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What is the probability that P>Q>0

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by goelmohit2002 » Thu Sep 17, 2009 4:40 am
Hi All,

if x^2 + y^2 = r^2 is the equation of the circle with centre as origin and radius as r.

Point (P,Q) is randomly selected inside the above circle. What is the probability that P>Q>0 ?

[spoiler]OA = 1/8. Can someone please tell how the answer is 1/8....since on line y=x, the above inequality that is P>Q>0 is not satisfied.
[/spoiler]

Thanks
Mohit
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by goelmohit2002 » Fri Sep 18, 2009 6:14 am
experts kindly share your opinion please !!!
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by caspermonday » Wed Sep 23, 2009 8:31 am
Beats me...

My intuition is that the probability should be less than 3/4 because there is one quarter in a circle where P that lies of X-axis is always greater than Q that lies on Y-axis (the quadrant where X>0, Y>0).

Evidently, that in quadrant where x<0, y>0, P can never exceed Q. Hence, probability is more that 1/4.

And in quadrants where x<0, y<0 and x>0, y>0 there are variants where P can be > Q. The problem is how to calculate the chances.

I keep on thinking. Thanks for the problem.
Any ideas, guys?
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by caspermonday » Wed Sep 23, 2009 11:24 am
Have a look at this. Guys seem to have solved your prob:
https://gmatclub.com/forum/probability-o ... 8e71ef576e
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