A familiar probability question

[mesarmath]

hi,

i have been studying for GRE subject

and i saw this question but i could not solve it

x , y and z are selected independently and at random from the interval [0,1], then the probability that x is bigger than y*z is ?

the answer is 3/4

but i want to know how? , i guess it should be solved by double integral.

thanks in advance for any help

edit: i just figured out that $\int_{0}^{1} \int_{0}^{1} (1-yz) dy dz = 3/4$
but i could not figure out how the area above the curve x=yz represents that probability geometrically ?

 

[tiny-tim]

Hi mesarmath!

… but i could not figure out how the area above the curve x=yz represents that probability geometrically ?

(it ain't a curve, it's a surface )

Because for each value y and z, the proportion of x > yz is the proportion below the surface, which is yz/1, and the proportion of x > yz is the proportion above the surface, which is (1 - yz)/1.

 

[mesarmath]

Hi mesarmath!


(it ain't a curve, it's a surface )

Because for each value y and z, the proportion of x > yz is the proportion below the surface, which is yz/1, and the proportion of x > yz is the proportion above the surface, which is (1 - yz)/1.

thanks

so we were looking for a volume,
curve was the thing that makes me confused

thanks again :)