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Re: Two points were chosen randomly on a number line [0,1]. The points [#permalink]
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gmatophobia wrote:

Therefore for the inequality to hold true, the sum of the two sides of the triangle should be greater than 0.5. For the sum of two triangles to be greater than 0.5, the two points should lie on opposite sides of the midpoint (\(\frac{1}{2}\)) as shown below -

--------- N ------------- X ---- \(\frac{1}{2}\) ---- Y ------- M -----------------

For the first point (say X), the probability that the point lies on one of the sides of 0.5 = \(\frac{1}{2}\)
For the second point (i.e. Y), the probability that the point lies on the other side of 0.5 = \(\frac{1}{2}\)

Net Probability = \(\frac{1}{2} * \frac{1}{2} = \frac{1}{4}\)


I think even if we get one point on one side of mid-point and second point on the other side , it won't be sufficient to form a triangle.
We need to add one more condition that maintains that the two points are not far apart from each other in a way that the distance between them is not greater than 0.5.
Like below example:


N -- X ---------------------- \(\frac{1}{2}\) --------------------- Y -- M

Does this makes sense ?
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Re: Two points were chosen randomly on a number line [0,1]. The points [#permalink]
Not too sure but would my guess be right if I say that this type of question won't come on the new focus GMAT?
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Re: Two points were chosen randomly on a number line [0,1]. The points [#permalink]
siddhant49 wrote:
Not too sure but would my guess be right if I say that this type of question won't come on the new focus GMAT?

­Not really, this is coordinate geometry and probability so such questions can be expected in the GFE.

 
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Two points were chosen randomly on a number line [0,1]. The points [#permalink]
gmatophobia wrote:
Bunuel wrote:
Two points were chosen randomly on a number line [0,1]. The points divide the line into three smaller lines. What is the probability that divide lines can make a triangle?

A. 1/2
B. 2/3
C. 1/4
D. 7/9
E. 5/9



 

Assume a number line as shown below -

--------- 0 ------------------------------ 1 -----------------

We have to select two points between 0 and 1, such that the three smaller line segments can form the sides of a triangle.

Let's mark two points on the number line

--------- N ------------- X -------- Y ------- M -----------------

N = 0; M = 1

The three line segments NX, XY, and YM should form the sides of a triangle.

As the line segments represent the sides of a triangle, the following inequalities should hold true -

NX < XY + YM
XY < NX + YM
YM < NX + XY

NX + YM + XY = 1

We can infer that if the sum of the values of any two sides is less than 0.5, the third side will be greater than 0.5 and hence the inequality will fail.

Therefore for the inequality to hold true, the sum of the two sides of the triangle should be greater than 0.5. For the sum of two triangles to be greater than 0.5, the two points should lie on opposite sides of the midpoint (\(\frac{1}{2}\)) as shown below -

--------- N ------------- X ---- \(\frac{1}{2}\) ---- Y ------- M -----------------

For the first point (say X), the probability that the point lies on one of the sides of 0.5 = \(\frac{1}{2}\)
For the second point (i.e. Y), the probability that the point lies on the other side of 0.5 = \(\frac{1}{2}\)

Net Probability = \(\frac{1}{2} * \frac{1}{2} = \frac{1}{4}\)

­I don't agree with this solution. This analysis misses the important constraints about the middle line segment formed Y - X.

The method of calculating the final probability is also erroneous I think, it should be such that the first point can lie ANYWHERE in the entire range i.e\(. (0, 1),\) and depending on which half of the line it is \((0, 0.5) or (0.5, 1)\) the other point must lie in THE OTHER half of the range.

So probability\( = 1*1/2 = 1/2 \)according to this method - which is incorrect.

As @Flower_Child pointed out, it misses the key constraint that the distance BETWEEN the two points must also be less than 0.5 since the length of the 2nd side \((Y-X) < (1-Y) + (X)\) which simplifies to \((Y-X) < 0.5\)


The idea is that we can choose the first point in 1 way anywhere from \((0, 1) \)exclusive, let's say it's unit from 0

Based on the value of X lying in 1st half or 2nd half of the segment we find the probable values of Y

Case 1: If X is on \((0, 0.5)\) then the second point can fall anywhere in the range \((0.5, X + 0.5)\)

To make calculations easier, let's divide the range by taking 0.1 as the least measurable unit. 

From here it is a bit of an on-the-spot method, and I couldn't find any generalized method to do this.

Here we go. So, if\( X = 0.2 \)then the range of Y will be \([0.6, 0.7]\) inclusive.

Similarly, if we want to find the average range of favorable Y based on the value of X from \(0.1 to 0.4,\) we can do as follows:

The sum of ranges of Y from \(0.5 = (0.6-0.5) + (0.7-0.5) + (0.8-0.5) + (0.9-0.5)\)
Average (total of the ranges/ no. of ranges) = \(1.0/4\)

Thus, the net prob for Y to form a triangle for any X between \((1 to 1/2) = 1/4\)

So, prob that X lies betn \(0 to 1/2\) and Y lies betn \(1/2 to (X-1/2) is 1/2 * 1/4 = 1/8\)

Case 2: Similarly we can find that the prob X lies between 1/2 to 1 and Y lies between \((X-1/2) to 1/2 = 1/2 * 1/4 = 1/8\)

Hence, net probability \(= 1/8 + 1/8 = 1/4\)­
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Two points were chosen randomly on a number line [0,1]. The points [#permalink]
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