Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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<math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} \mathrm{(E)} = \left(\dfrac{1}{6}\right)^{10}}</math> | <math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} \mathrm{(E)} = \left(\dfrac{1}{6}\right)^{10}}</math> | ||
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== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2005|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:23, 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}
$\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} \mathrm{(E)} = \left(\dfrac{1}{6}\right)^{10}}$ (Error compiling LaTeX. Unknown error_msg)
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 |
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