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

Revision as of 18:02, 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 == (Really quick) The only way to get a prime number is to roll $1$ on all other $11$ dices and roll $1$ prime number there are $3$ different prime numbers up to $6$ and $12$ positions for each prime number. we can easily find the total number of cases which is $\frac{1}{6^{12}$ (Error compiling LaTeX. Unknown error_msg) because there are $6$ faces and $12$ dices we multiply $3$ by $12$ which gets us $36$ and multiply $36$ by $\frac{1}{6^{12}$ (Error compiling LaTeX. Unknown error_msg) and we get $\frac{36}{6^{12}$ (Error compiling LaTeX. Unknown error_msg) so therefore the answer is \boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}.$

-LittleWavelet

@@ Line 18: / Line 18: @@
 -Benedict T (countmath1)
+== Solution 4 == (Really quick)
+The only way to get a prime number is to roll <math>1</math> on all other <math>11</math> dices and roll <math>1</math> prime number there are <math>3</math> different prime numbers up to <math>6</math> and <math>12</math> positions for each prime number. we can easily find the total number of cases which is <math>\frac{1}{6^{12}</math> because there are <math>6</math> faces and <math>12</math> dices we multiply <math>3</math> by <math>12</math> which gets us <math>36</math> and multiply <math>36</math> by <math>\frac{1}{6^{12}</math> and we get <math>\frac{36}{6^{12}</math> so therefore the answer is \boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}.$
+-LittleWavelet
 == See Also ==
 {{AMC10 box|year=2005|ab=B|num-b=11|num-a=13}}
 {{MAA Notice}}

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