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Senior Manager  Joined: 02 Feb 2004
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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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Question Stats: 58% (02:19) correct 42% (02:19) wrong based on 561 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) $$\frac{1}{4}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{5}$$
(E) $$\frac{2}{3}$$

PLS DRAW & ILLUSTRATE:

Attachments 0012.jpg [ 9.03 KiB | Viewed 12884 times ]

Originally posted by mirhaque on 15 Apr 2005, 04:45.
Last edited by Bunuel on 20 Feb 2012, 23:35, edited 2 times in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 64249
Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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

Hope it's clear.
Attachment: graph.php.png [ 15.81 KiB | Viewed 13260 times ]

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##### General Discussion
Intern  Joined: 28 Feb 2011
Posts: 32
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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1
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.

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

Anu
Math Expert V
Joined: 02 Sep 2009
Posts: 64249
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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1
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.
_________________
Manager  Joined: 25 Oct 2013
Posts: 138
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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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.

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.
VP  Joined: 06 Sep 2013
Posts: 1494
Concentration: Finance
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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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 Math Expert V
Joined: 02 Sep 2009
Posts: 64249
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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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 _________________
VP  Joined: 06 Sep 2013
Posts: 1494
Concentration: Finance
Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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OK sorry now i get it. For a second i forgot that pi = 3.14159

Thanks
Cheers
J Posted from my mobile device
Senior Manager  D
Joined: 17 Mar 2014
Posts: 451
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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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.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 64249
Re: PLS DRAW & ILLUSTRATE  [#permalink]

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1
6
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.

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-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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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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 64249
Re: Set T consists of all points (x, y) such that x^2+y^2=1. If  [#permalink]

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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.

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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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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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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_________________ Re: Set T consists of all points (x, y) such that x^2+y^2=1. If   [#permalink] 10 May 2020, 07:50

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