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 12"

(→‎Solution 2)
(→‎Solution 2)
Line 11: Line 11:
 
<math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>
 
<math>\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}</math>
  
−
<math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}}</math>
+
<math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} \mathrm{(E)} </math> \mathrm{(E)}
  
 
\left(\dfrac{1}{6}\right)^{10}}
 
\left(\dfrac{1}{6}\right)^{10}}

Revision as of 08:20, 1 June 2021

Problem

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$\mathrm{(A)} \left(\frac{1}{12}\right)^{12} \qquad \mathrm{(B)} \left(\frac{1}{6}\right)^{12} \qquad \mathrm{(C)} 2\left(\frac{1}{6}\right)^{11} \qquad \mathrm{(D)} \frac{5}{2}\left(\frac{1}{6}\right)^{11} \qquad \mathrm{(E)} \left(\frac{1}{6}\right)^{10}$

Solution

In order for the product of the numbers to be prime, $11$ of the dice have to be a $1$, and the other die has to be a prime number. There are $3$ prime numbers ($2$, $3$, and $5$), and there is only one $1$, and there are $\dbinom{12}{1}$ ways to choose which die will have the prime number, so the probability is $\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} = \dfrac{1}{2}\times\left(\dfrac{1}{6}\right)^{11}\times12=\left(\dfrac{1}{6}\right)^{11}\times6=\boxed{\mathrm{(E)}\ \left(\dfrac{1}{6}\right)^{10}}$.

Solution 2

There are three cases where the product of the numbers is prime. One die will show $2$, $3$, or $5$ and each of the other $11$ dice will show a $1$. For each of these three cases, the number of ways to order the numbers is $\dbinom{12}{1}$ = $12$ . There are $6$ possible numbers for each of the $12$ dice, so the total number of permutations is $6^{12}$. The probability the product is prime is therefore \frac{3\cdot 12}{6^{12}

$\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$

$\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} \mathrm{(E)}$ \mathrm{(E)}

\left(\dfrac{1}{6}\right)^{10}}

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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_12&oldid=154737"