Originally Posted by

**Moo** Hello,

In fact, you're looking for this probability :

$\displaystyle \mathbb{P}(0<x<M)$ (where M is the maximum **area** in sq. ft. of the pizza), which is, by definition of the pdf :

$\displaystyle \mathbb{P}(0<x<M)=\int_0^M f(x) ~dx$

Assuming that the box is a **square**, you're correct in finding the value $\displaystyle M=0.25 \pi$

So now just calculate $\displaystyle \int_0^{0.25 \pi} 6(x-x^2) ~dx$ (because $\displaystyle 0.25 \pi <1$, so you're in an interval contained in (0,1) and hence $\displaystyle f(x)=6(x-x^2)$)

and you'll have the proportion (or probability) you're looking for.