Printable

2005 AMC 10B Problem 12

Problem only Back to the problem

2005 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Probability · difficulty 2 of 5 · about 3 min · position P11-15

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

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

Answer

Key insight

A prime product forces eleven 1's and a single die showing 2, 3 or 5; count 12 positions times 3 primes over 6^12.

Solution

A product of twelve positive integers is prime only if every factor except one equals $1$ , and the remaining factor is itself prime. On a die the primes are $2, 3, 5$ .

So a favorable roll consists of eleven $1$ 's and one prime. Choose which of the $12$ dice shows the prime ( $12$ ways) and which prime it shows ( $3$ ways): $12 \cdot 3 = 36$ favorable outcomes.

All $6^{12}$ ordered outcomes are equally likely, so
$P = \frac{36}{6^{12}} = \frac{6^2}{6^{12}} = \frac{1}{6^{10}}.$

The answer is $\boxed{\textbf{(E)}\ \left(\frac{1}{6}\right)^{10}}$ .

Why this works

"The product is prime" is a very rigid condition: it pins down every die except one. Once the favorable outcomes are described structurally, counting them is a product of two small choices. Keep the count in powers of $6$ so it simplifies against the denominator without arithmetic.

The trap

Forgetting the factor of 12 for which die shows the prime, or counting 1 as a prime.

Common mistakes

Forgetting the factor of 12 for which die shows the prime, or counting 1 as a prime.
Leaving the count as $\frac{3 \cdot 12}{6^{12}}$ and failing to notice $36 = 6^2$ , then picking a choice with the wrong exponent.

Techniques

Set up the equation/formula and compute; no special trick needed