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

Difference between revisions of "2005 AMC 10B Problems/Problem 9"

(→‎Solution)
(→‎Solution)
Line 17: Line 17:
 
Adding these two probabilities will give us our final answer. <math>\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}</math>
 
Adding these two probabilities will give us our final answer. <math>\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}</math>
  
−
If you run out of time, you can see it is obviously greater than <math>1/2</math> so it narrows guesses.
+
If you run out of time, you can see it is obviously greater than <math>\frac{1}{2}</math> so it narrows guesses.
  
 
==Solution 2 (Complementary Counting)==
 
==Solution 2 (Complementary Counting)==

Revision as of 21:16, 31 May 2021

Problem

One fair die has faces $1$, $1$, $2$, $2$, $3$, $3$ and another has faces $4$, $4$, $5$, $5$, $6$, $6$. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3}$

Solution

In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.

Case 1: The first die is odd and the second die is even.

The probability of this happening is $\dfrac{4}{6}\times\dfrac{4}{6}=\dfrac{16}{36}=\dfrac{4}{9}$

Case 2: The first die is even and the second die is odd.

The probability of this happening is $\dfrac{2}{6}\times\dfrac{2}{6}=\dfrac{4}{36}=\dfrac{1}{9}$

Adding these two probabilities will give us our final answer. $\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}$

If you run out of time, you can see it is obviously greater than $\frac{1}{2}$ so it narrows guesses.

Solution 2 (Complementary Counting)

We see can see that there are $2$ ways to get an even number, and other ways get us an odd. Therefore, if we subtract the $P(\text{even})$, we will get $P(\text{odd})$. We can get an even either by getting $2$ evens or $2$ odds. Both cases give $\frac{2}{9}$, so $P(\text{odd})$ is $1-2 \cdot \frac{2}{9}=\boxed{\text{(D)} \frac{5}{9}}$

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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=2005_AMC_10B_Problems/Problem_9&oldid=154619"