In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)

First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

\(x^2+y^2=1\) is an equation of a circle also centered at the origin, with radius 1 and the \(area=\pi{r^2}=\pi\).

We are told that the point is IN the square and want to calculate the probability that it's outside the circle (\(x^2+y^2>1\) means that the point is outside the given circle):



P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=\(4-\pi\)
Total number of possible outcomes is the area of the square (as given that the point is in the square) =\(4\)

\(P=\frac{4-\pi}{4}=1-\frac{\pi}{4}\)

Answer: A.

Hope it's clear.