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Set T consists of all points (x, y) such that x^2 + y^2 =1.

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Manager
Manager
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Joined: 14 May 2005
Posts: 85
Location: San Francisco
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Kudos [?]: 2 [0], given: 0

Set T consists of all points (x, y) such that x^2 + y^2 =1. [#permalink] New post 14 Oct 2006, 13:50
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Set T consists of all points (x, y) such that x^2 + y^2 =1. If point (a, b) is selected from set T at random, what is the probability that b > a + 1?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 3/5
(E) 2/3
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Joined: 01 May 2006
Posts: 1814
Followers: 8

Kudos [?]: 91 [0], given: 0

 [#permalink] New post 14 Oct 2006, 14:24
For me (A),

In an XY plan, x^2+y^2 = 1 represents a circle with a radius of 1 and the center 0(0,0).

The area of circle that is matching the inequality b > a+1 is in the cadran II. It's 1/4 of the circle (not completly actually but anyway).

Thus, the probability to select 1 point on this arc is 1/4.
Manager
Manager
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Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 09 Nov 2006, 21:53
Fig, thanks.

I think that the wordings of the question are wrong. The question must be that what is the probability that b >= a + b. Otherwise, the probability must be slightly less than ¼, since the two boundary points (0, 1) & (-1, 0) cannot be part of the selection. This question is from one of the Challenges.
  [#permalink] 09 Nov 2006, 21:53
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Set T consists of all points (x, y) such that x^2 + y^2 =1.

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