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Set T consists of all points (x, y) such that x^2+y^2=1. If [#permalink] New post 15 Apr 2005, 05:45
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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:
[Reveal] Spoiler: OA

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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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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 21 Feb 2012, 00:36
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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:


Look at the diagram below.
Attachment:
graph.php.png
graph.php.png [ 15.81 KiB | Viewed 1247 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.

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: PLS DRAW & ILLUSTRATE [#permalink] 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] New post 21 Feb 2012, 12:25
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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 24 Mar 2012, 11:39
anuu 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.



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?
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 24 Mar 2012, 14:22
Re: PLS DRAW & ILLUSTRATE   [#permalink] 24 Mar 2012, 14:22
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