2010 AMC 12B Problem 18

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}

Answer: C

Concepts:geometric probability random walk

Difficulty rating: 2030

Small hint:

Anchor the middle jump on a fixed segment; the start and end depend on two independent angles

Big hint:

Reduce the condition to a region in the plane of those two angles, then compare areas

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{\frac{\pi^2}{2}}{2\pi^2}=\dfrac14.

Thus, the correct answer is C.