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Re M2012 [#permalink]
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16 Sep 2014, 01:08
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 ycoordinate > xcoordinate + 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: M2012 [#permalink]
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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: 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 ycoordinate > xcoordinate + 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: M2012 [#permalink]
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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: 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 ycoordinate > xcoordinate + 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: M2012 [#permalink]
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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



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Re: M2012 [#permalink]
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Re: M2012 [#permalink]
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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: M2012 [#permalink]
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20 Mar 2016, 05:09



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Re: M2012 [#permalink]
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18 Jul 2016, 12:39
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: M2012 [#permalink]
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19 Jul 2016, 00:17



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M2012 [#permalink]
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Updated on: 19 Jul 2016, 00:57
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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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 M2012 [#permalink]
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31 Aug 2016, 04:42
I think this is a highquality question and I agree with explanation.



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Re: M2012 [#permalink]
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06 Oct 2016, 02:08
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: M2012 [#permalink]
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06 Oct 2016, 04:55



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Re: M2012 [#permalink]
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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 arclength  L
Inscribed angle = (90*L)/pi*radius > looking at the graph, ins.angle =90 & r = 1,. Then L arclength 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: M2012 [#permalink]
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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 M2012 [#permalink]
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22 Sep 2017, 02:00
I think this is a highquality 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: M2012 [#permalink]
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