## GR9367 40

Forum for the GRE subject test in mathematics.
Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### GR9367 40

GR9367, Question 40

x,y,z are selected independently and at random from interval [0,1]. Find the probability of x >= yz.

I googled and found some explanation in this forum and on the internet regarding this problem. Usually, this is done by triple integrating the triple probability density function.

There was a suggestion that goes: "E(y) = E(z) = 1/2, and E(yz) = E(y)E(z) = 1/4. So the answer is P(x >= 1/4) = 3/4", which I don't think is correct. (or am I missing something?)

Is there a simpler way to deduce the answer of 3/4?

Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: GR9367 40

This is solved:

P(x >= yz) = 1 - yz for all y,z.

So we just double integrate (1 - yz) over the area 0 < y < 1 and 0 < z < 1.

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