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M20 #12

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M20 #12 [#permalink] New post 18 Nov 2008, 03:30
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 \gt a + 1 ?

(C) 2008 GMAT Club - m20#12

* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{3}{5}
* \frac{2}{3}
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Re: M20 #12 [#permalink] New post 18 Nov 2008, 22:44
Is it 1/4?

b > a + 1 is represented by the part of circle that is in second quadrant.
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Re: M20 #12 [#permalink] New post 22 Nov 2008, 19:17
i think its quadrant I and II. so 1/2
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Re: M20 #12 [#permalink] New post 23 Nov 2008, 11:30
scthakur wrote:
Is it 1/4?

b > a + 1 is represented by the part of circle that is in second quadrant.

Hey, the OA is A 1/4.
I got that but just by pure guess work.
I could not understand how the statement b>a+1 covers a quarter of the entire circle. Could you please explan how you figured that out?
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Re: M20 #12 [#permalink] New post 23 Nov 2008, 18:30
scthakur wrote:
Is it 1/4?

b > a + 1 is represented by the part of circle that is in second quadrant.


I agree its quadarant II, in quadarant I it would be off the region covered by the circle..
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Re: M20 #12 [#permalink] New post 23 Nov 2008, 19:55
ventivish wrote:
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 \gt a + 1 ?

(C) 2008 GMAT Club - m20#12

* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{3}{5}
* \frac{2}{3}



I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a.
For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 1/4.
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Re: M20 #12 [#permalink] New post 17 Dec 2008, 08:45
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GMAT TIGER wrote:
ventivish wrote:
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 \gt a + 1 ?

(C) 2008 GMAT Club - m20#12

* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{3}{5}
* \frac{2}{3}



I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a.
For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 1/4.

GMATTIGER, the point from your example doesn't belong to Set T. a^2+b^2 \ne 1 in your example.

I'm attaching an image to make it clearer. Line AB has equation y=x+1, just like b = a +1. Any points above the line AB satisfy the inequality b \gt a +1. Thus we have \frac{1}{4} of all the points from Set T that satisfy the inequality.
Attachments

m20-12.png
m20-12.png [ 4.85 KiB | Viewed 1440 times ]


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Re: M20 #12 [#permalink] New post 05 Apr 2012, 22:41
dzyubam wrote:
GMAT TIGER wrote:
ventivish wrote:
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 \gt a + 1 ?

(C) 2008 GMAT Club - m20#12

* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{3}{5}
* \frac{2}{3}



I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a.
For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 1/4.

GMATTIGER, the point from your example doesn't belong to Set T. a^2+b^2 \ne 1 in your example.

I'm attaching an image to make it clearer. Line AB has equation y=x+1, just like b = a +1. Any points above the line AB satisfy the inequality b \gt a +1. Thus we have \frac{1}{4} of all the points from Set T that satisfy the inequality.



This a really good explaination......
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Re: M20 #12 [#permalink] New post 06 Apr 2012, 06:20
Expert's post
ventivish wrote:
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 \gt a + 1 ?

(C) 2008 GMAT Club - m20#12

* \frac{1}{4}
* \frac{1}{3}
* \frac{1}{2}
* \frac{3}{5}
* \frac{2}{3}


Look at the diagram below.
Attachment:
graph.php.png
graph.php.png [ 15.81 KiB | Viewed 865 times ]

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

This question is also discussed here: m20-76028.html and here: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html

Hope it helps.
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Re: M20 #12   [#permalink] 06 Apr 2012, 06:20
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