Question

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

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

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

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

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

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

Solution:

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 $12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} = 6 \cdot \left(\dfrac16\right)^{11} = \left(\dfrac16\right)^{10}.$

Thus, E is the correct answer.

2005 AMC 10B Problem 12

Solution:

Problem 12 in Other Years