Difference between revisions of "2008 AMC 10B Problems/Problem 20"
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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>. | ||
The possible odd sums are <math>3, 5, 7, 9, 11</math>. | The possible odd sums are <math>3, 5, 7, 9, 11</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 | + | |
| + | 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. | ||
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| + | Without writing out a table, we can see that there are 2 ways to make <math>3</math>, and 2 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>. | <math>\frac{1}{2}-\frac{4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}</math>. | ||
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| + | ==Solution 3 == | ||
| + | The outcome of the first dice can be <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math>. For each of these <math>4</math> cases, we can find the possible outcomes for the second dice that makes the sum of the top two numbers <math>5</math>, <math>7</math>, or <math>9</math>, and then calculate the respective probabilities. | ||
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| + | '''Case 1 - the first dice is 1:''' the outcome of the second dice can be <math>4</math>, <math>6</math>, or <math>8</math>. There is a <math>\frac{1}{6}</math> probability of rolling a <math>1</math> with the first dice and a <math>\frac{1}{2}</math> probability of rolling a <math>4</math>, <math>6</math>, or <math>8</math> with the second dice. <math>\frac{1}{6}*\frac{1}{2} = \frac{1}{12}.</math> | ||
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| + | '''Case 2 - the first dice is 2:''' the outcome of the second dice can be <math>3</math> or <math>5</math>. There is a <math>\frac{1}{3}</math> probability of rolling a <math>2</math> with the first dice and a <math>\frac{1}{3}</math> probability of rolling a <math>3</math> or <math>5</math> with the second dice. <math>\frac{1}{3}*\frac{1}{3} = \frac{1}{9}.</math> | ||
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| + | '''Case 3 - the first dice is 3:''' the outcome of the second dice can be <math>4</math> or <math>6</math>. There is a <math>\frac{1}{3}</math> probability of rolling a <math>3</math> with the first dice and a <math>\frac{1}{3}</math> probability of rolling a <math>4</math> or <math>6</math> with the second dice. <math>\frac{1}{3}*\frac{1}{3} = \frac{1}{9}.</math> | ||
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| + | '''Case 4 - the first dice is 4:''' the outcome of the second dice can be <math>1</math>, <math>3</math>, or <math>5</math>. There is a <math>\frac{1}{6}</math> probability of rolling a <math>4</math> with the first dice and a <math>\frac{1}{2}</math> probability of rolling a <math>1</math>, <math>3</math>, or <math>5</math> with the second dice. <math>\frac{1}{6}*\frac{1}{2} = \frac{1}{12}.</math> | ||
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| + | <math>\frac{1}{12} + \frac{1}{9} + \frac{1}{9} + \frac{1}{12} = \frac{14}{36} = \frac{7}{18}</math>, so the answer is <math>\boxed{\textbf{(B) } \frac{7}{18}}</math>. ~azc1027 | ||
==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}} | ||
Latest revision as of 16:44, 14 June 2023
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
One approach 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 
.
The possible odd sums are 
.
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 2 ways to make , and 2 ways to make , for a probability of 
.
.
Solution 3
The outcome of the first dice can be , , , or . For each of these cases, we can find the possible outcomes for the second dice that makes the sum of the top two numbers , , or , and then calculate the respective probabilities.
Case 1 - the first dice is 1: the outcome of the second dice can be , , or . There is a 
probability of rolling a with the first dice and a probability of rolling a , , or with the second dice. 
Case 2 - the first dice is 2: the outcome of the second dice can be or . There is a 
probability of rolling a with the first dice and a probability of rolling a or with the second dice. 
Case 3 - the first dice is 3: the outcome of the second dice can be or . There is a probability of rolling a with the first dice and a probability of rolling a or with the second dice.
Case 4 - the first dice is 4: the outcome of the second dice can be , , or . There is a probability of rolling a with the first dice and a probability of rolling a , , or with the second dice.
, so the answer is 
. ~azc1027
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. 
