2011 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2011 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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Concepts:dice (probability)circle inequality

Difficulty rating: 1370

10.

A pair of standard $6$ -sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?

$\dfrac{1}{36}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{5}{18}$

Solution:

For diameter $d,$ area $\lt$ circumference means $\dfrac{\pi d^2}{4} \lt \pi d,$ i.e. $d \lt 4.$ Since $d \ge 2,$ this needs a sum of $2$ or $3.$

A sum of $2$ has probability $\dfrac{1}{36}$ and a sum of $3$ has probability $\dfrac{2}{36},$ totaling $\dfrac{3}{36} = \dfrac{1}{12}.$

Thus, the correct answer is B.

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