[SusanCher89]

Chose a point at random in a square with sides 0<x<1 and 0<y<1. Let X be the x coordinate and Y be the y coordinate of the point chosen. Find the conditional probability P(y<1/2 / y>x).

[Unknown008]

Well, you can do it fairly easily with the help of a sketch. On graph paper, your square will have a vertice on the origin and two vertices on the x and y axes. Have dotted lines on the lines y=1 and x=1. Then, graph the line y=1/2 and y=x. I guess your '/' in your probability means 'and'. Then, find the area below the line y=1/2, which is above the line y=x.

Below y = 1/2 because the value has to be less that 1/2.
Above y = x because the values has to be larger than x.

Put the resulting area over the area of the square to have your probability.

[Unknown008]

Well, you can do it fairly easily with the help of a sketch. On graph paper, your square will have a vertice on the origin and two vertices on the x and y axes. Have dotted lines on the lines y=1 and x=1. Then, graph the line y=1/2 and and y=x. I guess your '/' in your probability means 'and'. Then, find the area below the line y=1/2, which is above the line y=x.

Below y = 1/2 because the value has to be less that 1/2.
Above y = x because the values has to be larger than x.

Put the resulting area over the area of the square to have your probability.

Oops, now that I learned about conditional probabilities, I see that his is wrong.
The way you would solve it using the method I described would give you 1/8, but the answer is actually 1/4.

This is because:



So, for anyone who'd like to solve it...