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Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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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>.
 
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>.
 
== Solution 2==
 
== Solution 2==
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There are three cases where the product of the numbers is prime. One dice 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 <math>\dbinom{12}{1}</math> = <math>12</math>. The total number of permutations is <math>6^12</math>. The probability the product is prime is therefore \frac{12}{3}
+
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 <math>\dbinom{12}{1}</math> = <math>12</math>. The total number of permutations is <math>6^12</math>. The probability the product is prime is therefore \frac{12}{3}
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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 06:42, 1 June 2021

Problem

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

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

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 $\dbinom{12}{1}$ = $12$. The total number of permutations is $6^12$. The probability the product is prime is therefore \frac{12}{3}

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AMC 10 Problems and Solutions

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