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GR9367 40

Posted: Wed Aug 22, 2012 12:35 pm
by Legendre
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?

Re: GR9367 40

Posted: Wed Aug 22, 2012 1:17 pm
by Legendre
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.