Question

Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\dfrac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\dfrac{1}{12}.$ What is the least possible number of faces on the two dice combined?

$16$

$17$

$18$

$19$

$20$

Answer: B

Concepts:dice (probability)optimization

Difficulty rating: 2120

Solution:

Let the dice have $a\le b$ faces. Since both have at least $6$ faces, a sum of $7$ occurs in exactly $6$ ways, so a sum of $10$ occurs in $6\div\tfrac34=8$ ways.

The number of ways to roll $10$ is $\min(a,9)$ $-\max(1,10-b)+1=8.$ A sum of $12$ has probability $\tfrac{1}{12},$ so it occurs in $\tfrac{ab}{12}$ ways.

Having $8$ outcomes for a sum of $10$ requires $a\ge8$ and $b\ge9,$ so necessarily $a+b\ge17.$ For $(a,b)=(8,9),$ a sum of $10$ has $8$ outcomes, while a sum of $12$ has the $6$ outcomes with the first die showing $3,4,\ldots,8.$ Since $6/(8\cdot9)=1/12,$ both conditions hold and the lower bound $17$ is attained.

Thus, the correct answer is B.

2021 AMC 12B Spring Problem 19

Answer: B

Difficulty rating: 2120

Solution:

Problem 19 in Other Years