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 "2002 AMC 12B Problems/Problem 16"
(Fixed category)
Line 27: Line 27:
 
{{AMC12 box|year=2002|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2002|ab=B|num-b=15|num-a=17}}
  
−
[[Category:Introductory Algebra Problems]]
+
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:53, 29 July 2020

Problem

Juan rolls a fair regular octahedral die marked with the numbers $1$ through $8$. Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3?

$\mathrm{(A)}\ \frac 1{12} \qquad\mathrm{(B)}\ \frac 13 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 7{12} \qquad\mathrm{(E)}\ \frac 23$

Solution

Solution 1

On both dice, only the faces with the numbers $3,6$ are divisible by $3$. Let $P(a) = \frac{2}{8} = \frac{1}{4}$ be the probability that Juan rolls a $3$ or a $6$, and $P(b) = \frac{2}{6} = \frac 13$ that Amal does. By the Principle of Inclusion-Exclusion,

\[P(a \cup b) = P(a) + P(b) - P(a \cap b) = \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2} \Rightarrow \mathrm{(C)}\]

Alternatively, the probability that Juan rolls a multiple of $3$ is $\frac{1}{4}$, and the probability that Juan does not roll a multiple of $3$ but Amal does is $\left(1 - \frac{1}{4}\right) \cdot \frac{1}{3} = \frac{1}{4}$. Thus the total probability is $\frac 14 + \frac 14 = \frac 12$.

Solution 2

The probability that neither Juan nor Amal rolls a multiple of $3$ is $\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}$; using complementary counting, the probability that at least one does is $1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}$.

Solution 3

The product will be a multiple of 3 if and only if at least one of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is $2/8 = 1/4$. The probability that Juan does not roll 3 or 6, but Amal does is $(3/4) (1/3) = 1/4$. Thus, the probability that the product of the rolls is a multiple of 3 is \[\frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}.\] ~aopsav (Credit to AoPS Alcumus)

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_16&oldid=129757"