2== | ||
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| + | We approach this through casework. We have a <math>\frac{1}{4} \cdot \frac{3}{4} \cdot 2</math> chance of earning <math>1</math> dollar. We have a <math>\frac{1}{4} \cdot \frac{1}{4}</math> chance of earning 2 dollars, and a <math>\frac{3}{4} \cdot \frac{3}{4}</math> chance of losing 1 dollar. Thus, our final answer is <math>\frac{3}{8} \cdot 1 + \frac{1}{16} \cdot 2 + \frac{9}{16} \cdot -1 = \boxed{\mathrm{(B) \ } -\frac{1}{16}}</math> -SuperJJ | ||
==See Also== | ==See Also== | ||
Revision as of 09:57, 17 February 2021
Contents
[hide]Problem 22
A player chooses one of the numbers 
through 
. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered through 
If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins dollar. If the number chosen appears on the bottom of both of the dice, then the player wins 
dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses dollar. What is the expected return to the player, in dollars, for one roll of the dice?
Solution
There are 
ways for your number to show up once, 
way for your number to show up twice, and 
ways for your number to not show up at all. Think of this as a set of 
numbers with six 
one 
and nine 
The average of this set is the expected return to the player.
![\[\frac{6(1)+1(2)-9(1)}{16} = \frac{6+2-9}{16} = \boxed{\mathrm{(B) \ } -\frac{1}{16}}\]](http://latex.artofproblemsolving.com/7/5/b/75be7467f74e920f61e20d6784fe11fd29f6872a.png)
Solution 2
We approach this through casework. We have a 
chance of earning dollar. We have a 
chance of earning 2 dollars, and a 
chance of losing 1 dollar. Thus, our final answer is 
-SuperJJ
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
