Toolkit 70Geometric Approach in Probability1 dimension0≤x≤10\le x\le 10≤x≤1P (13<x<12)=12−13=16P\!\left(\tfrac{1}{3}<x<\tfrac{1}{2}\right)=\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{1}{6}P(31<x<21)=21−31=612 dimensions0≤x,y≤10\le x,y\le 10≤x,y≤1P(∣x−y∣≥12)=2×12×122=14P(|x-y|\ge \tfrac{1}{2})=2\times\frac{\tfrac{1}{2}\times\tfrac{1}{2}}{2}=\tfrac{1}{4}P(∣x−y∣≥21)=2×221×21=413 dimensions0≤x,y,z≤10\le x,y,z\le 10≤x,y,z≤1P(x+y+z≤1)=12×12×123=16P(x+y+z\le 1)=\frac{\tfrac{1}{2}\times\tfrac{1}{2}\times\tfrac{1}{2}}{3}=\tfrac{1}{6}P(x+y+z≤1)=321×21×21=61Related ProblemsCoreAMC 10B/12B 2024 (Problem 14/9)AMC 12A 2024 (Problem 20)AMC 10A 2025 (Problem 25)
