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Manager
Joined: 14 May 2005
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Set T consists of all points (x, y) such that x^2 + y^2 =1. [#permalink]
 14 Oct 2006, 14:50
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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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SVP
Joined: 01 May 2006
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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.
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Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
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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.
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