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Set T consists of all points (x,y) such that x^2+y^2 =1 [#permalink]
 28 Nov 2010, 08:54
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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 I looked at explanation, I understand how the line will pass through points (-1,0) and (0,1) .. but now when I graph the circle with radius 1 , the line will cut only a little portion of the top left quarter of the circle.. While the answer explanation says it will cut of the Whole top left quarter of the circle.. thus prob should of b>a+1 should be 1/4..... Can anyone please help me understand this.. Thanks , I appreciate it.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
28 Nov 2010, 09:20
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tradinggenius 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>a+1 ?
a) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3
I looked at explanation, I understand how the line will pass through points (-1,0) and (0,1) .. but now when I graph the circle with radius 1 , the line will cut only a little portion of the top left quarter of the circle.. While the answer explanation says it will cut of the Whole top left quarter of the circle.. thus prob should of b>a+1 should be 1/4..... Can anyone please help me understand this..
Thanks ,
I appreciate it. First of all:As for the question:Look at the diagram below. Attachment: 
graph.php.png [ 15.81 KiB | Viewed 2428 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. Hope it's clear.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
28 Nov 2010, 09:37
Yup , I understand now.. I dint realize the points were on the circumference.. Thanks.. and sorry for posting here.. Shall not happen in the future.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
28 Nov 2010, 09:52
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
29 Nov 2010, 05:24
Bunuel wrote: tradinggenius 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>a+1 ?
a) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3
I looked at explanation, I understand how the line will pass through points (-1,0) and (0,1) .. but now when I graph the circle with radius 1 , the line will cut only a little portion of the top left quarter of the circle.. While the answer explanation says it will cut of the Whole top left quarter of the circle.. thus prob should of b>a+1 should be 1/4..... Can anyone please help me understand this..
Thanks ,
I appreciate it. First of all:As for the question:Look at the diagram below. Attachment: graph.php.png 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. Hope it's clear. Good one Bunuel. I liked ur graphical approach which is exactly expected by GMAT test makers. Kudos for u. Regards, Murali.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
20 Sep 2011, 05:39
Can somebody please help me with this? Here's how i approached. Question Find probability of b>a+1 or b-a>1 Squaring both sides (b-a)^2 > 1^2 b^2+a^2-2ab > 1 -------(1) Now point (a,b) belongs to Set T which is a set of all points satisfying equation x^2+y^2=1. Hence point(a,b) will satisfy the equation as well a^2+b^2=1 -------(2) Substituting (2) in (1) 1-2ab>1 0>2ab or Find the probability of ab < 0 Now ab < 0 only in 2 quadrant, II and IV. Hence probability = 2/4=1/2. Where am i going wrong? . Thanks.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
20 Sep 2011, 08:04
jamifahad wrote: Can somebody please help me with this? Here's how i approached.
Question Find probability of b>a+1
or b-a>1
Squaring both sides
(b-a)^2 > 1^2
b^2+a^2-2ab > 1 -------(1)
Now point (a,b) belongs to Set T which is a set of all points satisfying equation x^2+y^2=1.
Hence point(a,b) will satisfy the equation as well
a^2+b^2=1 -------(2)
Substituting (2) in (1)
1-2ab>1
0>2ab
or Find the probability of ab < 0
Now ab < 0 only in 2 quadrant, II and IV.
Hence probability = 2/4=1/2.
Where am i going wrong? .
Thanks. The problem here is conceptual. Say, I have a set S = {-3, -2, -1, 0, 1, 2, 3} Question: What is the probability that an element x, taken from set S is greater than 0? What is the probability that x > 0? Is it the same as: What is the probability that x^2 > 0? No. Answer in the first case is 3/7 Answer in the second case is 6/7 Think twice before squaring an inequality. It is not given that b is greater than a+1. They have asked the probability that b is greater than a+1. The graphical approach is almost always better and quicker.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink]
20 Sep 2011, 10:13
VeritasPrepKarishma wrote: The problem here is conceptual. I see your point. Probability is tricky like when dealing with ratio of two elements and probability of picking 2 elements one after another.  Also, points cannot be in IV quadrant because we need points for which b>a+1. For IV quadrant a is +ve and b is -ve. a+1 will be a positive number. hence for b>a+1, b has to be a positive number which b is not in IV quadrant. Thanks, karishma.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -
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20 Sep 2011, 10:13
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