Probability A Tough One-Please Help!!

[apoorva.srivastva]

In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point
falls into the square region, what is the probability that the ordinates of the point
(x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

Kindly help with the reasoning!!

[thephoenix]

IMO B,

[satish.nagdev]

IMO C, for any x>=0.75 and y>=0.75 equation is satisfied, the square has side of length 2 so area will be 4, so considering given options I considered a circle with center (0,0) and radius 1. any point out of this circle (area = pi * r^2, here r=1) will satisfy the equation
hence 4 - pi

[rohan_vus]

IMO A..

Area of square = 2 * 2 = 4.
Area outside circle ( x^2 + y^2 = 1) = 4 - pi
So probability = Area outside circle but within square/ Area within square
==> 4 - pi/4 ==> 1 - pi/4

[apoorva.srivastva]

OA is A

[sanju09]

In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point
falls into the square region, what is the probability that the ordinates of the point
(x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

Kindly help with the reasoning!!

It's a typical problem of geometric probability in which the sample space is given by the total area on target, which is a square of side 2 in our case, hence total area = 4; and the favorable area would definitely be out of the circle obtained within the square bearing the equation x^2 + y^2 = 1. Hence, in our figure (see attachment), the red region is containing all points within the square ABCD that satisfy x^2 + y^2 > 1.

Go with A.