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