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

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

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.

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.

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.

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."



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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.

gmatbusters, Thanks! Learning something from you every day!

Thanks!!!

I will make sure learning continues

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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