Difference between revisions of "2007 AMC 10B Problems/Problem 22"
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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 | + | 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== |
Revision as of 14:43, 4 June 2021
Contents
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.
Solution 2
We approach this through casework. We have a chance of earning
dollar. We have a
chance of earning
dollars, and a
chance of losing
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 |
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