Geometry
Probability and statistics
Geometric probability (area-based)
20 practice questions
2 video lessons
Theory + worked examples
Theory
Geometric probability uses a ratio of measures for a uniformly random point:
\[P=\dfrac{\text{favorable measure}}{\text{total measure}},\]
where the measure is length on a segment, area in a region, or volume in a solid.
It assumes the point is equally likely anywhere in the whole region — a uniform distribution.
The chance of landing in the circle is the area ratio \(\pi/4\).
Geometric probability as a ratio of measures.
The area (or length) ratio:
\[P=\dfrac{\text{favorable area}}{\text{total area}}\]
Use the same kind of measure top and bottom — area over area, or length over length.
How to find a geometric probability
- Identify the total region and the favorable region.
- Measure each (length, area, or volume).
- Divide favorable by total.
- Keep the measures the same type and simplify.
Example 1 — Circle in a square
A dart lands at random in a square of side \(2r\) with an inscribed circle of radius \(r\). Find \(P(\text{inside the circle})\).
Solution
Divide the circle's area by the square's area.
| \(P\) | \(=\) | \(\dfrac{\pi r^2}{(2r)^2}\) |
| \(=\) | \(\dfrac{\pi r^2}{4r^2}=\dfrac{\pi}{4}\approx0.785\) |
Example 2 — Length on a segment
A point is chosen at random on a \(10\)-cm segment. Find the probability it lands within \(3\) cm of the left end.
Solution
Use the ratio of lengths.
| \(P\) | \(=\) | \(\dfrac{3}{10}=0.3\) |
Example 3 — Shaded region
A \(10\times10\) board has a \(4\times4\) shaded square. Find \(P(\text{a random point is shaded})\).
Solution
Divide the shaded area by the total area.
| \(P\) | \(=\) | \(\dfrac{4\times4}{10\times10}\) |
| \(=\) | \(\dfrac{16}{100}=0.16\) |
Example 4 — Annulus (ring)
A target has an outer radius \(6\) and a bullseye radius \(2\). Find \(P(\text{hitting the bullseye})\), assuming a random hit on the target.
Solution
Divide the bullseye area by the whole target area.
| \(P\) | \(=\) | \(\dfrac{\pi(2)^2}{\pi(6)^2}\) |
| \(=\) | \(\dfrac{4}{36}=\dfrac19\) |
Common pitfalls
Match the measures: area over area, not area over length.
Use the correct total region, not just part of it.
Assumes uniform randomness — equally likely everywhere.
Frequently asked questions
What is geometric probability?
The probability a random point lands in a region, found as the ratio of the favorable measure to the total measure.
How do you compute it?
Divide the favorable area (or length or volume) by the total.
What assumption does it make?
That the point is uniformly random — equally likely anywhere in the region.
What is the probability of a dart in an inscribed circle?
\(\dfrac{\pi r^2}{(2r)^2}=\dfrac{\pi}{4}\approx0.785\).
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