geometric-probabilityprobability
Two numbers $x$ and $y$ are chosen at random within the unit interval $[0, 1]$. What is the probability that the sum of these is greater than one and that the sum of their squares is less than one?
The following must be complied with: $x+y>1$ and $x^2+y^2<1$. We plot, and there is an area between these two. The problem I have is in calculating that area.
Selecting two points $x$ and $y$ from the interval $[0, 1]$ is identical to selecting a single point from the unit square. For the sake of clarity, let us calculate the probability that $y/x$ is closest to an even integer, because that's just the slope. By symmetry, that is equal to the corresponding probability for the ratio $x/y$.
Divide the unit square into two portions, one below the line $y = x$, and one above it. In the bottom portion, points that qualify are below the line $y = x/2$; this section has area $A_l = 1/4$.
In the upper portion, points that qualify are in successively smaller (inverted) triangles with apex at the origin, and bases along the segment from $(0, 1)$ to $(1, 1)$. These bases run from $2/3$ down to $2/5$, then $2/7$ down to $2/9$, then $2/11$ down to $2/13$, etc. Their collective area is therefore
$$ A_u = \frac{1}{2} \Bigl( \frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\cdots \Bigr) = \frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\cdots = 1-\frac{\pi}{4} $$
using this well-known series.
The combined area is therefore $A_l+A_u = (5-\pi)/4 \doteq 0.46460$. Here's a diagram of the configuration:
Comment: You have a path to the answer; this is just a way to confirm you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $\mathsf{Unif}(0, 1).$ And you seek $P(X+Y \le 1, XY \le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting the answer by simulation is sometimes called 'Monte Carlo integration'. It is mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions). You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green 'nibble' from 1/2. (As @copperhat has suggested.)
Best Answer
You just have to compute the area of your blue region. It equals $$\int_0^1 \sqrt{1-x^2}\,{\rm d}x -\frac{1}{2}.$$And you don't even need to do calculus for that, geometrically you know that $$\int_0^1 \sqrt{1-x^2}\,{\rm d}x = \frac{1}{4} \cdot \pi \cdot 1^2 = \frac{\pi}{4}.$$The probability is just $$\frac{\pi}{4} - \frac{1}{2} = \frac{\pi-2}{4}.$$