Unit 13: Probability and Geometric Modeling · ~15–30 min
Objective
Find probabilities using length and area models, comparing a favorable measure to a total measure.
Probability without counting outcomes
So far, probability has meant counting outcomes: favorable outcomes over total outcomes. But some situations involve outcomes that aren't countable at all — a point landing anywhere along a segment, or a dart landing anywhere inside a region. Geometric probability handles this by comparing measures — lengths or areas — instead of counts.
Length probability model
When outcomes are equally likely to fall anywhere along a segment, the probability of landing in a particular part of it is the ratio of that part's length to the whole segment's length:
P(event) = (length of favorable segment) ÷ (length of whole segment)
A point chosen at random on the segment [0, 10] falls in the gold region with probability 4/10 = 2/5.
Area probability model
The same idea extends to two dimensions: if a point lands randomly anywhere inside a region, the probability it lands inside a smaller favorable region is the ratio of the two areas:
P(event) = (area of favorable region) ÷ (area of total region)
This is the model behind dartboards, target games, and "random point in a shape" problems — circles inside squares, sectors of circles, or one region carved out of another.
A dart landing randomly anywhere in the square is more likely to land outside the circle than inside it.
Worked Example 1 · Length model
ProblemA point is chosen at random on segment AB, which has length 24 cm. Point C is on the segment with AC = 9 cm. Find the probability the point lands within 9 cm of A.
1.The favorable region is segment AC, with length 9. The whole segment AB has length 24.
2.P = 9/24 = 3/8.
P(within 9 cm of A) = 3/8
Worked Example 2 · Area model with circles
ProblemA circular dartboard has radius 10 in. A bullseye circle at its center has radius 2 in. Assuming a dart lands randomly somewhere on the board, find the probability it lands in the bullseye.
1.Area of bullseye = π(2)² = 4π. Area of whole board = π(10)² = 100π.
2.P = 4π / 100π = 4/100 = 1/25.
P(bullseye) = 1/25 = 0.04
Worked Example 3 · Circle inscribed in a square
ProblemA square dartboard has side length 8 in, with an inscribed circle of radius 4 in touching all four sides. Find the probability a randomly landing dart lands inside the circle.
1.Area of circle = π(4)² = 16π. Area of square = 8² = 64.
2.P = 16π/64 = π/4 ≈ 0.785.
P(inside circle) = π/4 ≈ 0.785
Guided practice
1.A point is chosen at random on segment PQ of length 15 cm. Point R is on the segment with PR = 6 cm. Find the probability the point lands between R and Q.Show answer
RQ = 15 − 6 = 9, so P = 9/15 = 3/5.
2.A rectangular garden is 12 ft by 6 ft. A 2 ft by 3 ft flower bed sits inside it. If a leaf falls randomly somewhere in the garden, find the probability it lands on the flower bed.Show answer
Garden area = 12(6) = 72. Flower bed area = 2(3) = 6. P = 6/72 = 1/12.
3.A circular spinner of radius 6 in has a shaded sector with central angle 45°. Find the probability the spinner lands in the shaded sector. (Hint: compare the sector's angle to the full 360°.)Show answer
Since a sector's area is proportional to its central angle, P = 45/360 = 1/8.
4.A number line segment runs from 0 to 40. Find the probability that a randomly chosen point lands strictly between 12 and 28.Show answer
Favorable length = 28 − 12 = 16. Total length = 40. P = 16/40 = 2/5.
5.Explain why geometric probability uses length or area ratios instead of counting outcomes.Show answer
When outcomes form a continuous region — like any point on a segment or inside a shape — there are infinitely many individual outcomes, so they can't be counted one by one. Comparing the measure (length or area) of the favorable region to the measure of the whole region gives a meaningful probability even though the outcomes themselves can't be listed.