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Point (A,B) is arbitrarily selected inside of a circle X^2 +

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Senior Manager
Senior Manager
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Joined: 30 Oct 2004
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Kudos [?]: 8 [0], given: 0

Point (A,B) is arbitrarily selected inside of a circle X^2 + [#permalink] New post 18 Sep 2005, 13:08
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A
B
C
D
E

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Point (A,B) is arbitrarily selected inside of a circle X^2 + Y^2 = 1.
What is the probability that A>B>0 ?
A) 1/8
B) 7/48
C) 1/6
D) 5/24
E) 1/4
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-Vikram

Senior Manager
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Location: Las Vegas, NV
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 [#permalink] New post 18 Sep 2005, 15:43
I'm going for 1/8

If I'm right I'll explain my reasoning
Senior Manager
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 [#permalink] New post 18 Sep 2005, 18:08
I also think it's 1/8.

1/4 of the circle lies in each quadrant, because the center is the origin. So there's a 1/4 chance that both A and B are positive.

Then, within that quadrant, half of the points have A>B and the other half have B>A.

So the probability logically should be 1/2 x 1/4 = 1/8.

What's the OA?
Senior Manager
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Joined: 30 Oct 2004
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 [#permalink] New post 18 Sep 2005, 18:14
Ah! Now that was a simple explanation. Thanks CLF.
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-Vikram

  [#permalink] 18 Sep 2005, 18:14
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Point (A,B) is arbitrarily selected inside of a circle X^2 +

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