Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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== Solution 2== | == Solution 2== | ||
There are three cases where the product of the numbers is prime. One die will show <math>2</math>, <math>3</math>, or <math>5</math> and each of the other <math>11</math> dice will show a <math>1</math>. For each of these three cases, the number of ways to order the numbers is <math>\dbinom{12}{1}</math> = <math>12</math> . There are <math>6</math> possible numbers for each of the <math>12</math> dice, so the total number of permutations is <math>6^{12}</math>. The probability the product is prime is therefore | There are three cases where the product of the numbers is prime. One die will show <math>2</math>, <math>3</math>, or <math>5</math> and each of the other <math>11</math> dice will show a <math>1</math>. For each of these three cases, the number of ways to order the numbers is <math>\dbinom{12}{1}</math> = <math>12</math> . There are <math>6</math> possible numbers for each of the <math>12</math> dice, so the total number of permutations is <math>6^{12}</math>. The probability the product is prime is therefore | ||
− | <math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} = \left(\dfrac{1}{6}\right)^{10} | + | <math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} =\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}</math>. |
~mobius247 | ~mobius247 |
Revision as of 14:40, 15 December 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 1
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
.
~mobius247
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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