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

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

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19 Nov 2007, 06:50
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

1/4
1/3
1/2
3/5
2/3
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19 Nov 2007, 08:06

x^2 + y^2 = 1 is a circle with the center at the origin(0,0) and a radius of 1.

You will find that in three of the 4 quarters of the circle, b > a+1 is just not possible. That leaves only one option.

Only the region to the left of the y-axis and above x-axis can satisfy b > a+1.

Therefore probability is 1/4.
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19 Nov 2007, 08:29
bs,

would you please be so kind to explain how you arrived with 1/4 from b> a+1?

I have a real issue with this problem and need a detailed explanation for this one.

take care
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19 Nov 2007, 09:16
tarek99 wrote:
bs,

would you please be so kind to explain how you arrived with 1/4 from b> a+1?

I have a real issue with this problem and need a detailed explanation for this one.
take care

To understand the logic, you will need to draw a circle with the center at origin(0,0) and a radius of 1.
You will find that the circle intersects the x-axis at (-1,0) and (1,0) and the y-axis at (0,1) and (0,-1).

Understand that -1<=a<=1 and -1<=b<=1. i.e. For the constraint to be satisfied, 'a' has to be negative and 'b' has to be positive. This is possible only in the second quadrant out of the four quadrants.. This is the shortest simplest explanation I can think of.

In case, you are unable to understand the above logic, here's an alternate explanation.

Assume the quadrants in the usual anti-clockwise manner.

To get a better understanding, you can use the values a = (+-) (1/ sqrt 2) and b = (+-) ( 1/sqrt 2) as test values. The signs of 'a' and 'b' will change according to the quadrant you are considering. For e.g: in the fourth quadrant, consider a = 1/(sqrt 2) and b = -1/(sqrt 2).

You will see that in the first quadrant (x>0, y>0), no value on the circle can possibly satisfy b > a+1.

The same holds for the third quadrant (x<0 , y<0). (Both 'a' and 'b' will be
negative)

And for the fourth quadrant (x>0,y<0) (where 'b' will always be less than 'a').

The given constraint can only be satisfied in the second quadrant(x<0,y>0). e.g: In the second quadrant, a = -1/(sqrt 2) and b = 1/(sqrt 2) can satisfy the given constraint.

This is a lengthy explanation but using this logic, this problem can be solved really fast as well. Once you have ruled out three quadrants, one look at the options leaves you with only one correct choice.
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19 Nov 2007, 14:11
jbs,

thanks, i really appreciate your effort. that really helped!

take care[/list]
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19 Nov 2007, 14:39
simple drawing
Attachments

GC55828.gif [ 6.05 KiB | Viewed 771 times ]

CEO
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20 Nov 2007, 09:41
nice explanation jbs.

OA is A.
20 Nov 2007, 09:41
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