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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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28 Nov 2010, 08:54
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56% (01:58) correct 44% (02:06) wrong based on 221 sessions

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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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28 Nov 2010, 09:20
6
1
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:

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

As for the question:

Look at the diagram below.
Attachment:

graph.php.png [ 15.81 KiB | Viewed 10550 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.

Hope it's clear.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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28 Nov 2010, 09:37
1
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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Posts: 59623
Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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28 Nov 2010, 09:52
3
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.

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 $$P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}$$.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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29 Nov 2010, 05:24
Bunuel 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:

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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.

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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Joined: 03 Mar 2010
Posts: 336
Schools: Simon '16 (M$) Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink] ### Show Tags 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. _________________ My dad once said to me: Son, nothing succeeds like success. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9859 Location: Pune, India Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 - [#permalink] ### Show Tags 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. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Senior Manager Joined: 03 Mar 2010 Posts: 336 Schools: Simon '16 (M$)
Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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

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12 Oct 2013, 02:23
The distance from a point (x0, y0) to a line ax+by+c=0 is given by the formula:

D=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}

=====>When the line is horizontal the formula transforms to: D=|Py-Ly|
Where: Py is the y-coordinate of the given point P; Ly is the y-coordinate of any point on the given vertical line L. | | the vertical bars mean "absolute value" - make it positive even if it calculates to a negative.
=====>When the line is vertical the formula transforms to: D=|Px-Lx|
Where: Px is the x-coordinate of the given point P; Lx is the x-coordinate of any point on the given vertical line L. | | the vertical bars mean "absolute value" - make it positive even if it calculates to a negative.

Buneul, can you explain arrow points(extracted from work book) with an example for each
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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06 Nov 2013, 08:02
You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

How come the portion is 1/4, please explain.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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06 Nov 2013, 20:01
schittuluri wrote:
You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

How come the portion is 1/4, please explain.

Note that set T is the set of points lying ON THE CIRCLE i.e. only on the circumference (not inside the circle). This is because the condition x^2 + y^2 = 1 is satisfied by the points lying on the circle only. Inside the circle, the points satisfy x^2 + y^2 < 1.

So our set T consists of the circumference of the circle. The given line cuts the circle such that a quarter of its CIRCUMFERENCE lies above the line. Now read the explanation given above again - you will notice the word 'circumference' used everywhere. By habit, we tend to think of area of the circle.

That is how we get 1/4.
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Re: GMATCLUB TEST , HARD QUESTIONS, PROBABILITY #8 -  [#permalink]

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07 Nov 2013, 02:32
schittuluri wrote:
You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

How come the portion is 1/4, please explain.

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 $$P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}$$.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html
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Re: Set T consists of all points (x,y) such that x^2+y^2 =1  [#permalink]

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08 Jul 2015, 22:50
1
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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html

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Re: Set T consists of all points (x,y) such that x^2+y^2 =1   [#permalink] 08 Jul 2015, 22:50
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