The Community Garden Probability Model
A community garden is a 20 m by 30 m rectangle divided into a 1 m grid, giving 600 unit cells. Organizers throw a frisbee that lands uniformly at random on the garden. A circular flower bed of radius 4 m sits with its center at the garden's center. Separately, a rectangular shed 6 m by 8 m sits in one corner, aligned with the grid. A volunteer claims the probability the frisbee lands in the flower bed is about 0.25 because the bed's diameter (8 m) is close to a quarter of the garden's width. Another volunteer wants the probability the frisbee lands in either the flower bed or the shed (they do not overlap) and also the probability it lands in the shed given that it did NOT land in the flower bed.
1. Compute the probability the frisbee lands in the circular flower bed. Is the 0.25 claim reasonable? Show the reasoning.
Read the worked answer
The frisbee lands uniformly over the 600 m² garden (20 × 30). Geometric probability = favorable area / total area.
Flower bed area = πr² = π · 4² = 16π ≈ 50.27 m².
P(bed) = 16π / 600 ≈ 50.27 / 600 ≈ 0.0838, about 8.4%.
The 0.25 claim is not reasonable. It confuses a linear comparison (diameter 8 m vs. width 20 m, ratio 0.4) with an area ratio. Probability for a uniformly random point depends on AREA, not length. The bed covers only about 8.4% of the garden, not 25%. Also note the bed (radius 4, centered) fits entirely inside the 20×30 garden, so no clipping adjustment is needed.
2. Find the probability the frisbee lands in the flower bed OR the shed, and the conditional probability it lands in the shed given it did not land in the flower bed.
Read the worked answer
Shed area = 6 × 8 = 48 m². The regions do not overlap, so:
P(bed OR shed) = (16π + 48) / 600 ≈ (50.27 + 48) / 600 = 98.27 / 600 ≈ 0.1638, about 16.4%.
Conditional probability: P(shed | not bed) = P(shed AND not bed) / P(not bed). Since shed and bed are disjoint, P(shed AND not bed) = P(shed) = 48/600 = 0.08. P(not bed) = 1 − 16π/600 ≈ 1 − 0.0838 = 0.9162.
P(shed | not bed) ≈ 0.08 / 0.9162 ≈ 0.0873, about 8.7%.
Interpretation: knowing the frisbee missed the flower bed slightly raises the chance it is in the shed (from 8.0% to 8.7%), because ruling out the bed redistributes that probability across the remaining area. The increase is small because the bed is a small fraction of the garden.