HELP geometric probability: area of a square and conditional probability

SusanCher89

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

No clue.

3. 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 teh 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 probabilty 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

Not "usually"- "always"

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

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 [itex]0\le x\le 1[/itex], [itex]0\le y\le 1[/itex]. 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?