2005 AMC 10B Problem 12

Question

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

\left(\dfrac{1}{12}\right)^{12}

\left(\dfrac{1}{6}\right)^{12}

2\left(\dfrac{1}{6}\right)^{11}

\dfrac52\left(\dfrac{1}{6}\right)^{11}

\left(\dfrac{1}{6}\right)^{10}

A product of positive integers is prime only when one factor is a prime and the rest are

1

Choose which die shows the prime, and which prime; the other

11

dice must show

1

The product is prime exactly when one die shows a prime (

2,

3,

or

5

) and the other eleven all show

1.

The probability that any single die is the prime one is

\dfrac36 = \dfrac12,

and each of the other eleven shows

1

with probability

\dfrac16.

Accounting for which of the twelve dice is prime, the probability is

\begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10}. \end{aligned}

Thus, E is the correct answer.

Problem 12 in other years