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

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Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

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

Set T consists of all points (x, y) such that x^2 + y^2 =1. [#permalink]  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
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

[#permalink]  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
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

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

[#permalink]  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 Â¼, 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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