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Re: M2012
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22 Sep 2017, 02:49
Bunuel wrote: Subham10 wrote: 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 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 yaxis at (1, 0) and (0, 1) respectively at the same exact points the circle cuts the axis. Got it. Yes, it was easy! Thanks & Regards



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Re: M2012
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23 Nov 2017, 13:25
Hi Bunuel,
A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (1,0) or the point (0,1) and none of them satisfies the condition. So shouldn't exact answer be a little less than 1/4.



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Re: M2012
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23 Nov 2017, 20:25
asthagupta wrote: Hi Bunuel,
A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (1,0) or the point (0,1) and none of them satisfies the condition. So shouldn't exact answer be a little less than 1/4. A point has no dimension, so it does not matter. The same way a line has only one dimension, length, but no area.
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Re: M2012
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30 Nov 2017, 11:06
Bunuel wrote: asthagupta wrote: Hi Bunuel,
A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (1,0) or the point (0,1) and none of them satisfies the condition. So shouldn't exact answer be a little less than 1/4. A point has no dimension, so it does not matter. The same way a line has only one dimension, length, but no area. Got your point boss Thanks



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Re: M2012
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23 Dec 2017, 00:34
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?? We have a point a,b such that ... \(b > a + 1\), we can simplify it as. \(b  a > 1\) We know b can max out at \(1\) since \(x^2 + y^2 = 1\) is a circle of radius \(1\). For \(b  a\) to be greater than \(1\), \(b\) has to be positive and \(a\) has to negative. This only happens in the 2nd quadrant which is contains 1 fourth of the total circle. So probability should be \(1/4\).
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Re: M2012
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27 Mar 2018, 00:40
The question states that b > a+1.
since we consider 1/4th of the circle in the second Quadrant. Aren't the points (1,0) and (0,1) also a part of 1/4th of the circumference. But these points do not satisfy b>a+1. In this case, the question should have been b>= a+1.
Is my analysis wrong?



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b>a+1 then ba>1 If we square, (ba)^2 > 1 a2 + b2  2ab > 1 1 2ab >1 [a2 + b2 =1] ab < 0. This means a and are of the opposite sign. So both second and 4th quadrants. So,i got 1/2. Can someone correct me where I went wrong ? The official solution looks ok to me. Confused. Bunuel : Can you help me here ?
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I got the issue with my approach, since i squared the straight line equation, i ended up with 2 possible parallel lines. Hence got 1/2. (ab)^2 < 1^2 (ab)^2 < (1)^2 Hence 1/2. I should have ignored the second line, or in other words, divided the probability by 2. Very good question.
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Re: M2012
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31 May 2018, 06:21
Bunuel wrote: 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. Hi Bunnel, I've made the exact graphical thought process as you and got the the answer Pi2/4Pi, which is not in the answer options. It's not clear to me why "Original question does not ask about the area, it asks about the portion of the circumference" as the question asks "Set T consists of all points (x,y) such that x2+y2=1. If point (a,b) is selected from set T at random, what is the probability that b>a+1b>a+1?" An area can be defined as an infinite number of point. Thus, for me the favorable cases should indeed be an area, since the Set T consists in ALL points that obey to a certain equation. Could you please clarify? Thanks a lot!



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Miracles86 wrote: Bunuel wrote: 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. Hi Bunnel, I've made the exact graphical thought process as you and got the the answer Pi2/4Pi, which is not in the answer options. It's not clear to me why "Original question does not ask about the area, it asks about the portion of the circumference" as the question asks "Set T consists of all points (x,y) such that x2+y2=1. If point (a,b) is selected from set T at random, what is the probability that b>a+1b>a+1?" An area can be defined as an infinite number of point. Thus, for me the favorable cases should indeed be an area, since the Set T consists in ALL points that obey to a certain equation. Could you please clarify? Thanks a lot! Set T consists of all the points on the circumference of the circle only (x^2 + y^2 = 1), NOT the all the points within the circle (x^2 + y^2 < 1), so we should find the probability that the point is ON the circumference but above y = x + 1, NOT the probability that the point is IN the circle but above y = x + 1. Hope it's clear.
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Re: M2012
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13 Dec 2018, 11:57
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. \(\frac{2}{3}\) Hi Bunuel, Please let me know if the following method is correct Given a^2 + b^2 = 1 and if b > a + 1 > b  a > 1 > (ba)^2 > (1)^2 > b^2 + a^2 2ab > 1 > 1 2ab > 1 > ab <0 Hence the question basically boils down to finding the probability of a*b < 0 Now for following 2 possibilities the equations stands correct 1. a < 0 and b > 0 2. a>0 and b < 0 And there are overall 8 possibilities. Since a and b each can take values of >0, <0 and =0 . therefore 2^3 = 8 hence P = 2/8 > 1/4







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