HELP geometric probability: area of a square and conditional probability

[SusanCher89]

Homework Statement

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).

Homework Equations

No clue.

The Attempt at a Solution

Apparently, according to the prof, the square need not be equilateral? And this is where I get stumped.

No clue here. Any help would be great.

 

[Mark44]

Try drawing a diagram of your square. BTW, a square is equlateral, so I don't know what your prof was talking about, or maybe you misunderstood him/her.

P(y < 1/2 | y > x) asks for the probability that a point's y coordinate is less than 1/2, given that the point is in the triangular region above and to the left of the line y = x. There is some geometry here that you can use.

 

[SusanCher89]

Thanks for your answer, Mark 44.

Let's assume that a square has equilateral sides (which it does, usually). That means that y=x! So P(y>x)= 0 Right!

Also, P(y<1/2) = .5, right??

I'm still pretty lost, any help is appreciated!

:)
Dania

 

[lanedance]

the sides of the square are given (and by definition equal), so i don't really understand the equilateral discussion...

anyway, the area of the square is 1
the probability of a point being in the square is 1

you shouldn't have to work too hard to convince yourself, that the probability of finding the point in a given region is in fact equal to the area of the region in this case

use that fact with the conditional probability equation to solve

 

[HallsofIvy]

Thanks for your answer, Mark 44.

Let's assume that a square has equilateral sides (which it does, usually).

Not "usually"- "always"

That means that y=x! So P(y>x)= 0 Right!

No, not right! (x, y) are coordinates of some point in the square, not the lengths of the sides.

Also, P(y<1/2) = .5, right??

If all points in the square are equally likely, yes. But you want the probability that y< 1/2 given that y> x so P(y< 1/2| y> x) is not necessarily 1/2.

[/quote]I'm still pretty lost, any help is appreciated!

:)
Dania[/QUOTE]
Draw a picture. To start with, of course, draw the square $0\le x\le 1$, $0\le y\le 1$. Now draw the line y= x. That will be a diagonal of the square. Requiring that y> x means we are in the upper half of that square, above the diagonal. Draw the line y= 1/2. Saying that y< 1/2 means we are below that line but still in the upper half of the square, above the diagonal. You should see that this area is a triangle. What is the area of that triangle? What percentage is it of the upper half of the square?