SusanCher89 Posts: 1, Reputation: 1 New Member #1 Oct 19, 2009, 03:09 PM
geometric probability, conditional probability with area of square
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 Posts: 8,076, Reputation: 723 Uber Member #2 Oct 20, 2009, 04:33 AM

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 Posts: 8,076, Reputation: 723 Uber Member #3 Mar 18, 2010, 07:25 AM
Originally Posted by 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:

$P(y<0.5 |y>x) = \frac{P(y<0.5 \cap y>x)}{P(y>x)}$

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

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