Q: Point (A,B) is randomly selected inside the circle x^2+y^2=1. What is the probability that A>B>0?
Ans:[spoiler]1/8[/spoiler]
I was trying to figure out the solution for this problem but could not get ans: 1/8. Since the radius of the circle is 1, only when point (A,B) is in 1st quadrant shall we get A and B >0. So shouldn't the probability be 1/4 ? How can we get the probability 1/8 to this problem?
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The equation for a circle with its center at the origin and a radius of r is as follows:Gmat Bond wrote:Q: Point (A,B) is randomly selected inside the circle x^2+y^2=1. What is the probability that A>B>0?
x² + y² = r².
Thus, x² + y² = 1 is a circle with its center at the origin and a radius of 1.
Every point (x, y) below the line y=x is such that y < x.
Thus, every point (A, B) in the shaded region above is such that A > B > 0.
The shaded region constitutes 1/8 of the circle.
Thus, if point (A, B) is randomly selected inside the circle, the probability that A > B > 0 is equal to [spoiler]1/8[/spoiler].
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