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n points are equally spaced on a circle, where n is an even number gre

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New post 16 Apr 2018, 03:29
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A
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C
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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. \((n-1)/6\)

B. \((n+2)/6\)

C. \(2/(3n+2)\)

D. \(3/(n-1)\)

E. \(6/(n+4)\)

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n points are equally spaced on a circle, where n is an even number gre  [#permalink]

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New post 16 Apr 2018, 04:28
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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. \((n-1)/6\)

B. \((n+2)/6\)

C. \(2/(3n+2)\)

D. \(3/(n-1)\)

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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IMG_20180416_164848.jpg [ 219.65 KiB | Viewed 1408 times ]


Now , if we put n =4 into remaining options.

C. \(2/(3n+2)\) = \(2/14 = 1/7\)

D. \(3/(n-1)\) = \(3/3 = 1\)

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  [#permalink]

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New post 16 Apr 2018, 05:49
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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 n-1 (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. \((n-1)/6\)

B. \((n+2)/6\)

C. \(2/(3n+2)\)

D. \(3/(n-1)\)

E. \(6/(n+4)\)
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Re: n points are equally spaced on a circle, where n is an even number gre  [#permalink]

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New post 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 n-1 (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. \((n-1)/6\)

B. \((n+2)/6\)

C. \(2/(3n+2)\)

D. \(3/(n-1)\)

E. \(6/(n+4)\)


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New post 16 Apr 2018, 08:16
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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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New post 16 Apr 2018, 08:20
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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  [#permalink]

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New post 16 Apr 2018, 08:45
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1
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= (n-2) points (since 2 points has been chosen for the diameter).
No of choices = \((n-2)C1 = (n-2)\)
Total no of favorable cases = \(n/2*(n-2)\)

Total no. of cases = 3 vertices of right angle ? chosen from n points=\(nC3\)
Thus,Prob(? with 3 vertices)=\(((n-2)n/2)/nC3\)
\(=3/(n-1)\)
Answer D
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New post 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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New post 17 Apr 2018, 04:35
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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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New post 17 Apr 2018, 05:38
gmatbusters, Thanks! Learning something from you every day! :)
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New post 17 Apr 2018, 06:19
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New post 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. \((n-1)/6\)

B. \((n+2)/6\)

C. \(2/(3n+2)\)

D. \(3/(n-1)\)

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/(n-1) ...

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.


POST-MORTEM (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(n-1)(n-2) 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): (n-2) choices (any point remaining is good) :: Total: n.1.(n-2) choices
First choice : n possibilities ("free") , second choice (scenario B): (n-2) possibilities (NOT to have the diameter yet) and third choice (in this scenario): 2 choices (diameter with first, or with second choice) :: Total: n.(n-2).2 choices
Scenarios A and B are mutually exclusive, hence we may add the total possibilities: 3n(n-2) possibilities
? = 3n(n-2) divided by C(n,3)*3! = 3/(n-1) as expected!
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