Difference between revisions of "2008 AMC 10B Problems/Problem 20"
Person1133 (talk | contribs) |
(→Solution 3) |
||
| (16 intermediate revisions by 6 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== | ||
| + | |||
| + | One approach 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>. | ||
| + | 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 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>. | ||
| + | |||
| + | ==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. | ||
| + | |||
| + | '''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> | ||
| + | |||
| + | '''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> | ||
| + | |||
| + | '''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> | ||
| + | |||
| + | '''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> | ||
| + | |||
| + | <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}} | ||
Latest revision as of 17:44, 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. 
