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Difference between revisions of "2005 AMC 10B Problems/Problem 12"

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Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
 
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
  
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<math>\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} </math>
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<math>\textbf{(A) } \left(\frac{1}{12}\right)^{12} \qquad \textbf{(B) } \left(\frac{1}{6}\right)^{12} \qquad \textbf{(C) } 2\left(\frac{1}{6}\right)^{11} \qquad \textbf{(D) } \frac{5}{2}\left(\frac{1}{6}\right)^{11} \qquad \textbf{(E) } \left(\frac{1}{6}\right)^{10} </math>
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== Solution ==
 
== Solution ==
 
In order for the product of the numbers to be prime, <math>11</math> of the dice have to be a <math>1</math>, and the other die has to be a prime number. There are <math>3</math> prime numbers (<math>2</math>, <math>3</math>, and <math>5</math>), and there is only one <math>1</math>, and there are <math>\dbinom{12}{1}</math> ways to choose which die will have the prime number, so the probability is <math>\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}}</math>.
 
In order for the product of the numbers to be prime, <math>11</math> of the dice have to be a <math>1</math>, and the other die has to be a prime number. There are <math>3</math> prime numbers (<math>2</math>, <math>3</math>, and <math>5</math>), and there is only one <math>1</math>, and there are <math>\dbinom{12}{1}</math> ways to choose which die will have the prime number, so the probability is <math>\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}}</math>.

Revision as of 14:38, 15 December 2021

Problem

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

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

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

~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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