If P = (X, Y) is a point in the given square, then X and Y are i.i.d random variables each with distribution
[tex]\displaystyle P(X = x) = \begin{cases}\dfrac14 & \text{if } -2 \le x \le 2 \\ 0 & \text{otherwise}\end{cases}[/tex]
and so the joint density of X and Y is
[tex]\displaystyle P(X = x, Y = y) = \begin{cases}\dfrac1{16} & \text{if }-2 \le x \le 2 \text{ and } -2 \le y \le 2 \\ 0 &\text{otherwise}\end{cases}[/tex]
We want to find P(X² + Y² ≤ 1). Points that satisfy this inequality lie in the set
R = {(x, y) : -1 ≤ x ≤ 1 and -√(1 - x²) ≤ y ≤ √(1 - x²)}
but we can more easily describe the region in polar coordinates by setting
x = r cos(t) and y = r sin(t)
so that the set R is identical to
R' = {(r, t) : 0 ≤ r ≤ 1 and 0 ≤ t ≤ 2π}
Integrate the joint density over R' :
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \iint_R \frac1{16} \, dx \, dy[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \iint_{R'} \frac r{16} \, dr \, dt[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \int_0^{2\pi} \int_0^1 \frac r{16} \, dr \, dt[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \int_0^{2\pi} \frac{1^2 - 0^2}{32} \, dt[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \frac1{32} \int_0^{2\pi} dt[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \frac{2\pi-0}{32}[/tex]
[tex]\displaystyle P(X^2 + Y^2 \le 1) = \boxed{\frac{\pi}{16}}[/tex]
