Difference between revisions of "2008 AMC 10B Problems/Problem 20"
(add solution 2) |
|||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| + | |||
| + | ===Solution 1=== | ||
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. | ||
| Line 20: | Line 22: | ||
<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>. | ||
| + | |||
| + | So we can find the probability of rolling <math>3</math> or <math>11</math> instead. Without writing out a table, we can see that there are two ways to make <math>3</math>, and two ways to make <math>11</math>, for a probability of <math>\frac{4}{36}</math>. | ||
| + | |||
| + | <math>\frac{1}{2}-\frac{4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}</math>. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2008|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:27, 1 February 2018
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
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. 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. 
