2005 AMC 10B Problem 12

P11-15~3 minPrint

Basic Probability Basic Counting Primes Set up the equation/formula and compute; no special trick needed

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

Show solution without answering

Key insight

A prime product needs eleven dice showing 1 and one die showing 2, 3 or 5: 12 times 3 = 36 good outcomes out of 6 to the 12th.

Solution

1. What does a prime product look like? A prime can only be split as $1 \times$ itself. So if twelve numbers multiply to a prime, eleven of them must be $1$ and the last one must be the prime. For example, $1 \cdot 1 \cdots 1 \cdot 5 = 5$ works, but $2 \cdot 3 = 6$ does not.
2. Primes on a die. The faces are $1$ to $6$ ; the primes are $2, 3, 5$ . (Remember, $1$ is not prime.)
3. Count good outcomes. Choose which die shows the prime: $12$ ways. Choose which prime it shows: $3$ ways. The other eleven dice must all be $1$ : only $1$ way. Total: $12 \cdot 3 = 36$ good outcomes.
4. Count all outcomes. Each die has $6$ faces, so there are $6^{12}$ outcomes, all equally likely.
5. Probability. Notice $36 = 6^2$ , so
$\frac{36}{6^{12}} = \frac{6^2}{6^{12}} = \frac{1}{6^{10}} = \left(\frac{1}{6}\right)^{10}.$

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

Why this works

"The product is prime" is a very strict condition: it fixes every die but one. Once you describe the good outcomes in words ("eleven $1$ 's and one prime"), counting them is two small choices multiplied together. Writing $36$ as $6^2$ lets it cancel neatly with $6^{12}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques