Hitting the bullseye by luck

Question

A dart lands at a uniformly random point on a circular board of radius 10 in. The central bullseye is the disk of radius 2 in. What is the probability the dart lands in the bullseye?

Accepted Answer

For a uniform point in a disk, probability is proportional to area, and area grows with the square of the radius. The bullseye and the board are concentric disks, so P(	ext{bullseye}) = \frac{\pi (2)^2}{\pi (10)^2} = \frac{4}{100} = \frac{1}{25} = 0.04 . Takeaway. Inside a disk, "within radius r" has probability \left(	frac{r}{R}ight)^2. The \pi cancels, and the square is why a bullseye of one fifth the radius covers only one twenty-fifth of the area.

Hitting the bullseye by luck

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