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?