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Difference between revisions of "2007 AMC 10B Problems/Problem 22"

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==Solution==
 
==Solution==
  
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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.
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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 sixteen numbers with six <math>1</math>'s, one <math>2</math>, and nine <math>-1</math>'s. 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==
 
==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
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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 <math>2</math> dollars, and a <math>\frac{3}{4} \cdot \frac{3}{4}</math> chance of losing <math>1</math> 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==
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{{AMC10 box|year=2007|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2007|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Combinatorics Problems]]

Latest revision as of 13:19, 6 January 2022

Problem 22

A player chooses one of the numbers $1$ through $4$. After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered $1$ through $4.$ If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins $1$ dollar. If the number chosen appears on the bottom of both of the dice, then the player wins $2$ dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses $1$ dollar. What is the expected return to the player, in dollars, for one roll of the dice?

$\textbf{(A) } -\frac{1}{8} \qquad\textbf{(B) } -\frac{1}{16} \qquad\textbf{(C) } 0 \qquad\textbf{(D) } \frac{1}{16} \qquad\textbf{(E) } \frac{1}{8}$

Solution

There are $2 \cdot 3 \cdot 1 = 6$ ways for your number to show up once, $1 \cdot 1 = 1$ way for your number to show up twice, and $3 \cdot 3 = 9$ ways for your number to not show up at all. Think of this as a set of sixteen numbers with six $1$'s, one $2$, and nine $-1$'s. The average of this set is the expected return to the player. \[\frac{6(1)+1(2)-9(1)}{16} = \frac{6+2-9}{16} = \boxed{\mathrm{(B) \ } -\frac{1}{16}}\]

Solution 2

We approach this through casework. We have a $\frac{1}{4} \cdot \frac{3}{4} \cdot 2$ chance of earning $1$ dollar. We have a $\frac{1}{4} \cdot \frac{1}{4}$ chance of earning $2$ dollars, and a $\frac{3}{4} \cdot \frac{3}{4}$ chance of losing $1$ dollar. Thus, our final answer is $\frac{3}{8} \cdot 1 + \frac{1}{16} \cdot 2 + \frac{9}{16} \cdot -1 = \boxed{\mathrm{(B) \ } -\frac{1}{16}}$ -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

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