Problem

Let $(X, Y)$ have the uniform distribution on $[0, 1] × [0, 1]$. Find the probability of $X+Y \geq 1/2$.

Solution

The boundary separating $X+Y$ can be plotted at $Y=1/2 - X$

The region we're after:

This can be viewed as the unit square minus the triangle beneath $Y=1/2 - X$.

$P(X + Y) \geq 1/2 = 1 - P(X+Y) < 1/2$

or 1 minus the area beneath the line. It's pretty easy to see visually that the area of this triangle will be $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}= \frac{1}{8}$ making the probability we're after $1- \frac{1}{8} = \frac{7}{8}$.

We can also compute this with an integral by setting the bounds appropriately:

$\int_0^{1/2} \int_0^{1/2 - X} f_{X,Y} dy dx$

where $f_{X,Y} = 1$

$\int_0^{1/2} \int_0^{1/2 - x} 1 dy dx$
=$\int_0^{1/2} 1/2 - x dx$
=$\frac{x}{2} - \frac{x^2}{2} \rvert_{0}^{1/2}$
=$\frac{1}{4} - \frac{1}{8}$
=$\frac{1}{8}$

And again, we subtract this from 1 to get the answer $\frac{7}{8}$.

Sources

  • 36-700 hw2 problem 5