Randomly choose two numbers between 0 and 1. What is the probability that their average is no greater than 2/3 and no less than 1/3?

Guest May 8, 2022

#1**+2 **

We can view this geometrically. Let x and y be the chosen numbers. We can plot (x, y) inside the unit square with vertices (0, 0) and (1, 1).

The constraints are \(\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23\) and we are basically finding the probability that the point (x, y) falls into the region described by the compound inequality. We can do so by considering the area of the region enclosed by \(\dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23\) and the unit square.

Let \(\Omega = \left\{(x, y) \in [0, 1]^2 : \dfrac13 \leq \dfrac {x + y}2 \leq \dfrac23\right\}\). Here's a plot of \(\Omega\):

So the probability is just \(\operatorname{Area}(\Omega) = 1 - 2\times \dfrac{\dfrac23 \times \dfrac23}2 = \dfrac59\)

MaxWong May 8, 2022