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Set T consists of all points (x, y) such that x^2+y^2=1. If

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

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New post Updated on: 21 Feb 2012, 00:35
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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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:

Attachments

0012.jpg
0012.jpg [ 9.03 KiB | Viewed 12034 times ]


Originally posted by mirhaque on 15 Apr 2005, 05:45.
Last edited by Bunuel on 21 Feb 2012, 00:35, edited 2 times in total.
Edited the question added the answer choices with OA
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Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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New post 21 Feb 2012, 00:36
21
25
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Image
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 \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}\).

Hope it's clear.
Attachment:
graph.php.png
graph.php.png [ 15.81 KiB | Viewed 12365 times ]

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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 21 Feb 2012, 12:01
Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.


Hi Bunnel,

Can you pls explain how did u calculate area of segment above the line (pi-2/4) for x^2+y^2<1

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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 21 Feb 2012, 12:25
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anuu wrote:
Hi Bunnel,

Can you pls explain how did u calculate area of segment above the line (pi-2/4) for x^2+y^2<1

Anu


Sure. Look at the diagram. The area above the line equals to 1/4th of the area of the circle (\(\frac{\pi{r^2}}{4}\)) minus the area of the isosceles right triangle made by radii (\(\frac{1}{2}*r*r\)): \(\frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4}\) (since given that \(r=1\)).

Hope it's clear.
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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 26 Nov 2013, 05:48
Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.


Great explanation! very elegant. thanks Bunuel.
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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 28 Dec 2013, 05:34
Bunuel wrote:
pkonduri wrote:
Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?


Check this: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html#p1047717

Hope it helps.


Hi Bunuel,

Still don't get how did you arrive at the conclusion that the area above the line within the circle is 1/4th of the circle for the original question

Would you kindly elaborate on this

Thanks
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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 28 Dec 2013, 06:22
jlgdr wrote:
Bunuel wrote:
pkonduri wrote:
Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?


Check this: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html#p1047717

Hope it helps.


Hi Bunuel,

Still don't get how did you arrive at the conclusion that the area above the line within the circle is 1/4th of the circle for the original question

Would you kindly elaborate on this

Thanks
Cheers!
J :)


Original question does not ask about the area, it asks about the portion of the circumference
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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New post 28 Dec 2013, 07:10
OK sorry now i get it. For a second i forgot that pi = 3.14159

Thanks
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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 08 Jun 2014, 21:42
Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.


Hi Bunuel,

Do we have more variations of such questions. If yes, could you share it i want to practice.

R/
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Re: PLS DRAW & ILLUSTRATE  [#permalink]

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New post 09 Jun 2014, 01:50
1
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ammuseeru wrote:
Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.


Hi Bunuel,

Do we have more variations of such questions. If yes, could you share it i want to practice.

R/
Ammu


I searched our questions banks and was able to find the following questions.

PROBABILITY AND GEOMETRY:


in-the-xy-plane-a-triangle-has-vertexes-0-0-4-0-and-88395.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
a-triangle-with-three-equal-sides-is-inscribed-inside-a-160874.html
a-game-at-the-state-fair-has-a-circular-target-with-a-radius-171756.html
a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html
an-x-y-coordinate-pair-is-to-be-chosen-at-random-from-the-146005.html
a-cylinder-has-a-base-with-a-circumference-of-20pi-meters-132132.html
a-circular-racetrack-is-3-miles-in-length-and-has-signs-post-106203.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
in-the-xy-plane-the-vertex-of-a-square-are-88246.html
a-searchlight-on-top-of-the-watch-tower-makes-3-revolutions-76069.html
point-p-a-b-is-randomly-selected-in-the-region-enclosed-by-160615.html

Hope this helps.
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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New post 08 Dec 2014, 14:30
Hi Bunuel,

I understand the approach but can you please let me know how did you draw that blue line for f(x)=x+1?

Thanks


Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.

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

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New post 08 Dec 2014, 14:34
sunita123 wrote:
Hi Bunuel,

I understand the approach but can you please let me know how did you draw that blue line for f(x)=x+1?

Thanks


Bunuel wrote:
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


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.

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.


The blue line is y = x + 1. The area above it is y > x + 1. So, all the points for which y-coordinate is greater than x-coordinate + 1.
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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New post 22 Jan 2018, 16:34
Hi All,

This question is fairly 'high-concept', but you can solve it with a drawing and a bit of logic.

When you graph the equation X^2 + Y^2 = 1, you will have a circle with a radius of 1 that is centered around the Origin. There are only four points on that circle that are integer values: (0,1), (1, 0), (0, -1) and (-1, 0). All of the other points are positive/negative fractional values. Knowing that, the ONLY way for the inequality Y > X + 1 to occur is when X is NEGATIVE and Y is POSITIVE. That outcome only occurs in the 2nd quadrant of the graph (which is 1/4 of the circle). While you might be unsure about whether every point in that quadrant would 'fit' the inequality or not, the answer choices ARE numbers - and the smallest of them is 1/4. Since there's no smaller possibility, 1/4 must be the answer.

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

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New post 22 Jan 2018, 17:46
mirhaque 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) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{3}{5}\)
(E) \(\frac{2}{3}\)


PLS DRAW & ILLUSTRATE:


Its very simple if we proceed in this way... we are given that for any point (a,b) selected from set T \(a^2 + b^2 = 1\). Also, as \(b - a > 1...... i.e. b^2 + a^2 - 2ab > 1\)
or \(ab < 0\). Now, lets consider this sample space Set T = \({(1/\sqrt{2},-1/\sqrt{2}) (1/\sqrt{2},1/\sqrt{2}) (-1/\sqrt{2},1/\sqrt{2}) (-1/\sqrt{2},-1/\sqrt{2})}\).. Now as \(ab < 0\), we can select only 2 points from this sample space of \(8.. i.e. 2/8 = 1/4\)... You can try with \(3/\sqrt{5},4/\sqrt{5}\) and write all the possible conditions for this number as above.
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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