2011 AMC 12A Problems/Problem 10

Revision as of 22:41, 9 February 2011 by Cortana (talk | contribs) (Solution)

Problem

Solution

It is clear that only a diameter of $2$ and $3$ would result in the circumference being larger than the radius.

For $2,$ the radius is $1$ so $2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi$ The area is $\pi r^2 \rightarrow \pi 1^2 \right arrow \pi$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) Thus, $2\pi > \pi$ so we need snake eyes or $2$ $1's$ and the probability is $\dfrac{1}{6} \cdot  \dfrac{1}{6} \rightarrow \dfrac{1}{36}$

By the same work using diameter of $3$, we find that the circumference is greater than the area. So $3$ would be found by rolling a $1$ and a $2$ so the probability is $\dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36}$.

Thus, those are the only two cases however there are $2$ ways to roll a $3$ with $2$ die so the probability is $\dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} \rightarrow \dfrac{3}{36} \rightarrow \dfrac{1}{12} \rightarrow \boxed{B}$

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 12 Problems and Solutions
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