Probability of rectangles area being less than 0.5 w/ total length of sides = 2 Question: A random point splits the interval [0,2] in two parts. Those two parts make up a rectagle. Calculate the probability of that rectangle having an area less than 0.5.
So, this is as far as I've gotten; I've set 
$X := length of one side$.
Then $Y := length of other side$. Both of the variables are uniformally distributed but Y is dependent on X. But if I set $Y = (2-X)$, then I only have to figure out the probability of $P(XY<0.5) = P(X(2-X)<0.5)$.
Here is where I can't go any further. Any ideas?
 A: If you solve $X(2-X)<0.5$ for $X$ you get $X>1+\frac{1}{\sqrt{2}}$ and $X<1-\frac{1}{\sqrt{2}}$
The cdf of $X$ is
$F(x)=\begin{cases}0, \ \ x<0 \\ \frac{x}{2}, \ \  \ 0\leq x\leq 2 \\  1, \ \ x>0\end{cases}$
It is $F(X>x)=1-F(X<x)$
Therefore you have to calculate $F(1-\frac{1}{\sqrt{2}})+(1-F(1+\frac{1}{\sqrt{2}}))$
Remark
The distibution of $X$ is symmetric around the mean 1. 
Thus $F(1-\frac{1}{\sqrt{2}})+(1-F(1+\frac{1}{\sqrt{2}}))=2\cdot F(1-\frac{1}{\sqrt{2}})$ and the result reduces to the expression $1-\frac{1}{\sqrt{2}}\approx 29.29\%$
A: Assume $X\sim U(0,2)$, that is the cut point is uniformly distributed between $0$ and $2$. Then if $X=x$ the area $A=a=x(2-x)$ and this is greater than a half between the roots of $x(2-x)=1/2$ which are $1\pm\sqrt{1/2}$.
The pdf of $X$ is $f_X(x)=1/2$ for $x\in (0,2)$ and zero otherwise. So the probability that$A\ge 1/2$ is:
$$
P(\mbox{A} \ge 1/2)=\int_{x=1-\sqrt{1/2}}^{1+\sqrt{1/2}}\frac{1}{2}\;dx=\frac{1}{\sqrt{2}}\approx 0.707
$$
So the probability that the area is less than $1/2$ is 
$$
P(\mbox{A}< 1/2)=1-P(\mbox{A}\ge 1/2)=1-\frac{1}{\sqrt{2}}\approx 0.293
$$
