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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 ==
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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>.
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== Solution 1 ==
 +
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{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}</math>.
  
 
== Solution 2==
 
== Solution 2==
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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 $\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} =
+
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}} =\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}</math>.
 +
 
 +
~mobius247
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 +
== Solution 3==
 +
 
 +
The only way to get a product of that is a prime number is to roll all ones except for such prime, e.g: <math>11</math> ones and <math>1</math> two. So we seek the probability of rolling <math>11</math> ones and <math>1</math> prime number. The probability of rolling <math>11</math> ones is <math>\frac{1}{6^{11}}</math> and the probability of rolling a prime is <math>\frac{1}{2}</math>, giving us a probability of <math>\frac{1}{6^{11}}\cdot\frac{1}{2}</math> of this outcome occuring. However, there are <math>\frac{12!}{11!\cdot{1!}}=12</math> ways to arrange the ones and the prime. Multiplying the previous probability by <math>12</math> gives us <math>\frac{1}{6^{11}}\cdot\frac{1}{2}\cdot{6}=\frac{1}{6^{10}}=\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}.</math>
 +
 
 +
-Benedict T (countmath1)
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 +
== Solution 4==
 +
The only way to get a prime number product is to roll \(1\) on all \(11\) dices and roll a prime number on the remaining one.
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 +
There are 3 prime numbers less than or equal to 6: \(2, 3, 5\). For each of the 12 dice, we can choose one to roll the prime number on, while the other 11 show 1. So, there are \(12 \times 3 = 36\) favorable outcomes.
 +
 
 +
Since each die has 6 faces and there are 12 dice, the total number of possible outcomes is \(6^{12}\).
 +
 
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Therefore, the probability is
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<cmath>\frac{36}{6^{12}} = \left(\frac{1}{6}\right)^{10}.</cmath>
 +
 
 +
<cmath>\boxed{\textbf{(E) } \left(\dfrac{1}{6}\right)^{10}}</cmath>
 +
-LittleWavelet
  
 
== 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}}

Latest revision as of 18:10, 6 August 2025

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 1

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 $\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} = \dfrac{1}{2}\times\left(\dfrac{1}{6}\right)^{11}\times12=\left(\dfrac{1}{6}\right)^{11}\times6=\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}$.

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

~mobius247

Solution 3

The only way to get a product of that is a prime number is to roll all ones except for such prime, e.g: $11$ ones and $1$ two. So we seek the probability of rolling $11$ ones and $1$ prime number. The probability of rolling $11$ ones is $\frac{1}{6^{11}}$ and the probability of rolling a prime is $\frac{1}{2}$, giving us a probability of $\frac{1}{6^{11}}\cdot\frac{1}{2}$ of this outcome occuring. However, there are $\frac{12!}{11!\cdot{1!}}=12$ ways to arrange the ones and the prime. Multiplying the previous probability by $12$ gives us $\frac{1}{6^{11}}\cdot\frac{1}{2}\cdot{6}=\frac{1}{6^{10}}=\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}.$

-Benedict T (countmath1)

Solution 4

The only way to get a prime number product is to roll \(1\) on all \(11\) dices and roll a prime number on the remaining one.

There are 3 prime numbers less than or equal to 6: \(2, 3, 5\). For each of the 12 dice, we can choose one to roll the prime number on, while the other 11 show 1. So, there are \(12 \times 3 = 36\) favorable outcomes.

Since each die has 6 faces and there are 12 dice, the total number of possible outcomes is \(6^{12}\).

Therefore, the probability is \[\frac{36}{6^{12}} = \left(\frac{1}{6}\right)^{10}.\]

\[\boxed{\textbf{(E) } \left(\dfrac{1}{6}\right)^{10}}\] -LittleWavelet

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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