Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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<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}}</math> |
\left(\dfrac{1}{6}\right)^{10}} | \left(\dfrac{1}{6}\right)^{10}} | ||
Revision as of 08:17, 1 June 2021
Contents
Problem
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
Solution
In order for the product of the numbers to be prime, 
of the dice have to be a 
, and the other die has to be a prime number. There are 
prime numbers (
, , and 
), and there is only one , and there are 
ways to choose which die will have the prime number, so the probability is 
.
Solution 2
There are three cases where the product of the numbers is prime. One die will show , , or and each of the other dice will show a . For each of these three cases, the number of ways to order the numbers is = 
. There are 
possible numbers for each of the dice, so the total number of permutations is 
. The probability the product is prime is therefore \frac{3\cdot 12}{6^{12}
\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. 
