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n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 03:29
Question Stats:
44% (02:39) correct 56% (02:39) wrong based on 57 sessions
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n points are equally spaced on a circle, where n is an even number greater than 3. If 3 of the n points are to be chosen at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle? A. \((n1)/6\) B. \((n+2)/6\) C. \(2/(3n+2)\) D. \(3/(n1)\) E. \(6/(n+4)\)
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n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 04:28
gmatbusters wrote: n points are equally spaced on a circle, where n is an even number greater than 3. If 3 of the n points are to be chosen at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?
A. \((n1)/6\)
B. \((n+2)/6\)
C. \(2/(3n+2)\)
D. \(3/(n1)\)
E. \(6/(n+4)\) OA is D As given in the question , possible value of n( even and >3) can be 4,6,8..... if we put n=8 in option A and n=6 in option B , probability is coming out to be greater than 1,which is not possible. So we can eliminate option A and B. Now we take n=4, so we have to choose any 3 points of these 4 to make triangle. For finding the probability , we have to find the total number of case for n=4 . it is given by \(C(4,3)\). That comes out be 4. If we draw all these 4 cases,we will find that all 4 are right angle triangle so probability is 1 for n=4. Attachment:
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Now , if we put n =4 into remaining options. C. \(2/(3n+2)\) = \(2/14 = 1/7\) D. \(3/(n1)\) = \(3/3 = 1\) E. \(6/(n+4)\)= \(6/8=3/4\) Only option D is giving probability =1, so OA is D
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n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 05:49
Here is my lazy approach The more "n" you have, the less is probability is, hence "n" should be in denominator. So we left with D and E options. Now it is time to choose between D and E. For me D "feels" right as n+4 (in E) has no meaning and decrease chances, and n1 (in D) makes sence as it increases chances, because points are evenly spreaded and each point has a chance to get a point in front of it, so there should be more chances than 3 out of "n". I know this is not a sientific approach, but sometimes a little bit of abstract thinking and logic can save time and nerves. A. \((n1)/6\) B. \((n+2)/6\) C. \(2/(3n+2)\) D. \(3/(n1)\) E. \(6/(n+4)\)



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Re: n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 07:12
How u eliminated C "more "n" you have, the less is probability is, hence "n" should be in denominator. So we left with D and E options." Hero8888 wrote: Here is my lazy approach more "n" you have, the less is probability is, hence "n" should be in denominator. So we left with D and E options. Now it is time to choose between D and E. For me D "feels" right as n+4 (in E) has no meaning and decrease chances, and n1 (in D) makes sence as it increases chances, because points are evenly spreaded and each point has a chance to get a point in front of it, so there should be more chances than 3 out of "n". I know this is not a sientific approach, but sometimes a little bit of abstract thinking and logic can save time and nerves. A. \((n1)/6\) B. \((n+2)/6\) C. \(2/(3n+2)\) D. \(3/(n1)\) E. \(6/(n+4)\) Posted from my mobile device
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n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 08:16
gmatbusters wrote: How u eliminated C
"more "n" you have, the less is probability is, hence "n" should be in denominator. So we left with D and E options." I meant the greater ''n'' is, so the only way it can influence this way is denominator. I have missed the option C, thanks, but the reason for C is the same as for E : what is the reason to think that you have to pick more than from given "n" points? So you can't have "+" there. I have mentioned, that this is not the best approach, I would say it's related to POE.



