Difference between revisions of "2008 AMC 10B Problems/Problem 20"
Person1133 (talk | contribs) |
|||
| (5 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | The faces of a cubical die are marked with the numbers 1, 2, 2, 3, 3, and 4. The faces of another die are marked with the numbers 1, 3, 4, 5, 6, and 8. What is the probability that the sum of the top two numbers will be 5, 7, or 9? | + | The faces of a cubical die are marked with the numbers <math>1</math>, <math>2</math>, <math>2</math>, <math>3</math>, <math>3</math>, and <math>4</math>. The faces of another die are marked with the numbers <math>1</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, and <math>8</math>. What is the probability that the sum of the top two numbers will be <math>5</math>, <math>7</math>, or <math>9</math>? |
| − | |||
| − | ==Solution== | + | <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== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==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 and just subtract that from <math>\frac{1}{2}</math>, which seems easier. 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}} | ||
Revision as of 16:46, 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. 
