Hi Max: Please take a look at this question:** https://web2.0calc.com/questions/pls-halp_27**

Guest Apr 16, 2022

#1**+2 **

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let \(D\) be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let \(\Omega\) be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and \(D\).

Basically, the probability is equal to \(\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}\).

I believe that is the answer.

Edit: Edited. This should be correct now.

MaxWong Apr 16, 2022

#1**+2 **

Best Answer

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let \(D\) be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let \(\Omega\) be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and \(D\).

Basically, the probability is equal to \(\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}\).

I believe that is the answer.

Edit: Edited. This should be correct now.

MaxWong Apr 16, 2022