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AMC 10 & 12

Geometric Probability

When outcomes are points chosen at random from a region, probability becomes a ratio of measures — length, area, or volume.

Ordinary probability counts favorable outcomes out of total outcomes. Geometric probability replaces counting with measuring, when the outcomes form a continuous region rather than a finite list:

P=measure of favorable regionmeasure of total regionP = \dfrac{\text{measure of favorable region}}{\text{measure of total region}}

where “measure” means length for problems on a segment, area for problems in a 2D region, and volume for problems in 3D — matching the dimension of the sample space.

The one thing to check carefully before writing this ratio down: make sure the favorable region is actually entirely contained within the total region, and that you've correctly identified both regions' boundaries — a surprising number of errors come from mismatched or mis-shaped regions rather than the arithmetic itself.

Worked example

Problem: A point is chosen at random inside a square of side length 4. A circle of radius 1 is inscribed at the center of the square. Find the probability the point lands inside the circle.

Solution: The total region is the square, with area 42=164^2=16. The favorable region is the circle, with area π(1)2=π\pi(1)^2=\pi. So:

P=π16P = \dfrac{\pi}{16}

Practice problems

1.A point is chosen randomly on a segment of length 10. Find the probability it lands within 3 units of the segment's midpoint.

2.A point is chosen randomly inside a circle of radius 6. Find the probability it lands within 2 units of the center.

3.A point is chosen randomly inside a square of side 6, which contains an inscribed circle (tangent to all four sides). Find the probability the point lands outside the circle.

4.Two points are chosen randomly and independently on a segment of length 1. Setting up coordinates (x,y)(x,y) in the unit square, describe (without fully solving) how you'd find the probability the two points are within 12\tfrac12 of each other.