# 2005 AMC 10B Problems/Problem 12

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 