Difference between revisions of "2008 AMC 10B Problems/Problem 20"
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<math>\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9</math> | <math>\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9</math> | ||
| − | + | ==Solution 1== | |
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The easiest way is to write a table of all <math>36</math> possible outcomes, do the sums, and count good outcomes. | The easiest way is to write a table of all <math>36</math> possible outcomes, do the sums, and count good outcomes. | ||
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<math>\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}</math>. | <math>\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}</math>. | ||
| − | + | ==Solution 2== | |
Each die is equally likely to roll odd or even, so the probability of an odd sum is <math>\frac{1}{2}</math>. | Each die is equally likely to roll odd or even, so the probability of an odd sum is <math>\frac{1}{2}</math>. | ||
Revision as of 15:46, 15 February 2021
Contents
Problem
The faces of a cubical die are marked with the numbers 
, 
, , 
, , and 
. The faces of another die are marked with the numbers , , , 
, 
, and 
. What is the probability that the sum of the top two numbers will be , 
, or 
?
Solution 1
The easiest way is to write a table of all 
possible outcomes, do the sums, and count good outcomes.
1 3 4 5 6 8 ------------------ 1 | 2 4 5 6 7 9 2 | 3 5 6 7 8 10 2 | 3 5 6 7 8 10 3 | 4 6 7 8 9 11 3 | 4 6 7 8 9 11 4 | 5 7 8 9 10 12
We see that out of possible outcomes give the sum of , the sum of , and the sum of , hence the resulting probability is
.
Solution 2
Each die is equally likely to roll odd or even, so the probability of an odd sum is 
.
So we can find the probability of rolling or 
instead and just subtract that from , which seems easier. Without writing out a table, we can see that there are two ways to make , and two ways to make , for a probability of 
.
.
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
