Question

Person A choses a random real number a from 0 to 2 , Person B choses a random real number b from 0 to 2

What is the probability that | a - b | > 1/3

in words: what is the probability that the absolute value of the difference is higher than 1/3

Answer 1

Question 24849Person A choses a random real number $a$ from 0 to 2 , Person B choses a random real number $b$ from 0 to 2 What is the probability that $|a - b| \geq 1/3$ .

Solution. In fact, we get some random point in the square $[0,2] \times [0,2]$, so here we have deal with geometric probability and the probability above is the probability that our point $(a,b)$ is not in the traingles(defined by vertexes) $(0,0), (0,1/3), (1/3,0), (1,1), (1,2/3), (2/3,1)$. We need to compute the area of the square without those triangles $4 - 2 \cdot 1/2(1/3)^2$ equals $E[\mathbb{I}_{|a-b| \geq 1/3}] = \int_{(a,b) \in [0,2]^2, |a-b| \geq 1/3} 1/4 \, da \, db = \frac{4 - 2 \cdot 1/2(1/3)^2}{4} = \frac{35}{36}$.