## probability

Forum for the GRE subject test in mathematics.
Posts: 24
Joined: Sun Oct 05, 2008 1:41 am

### probability

I know these problems were posted before, but I found a more intuitive approach to solving them that I think might be helpful...

9747: Let x and y be uniformly distributed random variables on [0,1] the probability that the distance between x and y is less than 1/2 is...

To solve this we note that the restriction requires |x-y| < 1/2, so that x - y < 1/2 or x-y > -1/2. These give y>x-1/2 and y<x+1/2. The solution is to graph the lines y=x-1/2 and y=x+1/2 in the square region [1,0]x[0,1] and calculate the area between them. The probability that |x-y| < 1/2 is equal to this area, which is 3/4.

9340: If x,y and z are selected at random from the interval [0,1], then the probability that x >= yz is...

In this case, we graph the region x=yz in the cube [1,0,0]x[0,1,0]x[0,0,1] and then compute the area in the region for which x > yz. This is the area above the curve x=yz, which is the volume of the cube minus the volume integral of yz with all bounds going from 0 to 1. The probability that x >= yz is therefore equal to 1 - 1/4 = 3/4.
Last edited by EmLasker on Sat Oct 18, 2008 1:42 pm, edited 1 time in total.

Kastro
Posts: 10
Joined: Mon Oct 13, 2008 10:36 pm
The solution to 9340 is 3/4, not 1/4.

That said, I seem to have arrived at the following paradox, so if someone could point out the error in the following, that would be great:

Let x,y and z be selected at random from the interval [0,1]

Then P(x>y) = 1/2 and P(x>z) = 1/2

Clearly, if x > y or x > z then x > yz (b/c y,z < 1). Also, note that P(x>y) and P(x>z) are independent.

Then P(x>y OR x>z) = P(x>y) + P(x>z) - P(x>y)P(x>z) = 1/2 + 1/2 - 1/4 = 3/4

But there also exist values for y and z such that x < y, x < z and still x > yz (e.g. x = 1/2, y=z=3/5). Since there is a non-zero probability of this occurring, this suggests that P(x>yz) > 3/4.

Is doing math at 4am causing me to miss something really simple? What's wrong here?