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

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Senior Manager
Joined: 30 Oct 2004
Posts: 286
Followers: 1

Kudos [?]: 12 [0], given: 0

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

-Vikram

Senior Manager
Joined: 15 Aug 2005
Posts: 257
Location: Las Vegas, NV
Followers: 2

Kudos [?]: 0 [0], given: 0

I'm going for 1/8

If I'm right I'll explain my reasoning
Senior Manager
Joined: 27 Aug 2005
Posts: 332
Followers: 2

Kudos [?]: 40 [0], given: 0

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
Joined: 30 Oct 2004
Posts: 286
Followers: 1

Kudos [?]: 12 [0], given: 0

Ah! Now that was a simple explanation. Thanks CLF.
_________________

-Vikram

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