Difference between revisions of "2008 AMC 10B Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | {{ | + | |
| + | The easiest way is to write a table of all <math>36</math> 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 <math>36</math> possible outcomes <math>4</math> give the sum of <math>5</math>, <math>6</math> the sum of <math>7</math>, and <math>4</math> the sum of <math>9</math>, hence the resulting probability is | ||
| + | <math>\frac{4+6+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}} | ||
Revision as of 16:33, 25 January 2009
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
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
.
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 | ||
