Day 5 · Three Rings¶

Edited by Charles

Probability · Area · Multinomial distribution

Problem¶

A circular target has three regions: an inner disk of radius 4 inches, a middle ring extending to radius 8, and an outer ring extending to radius 12. Arrows land independently and uniformly over the target's area.

With three arrows, what is the probability of exactly one hit in each region?
With \(3n\) arrows, what is the probability of exactly \(n\) hits in each region?

Core idea

Hit probabilities depend on area, not radial width. Then count the possible orders of the hits.

Short solution¶

1. Calculate the three probabilities. The total area is \(144\pi\):

\[p_1=\frac{4^2}{12^2}=\frac19,\qquad p_2=\frac{8^2-4^2}{12^2}=\frac13,\qquad p_3=\frac{12^2-8^2}{12^2}=\frac59.\]

2. Include every ordering of three arrows. One particular order, inner then middle then outer, has probability \(p_1p_2p_3\). There are \(3!=6\) such orders.

Part 1

\[\boxed{P=3!\cdot\frac19\cdot\frac13\cdot\frac59 =\frac{10}{81}\approx12.35\%}\]

3. Generalize to \(3n\) arrows. The number of arrangements containing \(n\) hits of each type is \((3n)!/(n!)^3\). Every such arrangement has the same probability.

Part 2

\[\boxed{P=\frac{(3n)!}{(n!)^3} \left(\frac19\right)^n\left(\frac13\right)^n\left(\frac59\right)^n}\]

Watch out

Equal radial widths do not imply equal areas. Multiplying the probabilities without the arrangement factor counts only one order.

Day 5 overview · Next question