Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2008 AMC 10B Problems/Problem 20

Revision as of 16:43, 14 June 2023 by Azc1027 (talk | contribs) (→‎Solution 3)

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
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_20&oldid=194406"