Toolkit 70

Geometric Approach in Probability

0\le x\le 1

P\!\left(\tfrac{1}{3}<x<\tfrac{1}{2}\right)=\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{1}{6}

0\le x,y\le 1

P(|x-y|\ge \tfrac{1}{2})=2\times\frac{\tfrac{1}{2}\times\tfrac{1}{2}}{2}=\tfrac{1}{4}

0\le x,y,z\le 1

P(x+y+z\le 1)=\frac{\tfrac{1}{2}\times\tfrac{1}{2}\times\tfrac{1}{2}}{3}=\tfrac{1}{6}

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