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Probability A Tough One-Please Help!!

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by apoorva.srivastva » Sun Dec 20, 2009 6:34 am
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!!
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by thephoenix » Sun Dec 20, 2009 7:26 am
IMO B,
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by satish.nagdev » Sun Dec 20, 2009 8:02 am
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
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by rohan_vus » Sun Dec 20, 2009 8:06 am
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
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by apoorva.srivastva » Sun Dec 20, 2009 8:14 am
OA is A
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by sanju09 » Thu Dec 24, 2009 4:36 am
apoorva.srivastva wrote: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.
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x^2 + y^2 is greater than 1.docx
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