2005 AMC 10B Problems/Problem 12

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

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2005 AMC 10B Problems/Problem 12

Contents

Problem

Solution

Solution 2

See Also