Dartboard Geometric Probability and Expected Value

Problem

A dartboard has three scoring zones: a bullseye of radius $2$ worth $50$ points, a middle ring from radius $2$ to $5$ worth $20$ points, and an outer ring from radius $5$ to $10$ worth $5$ points; find the probability of landing in each zone and the expected score per throw.

Step 1: Compute the area ratios

The probability of hitting a zone is its area divided by the total board area.

For the bullseye,

$$\frac{\pi (2)^2}{\pi (10)^2}=\frac{4}{100}=\frac{1}{25}.$$

For the middle ring,

$$\frac{\pi (5)^2-\pi (2)^2}{\pi (10)^2}=\frac{25-4}{100}=\frac{21}{100}.$$

For the outer ring,

$$\frac{\pi (10)^2-\pi (5)^2}{\pi (10)^2}=\frac{100-25}{100}=\frac{75}{100}.$$

Step 2: Calculate the expected score

Multiply each score by its probability and add the results:

$$50\left(\frac{4}{100}\right)+20\left(\frac{21}{100}\right)+5\left(\frac{75}{100}\right).$$

That gives

$$2+4.2+3.75=\frac{995}{100}=9.95.$$

Answer

The probabilities are $\frac{4}{100}$, $\frac{21}{100}$, and $\frac{75}{100}$, and the expected score per throw is $9.95$ points.