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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 \(b \gt a + 1\) ?
(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]
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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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Re: M20 #12 [#permalink]
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22 Nov 2008, 20:17
i think its quadrant I and II. so 1/2



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Re: M20 #12 [#permalink]
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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



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Re: M20 #12 [#permalink]
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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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Re: M20 #12 [#permalink]
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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 QdII 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]
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17 Dec 2008, 09:45
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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 QdII 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.
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Re: M20 #12 [#permalink]
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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 QdII 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]
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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 1352 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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). This question is also discussed here: m2076028.html and here: settconsistsofallpointsxysuchthatx2y21if15626.htmlHope it helps.
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