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M20-12

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M20-12 [#permalink]

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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 \(b \gt a + 1\)?

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

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Re M20-12 [#permalink]

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New post 16 Sep 2014, 01:08
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Official Solution:

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. \(\frac{2}{3}\)


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 \(r=\sqrt{1}=1\).

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 \(b \gt a+1\)? 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 \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).


Answer: A
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Re: M20-12 [#permalink]

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New post 30 Oct 2014, 16:24
Bunuel wrote:
Official Solution:

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. \(\frac{2}{3}\)


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 \(r=\sqrt{1}=1\).

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 \(b \gt a+1\)? 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 \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).


Answer: A


Is there any other simple solution??
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Re: M20-12 [#permalink]

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New post 31 Oct 2014, 04:56
awal_786@hotmail.com wrote:
Bunuel wrote:
Official Solution:

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. \(\frac{2}{3}\)


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 \(r=\sqrt{1}=1\).

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 \(b \gt a+1\)? 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 \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).


Answer: A


Is there any other simple solution??


I, personally, don't know any simpler one. The one presented should be simple enough after some practice.

Check other Probability and Geometry questions in our Special Questions Directory.

Hope it helps.
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Re: M20-12 [#permalink]

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New post 31 Oct 2014, 08:11
I understood how you made this circle, but I can't understand how you made this blue line and computed probability from this :shock:
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Re: M20-12 [#permalink]

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New post 31 Oct 2014, 09:50
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awal_786@hotmail.com wrote:
I understood how you made this circle, but I can't understand how you made this blue line and computed probability from this :shock:


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: M20-12 [#permalink]

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New post 20 Mar 2016, 04:49
Hi Bunuel,

Shouldn't it be 1/4 of the area rather than circumference ?

Please confirm.

Regards
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Re: M20-12 [#permalink]

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New post 20 Mar 2016, 05:09
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vinnik wrote:
Hi Bunuel,

Shouldn't it be 1/4 of the area rather than circumference ?

Please confirm.

Regards


Original question does not ask about the area, it asks about the portion of the circumference

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: M20-12 [#permalink]

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Re: M20-12 [#permalink]

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New post 18 Jul 2016, 12:39
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Bunuel Hi,
I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Thanks
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Re: M20-12 [#permalink]

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New post 19 Jul 2016, 00:17
sathishm07 wrote:
Bunuel Hi,
I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Thanks



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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M20-12 [#permalink]

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New post Updated on: 19 Jul 2016, 00:57
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for choosing left and right
you can substitute origin(0,0) in the equation y>x+1
=> 0>1 which is not correct .
Hence origin doesn't satisfy the equation thus it is on the side not containing the origin.

Originally posted by ShravyaAlladi on 19 Jul 2016, 00:35.
Last edited by ShravyaAlladi on 19 Jul 2016, 00:57, edited 1 time in total.
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M20-12 [#permalink]

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New post 19 Jul 2016, 00:56
sathishm07 wrote:
I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Thanks


for choosing left and right
you can substitute origin(0,0) in the equation y>x+1
=> 0>1 which is not correct .
Hence origin doesn't satisfy the equation thus it is on the side not containing the origin
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Re M20-12 [#permalink]

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New post 31 Aug 2016, 04:42
I think this is a high-quality question and I agree with explanation.
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Re: M20-12 [#permalink]

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New post 06 Oct 2016, 02:08
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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 \(b \gt a + 1\)?

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

Sir,

Do you have few more quetsions list where Geomtry is Clubbed with Probability.
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Re: M20-12 [#permalink]

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New post 06 Oct 2016, 04:55
RichaChampion wrote:
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 \(b \gt a + 1\)?

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

Sir,

Do you have few more quetsions list where Geomtry is Clubbed with Probability.


Check other Probability and Geometry questions in our Special Questions Directory.
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Re: M20-12 [#permalink]

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New post 07 Dec 2016, 05:53
Dear Bunnuel,

Thank you for the graph and explanation. However, could u explain why this approach is incorrect?

r = 1 Circum = 2pi.

Any point that satisfies b> a+1 can be translated into if y=x+1 and we see this covers 1/4 of whole Circum.

However, can I use Prob=n/N approach where N is whole circumference. n - is represented by the arc-length - L

Inscribed angle = (90*L)/pi*radius ------> looking at the graph, ins.angle =90 & r = 1,. Then L -arc-length is pi. Thus I have Prob = n/N = pi/2pi = 1/2

Pls let me know why this approach did not work ?


Thank you
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Re: M20-12 [#permalink]

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New post 18 Jan 2017, 11:54
x^2+y^2=1. therefore it is a circle with radius 1. (put x=0 , y=+1/-1 similarly, y=0, x=+1/-1. gives the points where the line will cross the axes). Draw the circle.

since we are concerned about points inside this circle so (x,y) should not exceed +1/-1.
b>a+1 is only possible when a is negative greater than -1. this gives positive b; for eg. if a=-0.5 then b>0.5.
so all points (a,b) will be in IInd quadrant.
X & Y axis divides the circle in 4 parts. out of the 4 (a,b) can lie on only 1 quadrant. therefor probability will be 1/4
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Re M20-12 [#permalink]

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New post 22 Sep 2017, 02:00
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "You can see that portion of the circle which is above the line is 1/4 of the circumference of the circle"--

How did u conclude this? Please show calculation
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Re: M20-12 [#permalink]

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New post 22 Sep 2017, 02:35
Subham10 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "You can see that portion of the circle which is above the line is 1/4 of the circumference of the circle"--

How did u conclude this? Please show calculation


This should be easy. I'll give you the hints and let you figure out the rest.

1. The circle is centred at the origin, so each quadrant has 1/4 of the circumference.
2. y = x + 1 cuts the x and y-axis at (-1, 0) and (0, 1) respectively at the same exact points the circle cuts the axis.
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Re: M20-12   [#permalink] 22 Sep 2017, 02:35

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