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Re: n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 08:20
You are right, I saw that u missed C, yes it is wrong option... We we should not win by accident, so don't leave any option unintentionally. Hero8888 wrote: gmatbusters wrote: How u eliminated C
"more "n" you have, the less is probability is, hence "n" should be in denominator. So we left with D and E options." I meant the greater ''n'' is, so the only way it can influence this way is denominator. I have missed the option C, thanks, but the reason for C is the same as for E : what is the reason to think that you have to pick more than from given "n" points? So you can't have "+" there. I have told that this is not the best approach, I would say it's related to POE.
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Re: n points are equally spaced on a circle, where n is an even number gre
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16 Apr 2018, 08:45
CONVENTIONAL APPROACH :To construct a right angle on a circle,we need the diameter D of the circle upon which the other 2 sides of the triangle are contingen. Since the points are equally spaced, no of diagonals will be n/2.To select one diagonal from n/2, no of choices = \((n/2)C1\) = n/2 Then to select the other 2 sides, we need to select a point from remaining points= (n2) points (since 2 points has been chosen for the diameter). No of choices = \((n2)C1 = (n2)\) Total no of favorable cases = \(n/2*(n2)\) Total no. of cases = 3 vertices of right angle ? chosen from n points=\(nC3\) Thus,Prob(? with 3 vertices)=\(((n2)n/2)/nC3\) \(=3/(n1)\) Answer D
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Re: n points are equally spaced on a circle, where n is an even number gre
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17 Apr 2018, 03:43
gmatbusters Appreciate your solution, I understand the procedure you followed but I do not get how it takes care of 'right triangle' clause if we choose each points in such way.



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Re: n points are equally spaced on a circle, where n is an even number gre
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17 Apr 2018, 04:35
The concept used is THE ANGLE IN A SEMICIRCLE IS ALWAYS 90 DEGREE. Thus when we select first two points as end point of diameter then the triangle formed will always be a right angled triangle. tll001 wrote: gmatbusters Appreciate your solution, I understand the procedure you followed but I do not get how it takes care of 'right triangle' clause if we choose each points in such way.
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Re: n points are equally spaced on a circle, where n is an even number gre
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17 Apr 2018, 05:38
gmatbusters, Thanks! Learning something from you every day!



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Re: n points are equally spaced on a circle, where n is an even number gre
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17 Apr 2018, 06:19
Thanks!!! I will make sure learning continues tll001 wrote: gmatbusters, Thanks! Learning something from you every day!
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n points are equally spaced on a circle, where n is an even number gre
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17 Sep 2018, 09:13
gmatbusters wrote: n points are equally spaced on a circle, where n is an even number greater than 3. If 3 of the n points are to be chosen at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?
A. \((n1)/6\)
B. \((n+2)/6\)
C. \(2/(3n+2)\)
D. \(3/(n1)\)
E. \(6/(n+4)\)
Let´s explore the simplest particular case: n = 4 Choosing any 3 points among A, B, C, D (see figure), it´s CERTAIN that we will have chosen a right triangle, therefore when n = 4 we expect 1 as our "target". (A) 3/6 is not 1, refuted (B) 6/6 is 1, this alternative is a "survivor" (C) 2/14 is not 1, refuted (D) 3/3 is 1, this alternative is a "survivor" (E) 6/8 is not 1, refuted Now let´s compare (B) (n+2)/6 and (D) 3/(n1) ... When n increases, (n+2)/6 increases and this is no good, hence (B) is refuted. (Reason: with greater values of n, the probability of getting a right triangle decreases!) The only "survivor" (D) is the right alternative choice! This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. POSTMORTEM (solving the general case to prove that (D) is really the right answer, without excluding the other choices, as we did before): There are C(n,3)*3! = n(n1)(n2) equiprobable possible choices if we consider the order in which the points are chosen in the circle (to help the "favorable counting")! A favorable situation is such that 2 of the 3 chosen points must be opposite to each other (to be a diameter of the circle). First choice : n possibilities ("free") , second choice (scenario A): 1 possibility (to have the diameter already) and third choice (in this scenario): (n2) choices (any point remaining is good) :: Total: n.1.(n2) choices First choice : n possibilities ("free") , second choice (scenario B): (n2) possibilities (NOT to have the diameter yet) and third choice (in this scenario): 2 choices (diameter with first, or with second choice) :: Total: n.(n2).2 choices Scenarios A and B are mutually exclusive, hence we may add the total possibilities: 3n(n2) possibilities ? = 3n(n2) divided by C(n,3)*3! = 3/(n1) as expected!
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n points are equally spaced on a circle, where n is an even number gre
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