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# m20 #12

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Intern
Joined: 04 Nov 2012
Posts: 9

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08 Jan 2013, 11:38
mourinhogmat1 wrote:
Great problem. I got it wrong as 1/2. I deduced that a is negative which left me with two quadrants worth of points. But failed to consider the line theory.

Likewise. Got the answer 1/2 assuming that the a<0 is only possible in 2 of 4 quads... .

Realized the mistake now and took note of the underlying concept.

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Intern
Joined: 14 Jul 2012
Posts: 15

Kudos [?]: 5 [0], given: 8

Location: United States
GPA: 3.3
WE: Information Technology (Computer Software)

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13 Jan 2013, 22:17
Okie, no great maths here,Lets look at this problem in this way -
we know its a circle with origin at 0,0 and radius as 1.Now, if you can visualise the circle,a cannot be positive to meet the condition in the question, because then b will have to be greater than 1 (which is more than the radius of the circle) and hence not possible.
Hence a has to be negative, again b cannot be negative because +1 to negative x will always make it greater than corresponding point of B.
Hence a is -ve and B is +ve.You can check this with trial -> a= -1.4 , then B would be 1.4 which satisfies the equation b>a+1.Hence one 1 quadrant out of four satisfies the equation,Hence answer 1/4. (A).
Hope this helps !!

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Intern
Joined: 23 Jun 2013
Posts: 44

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15 Mar 2014, 22:08
Bunuel wrote:
dczuchta wrote:
Would someone please explain this problem to me? Thank you.

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 $$b \gt a + 1$$ ?

(A) $$\frac{1}{4}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{5}$$
(E) $$\frac{2}{3}$$

[Reveal] Spoiler: OA
A

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

I did not understand this part "You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4." Above the line and inside the boundary of the circle is just small part. How is it 1/4 of the circle

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Joined: 02 Sep 2009
Posts: 42618

Kudos [?]: 135762 [0], given: 12708

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16 Mar 2014, 04:07
Nilabh_s wrote:
Bunuel wrote:
dczuchta wrote:
Would someone please explain this problem to me? Thank you.

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 $$b \gt a + 1$$ ?

(A) $$\frac{1}{4}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{5}$$
(E) $$\frac{2}{3}$$

[Reveal] Spoiler: OA
A

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

I did not understand this part "You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4." Above the line and inside the boundary of the circle is just small part. How is it 1/4 of the circle

Please re-read the solution. Above the line is 1/4 th of the circumference not the area of the circle.
_________________

Kudos [?]: 135762 [0], given: 12708

Re: m20 #12   [#permalink] 16 Mar 2014, 04:07

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# m20 #12

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