2010 AMC 12B Problems/Problem 18

Revision as of 23:12, 25 January 2012 by Jhggins (talk | contribs) (Solution)

Problem

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

$\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}$

Solution

The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius 2 from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this larger circle, so you find the ratio of the two areas, which is $\boxed{\text{(C)} \frac {1}{4}}$.

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions