Question

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

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Solution:

This is a continuous (geometric) probability. Anchor the second jump from $P=(0,0)$ to $Q=(1,0),$ and let $\alpha,\beta$ be the directions of the first and third jumps, so the start is $A=(\cos\alpha,\sin\alpha)$ and the end is $B=(1+\cos\beta,\sin\beta).$

Taking $0\le\alpha\le\pi$ and $0\le\beta\le2\pi,$ the requirement $AB\le1$ holds exactly when $\alpha\le\beta\le\pi.$

In the $\alpha\beta$ -rectangle of area $2\pi^2,$ the favorable region is a triangle of area $\tfrac{\pi^2}{2},$ so the probability is $\dfrac{\pi^2/2}{2\pi^2}=\dfrac14.$

Thus, the correct answer is C.

2010 AMC 12B Problem 18

Solution:

Problem 18 in Other Years