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by knight247 » Thu Sep 29, 2011 11:58 am
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 ?

OA [spoiler]1/8[/spoiler]
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by GmatMathPro » Thu Sep 29, 2011 12:13 pm
Draw the graph of the circle in the coordinate plane and draw the line y=x. We want to know the probability that both the x and y coordinates are positive and the x-coordinate is greater than the y-coordinate. Only the points in the first quadrant have their x and y coordinates both positive, and only the points to the RIGHT of the line y=x have bigger x-coordinates than y-coordinates. If you draw the picture it's really easy to see that this is 1/8 of the circle.
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by knight247 » Thu Sep 29, 2011 12:25 pm
Hey Pete,
Thanks for ur response Bro. But what about a point in the 4th quadrant? Any point in the 4th quadrant would have +ve x coordinate and -ve y coordinate. So wouldn't that also count as x>y?
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by GmatMathPro » Thu Sep 29, 2011 12:27 pm
Yes, but remember we want points (P,Q) with P>Q>0, which means P and Q are BOTH positive. In the 4th quadrant only P would be positive, but Q would be negative.
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by knight247 » Thu Sep 29, 2011 12:30 pm
Whoops..Ur right bro. I overlooked the >0 part. A light bulb just lit over my head...just like in ur profile pic...lolzzz...Thanks
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by GmatMathPro » Thu Sep 29, 2011 12:43 pm
Haha. Happy to help, my man.
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by GMATGuruNY » Thu Sep 29, 2011 1:25 pm
For an illustrated solution to a very similar problem, check here:

https://www.beatthegmat.com/probability-t90694.html
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