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Re M1613 [#permalink]
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15 Sep 2014, 23:58
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Official Solution:If a point is arbitrarily selected inside a circle of radius \(R\), what is the probability that the distance from this point to the center of the circle will be greater than \(\frac{R}{2}\) ? A. \(\frac{1}{2}\) B. \(\frac{3}{4}\) C. \(\frac{7}{8}\) D. \(\frac{1}{4}R^2\) E. \(\frac{3}{4}R^2\) For the point to be further than \(\frac{R}{2}\) from the center, it has to lie on the disk confined between the circumference of radius \(R\) and the circumference of radius \(\frac{R}{2}\). The required probability will equal the ratio of the area of this disk to the area of the big circle. The area of the disk = \(\pi R^2  \pi (\frac{R}{2})^2 = \frac{3}{4} \pi R^2\). The area of the big circle = \(\pi R^2\). The ratio = \(\frac{3}{4}\). Answer: B
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Re: M1613 [#permalink]
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25 Dec 2014, 02:17
Bunuel wrote: Official Solution:
If a point is arbitrarily selected inside a circle of radius \(R\), what is the probability that the distance from this point to the center of the circle will be greater than \(\frac{R}{2}\) ?
A. \(\frac{1}{2}\) B. \(\frac{3}{4}\) C. \(\frac{7}{8}\) D. \(\frac{1}{4}R^2\) E. \(\frac{3}{4}R^2\)
For the point to be further than \(\frac{R}{2}\) from the center, it has to lie on the disk confined between the circumference of radius \(R\) and the circumference of radius \(\frac{R}{2}\). The required probability will equal the ratio of the area of this disk to the area of the big circle. The area of the disk = \(\pi R^2  \pi (\frac{R}{2})^2 = \frac{3}{4} \pi R^2\). The area of the big circle = \(\pi R^2\). The ratio = \(\frac{3}{4}\).
Answer: B I took R = 14 for bigger circle and calculated. It comes to 47 pi / 196 pi. What am I doing wrong?
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Re: M1613 [#permalink]
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25 Dec 2014, 02:19
aimtoteach wrote: Bunuel wrote: Official Solution:
If a point is arbitrarily selected inside a circle of radius \(R\), what is the probability that the distance from this point to the center of the circle will be greater than \(\frac{R}{2}\) ?
A. \(\frac{1}{2}\) B. \(\frac{3}{4}\) C. \(\frac{7}{8}\) D. \(\frac{1}{4}R^2\) E. \(\frac{3}{4}R^2\)
For the point to be further than \(\frac{R}{2}\) from the center, it has to lie on the disk confined between the circumference of radius \(R\) and the circumference of radius \(\frac{R}{2}\). The required probability will equal the ratio of the area of this disk to the area of the big circle. The area of the disk = \(\pi R^2  \pi (\frac{R}{2})^2 = \frac{3}{4} \pi R^2\). The area of the big circle = \(\pi R^2\). The ratio = \(\frac{3}{4}\).
Answer: B I took R = 14 for bigger circle and calculated. It comes to 47 pi / 196 pi. What am I doing wrong? You should show your work...
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Hey, I used a different, more visual approach. 1) I drew the circle. 2) I drew the 2 diameters throught the center of the circle, the intersection of which created a cross that divided tha circle in 4. 3) I halved all of the diameters (marking the radii in fact) and drew the lines that went through their mid points. 4) 16 sections were created. 5) Out of the 16 sections, 4 marked the areas equal or less than half of the radii (and created a square around the center). 6) 12 of the sections were more than half radius away from the center of the circle In total: 12 / 16 = 3 /4 Does this make sense?
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Last edited by pacifist85 on 21 Jan 2015, 02:45, edited 1 time in total.



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Re: M1613 [#permalink]
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21 Jan 2015, 02:38
pacifist85 wrote: Hey,
I used a different, more visual approach.
1) I drew the circle. 2) I drew the 2 diagonals throught the center of the circle, the intersection of which created a cross that divided tha circle in 4. 3) I halved all of the diagonals (marking the radii in fact) and drew the lines that went through their mid points. 4) 16 sections were created. 5) Out of the 16 sections, 4 marked the areas equal or less than half of the radii (and created a square around the center). 6) 12 of the sections were more than half radius away from the center of the circle
In total: 12 / 16 = 3 /4
Does this make sense? I guess by diagonals you mean diameter... Still your approach is nor clear to me. It would be better if you could illustrate it. Anyway, I think the shortest approach is given above and I'd advice to study it.
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Re: M1613 [#permalink]
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21 Jan 2015, 02:47
Bunuel wrote: pacifist85 wrote: Hey,
I used a different, more visual approach.
1) I drew the circle. 2) I drew the 2 diagonals throught the center of the circle, the intersection of which created a cross that divided tha circle in 4. 3) I halved all of the diagonals (marking the radii in fact) and drew the lines that went through their mid points. 4) 16 sections were created. 5) Out of the 16 sections, 4 marked the areas equal or less than half of the radii (and created a square around the center). 6) 12 of the sections were more than half radius away from the center of the circle
In total: 12 / 16 = 3 /4
Does this make sense? I guess by diagonals you mean diameter... Still your approach is nor clear to me. It would be better if you could illustrate it. Anyway, I think the shortest approach is given above and I'd advice to study it. Yes indeed, i meant to say diameters. I uploadd an image, i hope it can be seen. By all means, if you think it might be confusing people delete it!



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Re: M1613 [#permalink]
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21 Jan 2015, 02:57
pacifist85 wrote: Hey,
I used a different, more visual approach.
1) I drew the circle. 2) I drew the 2 diameters throught the center of the circle, the intersection of which created a cross that divided tha circle in 4. 3) I halved all of the diameters (marking the radii in fact) and drew the lines that went through their mid points. 4) 16 sections were created. 5) Out of the 16 sections, 4 marked the areas equal or less than half of the radii (and created a square around the center). 6) 12 of the sections were more than half radius away from the center of the circle
In total: 12 / 16 = 3 /4
Does this make sense? That's not correct. Check the image below: Attachment: Untitled.png The question ask about the probability that the distance between a point selected and the center will be greater than R/2, so the probability of a point being in yellow disc. Hope it's clear.
>> !!!
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M1613 [#permalink]
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21 Jan 2015, 03:11
I am sorry I cannot see the difference...
I mean, I realize that I used squares, and they are not equal. However, there is a bit more in the central square than what I need and a bit less in the smaller "corner" squares that what I need. Which in a sense "cancel out" this error.
However, the probability I was trying to find was the one after 1/2 of the radii (so the yellow region as you indicated).
That said, I know it is not a "math" way to solve this, but I don't know if it is a wrong way to perceive it..
Anyway, I did read the proposed solution and found it quite helpful! Thank you.



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Re: M1613 [#permalink]
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03 Jul 2017, 07:11
I marked the answer as 1/2 on thinking we need to select one region out of two regions.
But, I realized, by doing so, I am assuming that the number of points within the 0 to R/2 disc is equal to the number of points within the R/2 to R disc.
But, we are NOT sure whether they are equal. So, the approach using area may be a better one.



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Re M1613 [#permalink]
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23 Aug 2017, 18:48
I think this is a highquality question and I agree with explanation.










