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Difference between revisions of "2008 AMC 10B Problems/Problem 20"
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==Solution 1==
 
==Solution 1==
  
−
The easiest way is to write a table of all <math>36</math> possible outcomes, do the sums, and count good outcomes.
+
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  3  4  5  6  8
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  4 |  5  7  8  9 10 12
 
  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  
+
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>.
 
<math>\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}</math>.
  
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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>.
  
−
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>.
+
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>.
 
<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}}
 
{{MAA Notice}}

Latest revision as of 16:44, 14 June 2023

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 $9$?

$\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9$

Solution 1

One approach is to write a table of all $36$ 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 $36$ possible outcomes, $4$ give the sum of $5$, $6$ the sum of $7$, and $4$ the sum of $9$, hence the resulting probability is $\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$.

Solution 2

Each die is equally likely to roll odd or even, so the probability of an odd sum is $\frac{1}{2}$. The possible odd sums are $3, 5, 7, 9, 11$.

So we can find the probability of rolling $3$ or $11$ instead and just subtract that from $\frac{1}{2}$, which seems easier.

Without writing out a table, we can see that there are 2 ways to make $3$, and 2 ways to make $11$, for a probability of $\frac{4}{36}$.

$\frac{1}{2}-\frac{4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$.

Solution 3

The outcome of the first dice can be $1$, $2$, $3$, or $4$. For each of these $4$ cases, we can find the possible outcomes for the second dice that makes the sum of the top two numbers $5$, $7$, or $9$, and then calculate the respective probabilities.

Case 1 - the first dice is 1: the outcome of the second dice can be $4$, $6$, or $8$. There is a $\frac{1}{6}$ probability of rolling a $1$ with the first dice and a $\frac{1}{2}$ probability of rolling a $4$, $6$, or $8$ with the second dice. $\frac{1}{6}*\frac{1}{2} = \frac{1}{12}.$

Case 2 - the first dice is 2: the outcome of the second dice can be $3$ or $5$. There is a $\frac{1}{3}$ probability of rolling a $2$ with the first dice and a $\frac{1}{3}$ probability of rolling a $3$ or $5$ with the second dice. $\frac{1}{3}*\frac{1}{3} = \frac{1}{9}.$

Case 3 - the first dice is 3: the outcome of the second dice can be $4$ or $6$. There is a $\frac{1}{3}$ probability of rolling a $3$ with the first dice and a $\frac{1}{3}$ probability of rolling a $4$ or $6$ with the second dice. $\frac{1}{3}*\frac{1}{3} = \frac{1}{9}.$

Case 4 - the first dice is 4: the outcome of the second dice can be $1$, $3$, or $5$. There is a $\frac{1}{6}$ probability of rolling a $4$ with the first dice and a $\frac{1}{2}$ probability of rolling a $1$, $3$, or $5$ with the second dice. $\frac{1}{6}*\frac{1}{2} = \frac{1}{12}.$

$\frac{1}{12} + \frac{1}{9} + \frac{1}{9} + \frac{1}{12} = \frac{14}{36} = \frac{7}{18}$, so the answer is $\boxed{\textbf{(B) } \frac{7}{18}}$. ~azc1027

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 10 Problems and Solutions

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