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# M20 #12

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Senior Manager
Joined: 20 Feb 2008
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Location: Bangalore, India
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18 Nov 2008, 04:30
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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Joined: 17 Jun 2008
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18 Nov 2008, 23:44
Is it 1/4?

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

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22 Nov 2008, 20:17
i think its quadrant I and II. so 1/2

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Senior Manager
Joined: 20 Feb 2008
Posts: 295

Kudos [?]: 51 [0], given: 0

Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross

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23 Nov 2008, 12:30
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?
Thankyou

Kudos [?]: 51 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3345

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23 Nov 2008, 19:30
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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23 Nov 2008, 20:55
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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Posts: 1216

Kudos [?]: 986 [3], given: 334

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17 Dec 2008, 09:45
3
KUDOS
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 [ 4.85 KiB | Viewed 1966 times ]

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Kudos [?]: 986 [3], given: 334

Manager
Joined: 10 Jan 2011
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Location: India
GMAT Date: 07-16-2012
GPA: 3.4
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05 Apr 2012, 23:41
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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-------Analyze why option A in SC wrong-------

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Math Expert
Joined: 02 Sep 2009
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06 Apr 2012, 07:20
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 [ 15.81 KiB | Viewed 1395 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.

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, 07:20
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# M20 #12

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