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# M20-12

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:08
Expert's post
20
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Difficulty:

95% (hard)

Question Stats:

54% (02:14) correct 46% (01:41) wrong based on 117 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 $$b \gt a + 1$$?

A. $$\frac{1}{4}$$
B. $$\frac{1}{3}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{5}$$
E. $$\frac{2}{3}$$
[Reveal] Spoiler: OA

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Math Expert
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16 Sep 2014, 00:08
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Expert's post
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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:

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 &gt; 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 $$P=\frac{1}{4}$$.

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30 Oct 2014, 15:24
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:

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 &gt; 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 $$P=\frac{1}{4}$$.

Is there any other simple solution??

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Math Expert
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31 Oct 2014, 03:56
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:

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 &gt; 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 $$P=\frac{1}{4}$$.

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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31 Oct 2014, 07:11
I understood how you made this circle, but I can't understand how you made this blue line and computed probability from this

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31 Oct 2014, 08:50
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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

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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20 Mar 2016, 03:49
Hi Bunuel,

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

Regards

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20 Mar 2016, 04:09
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vinnik wrote:
Hi Bunuel,

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

Regards

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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20 Mar 2016, 04:11
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vinnik wrote:
Hi Bunuel,

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

Regards

Check other Probability and Geometry questions in our Special Questions Directory.
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18 Jul 2016, 11:39
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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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18 Jul 2016, 23:17
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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18 Jul 2016, 23:35
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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.

Last edited by ShravyaAlladi on 18 Jul 2016, 23:57, edited 1 time in total.

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18 Jul 2016, 23:56
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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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31 Aug 2016, 03:42
I think this is a high-quality question and I agree with explanation.

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06 Oct 2016, 01:08
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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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06 Oct 2016, 03:55
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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07 Dec 2016, 04:53
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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18 Jan 2017, 10:54
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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22 Sep 2017, 01:00
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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22 Sep 2017, 01:35
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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