Last visit was: 21 May 2024, 16:38 It is currently 21 May 2024, 16:38
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# Two points were chosen randomly on a number line [0,1]. The points

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625633 [14]
Given Kudos: 81918
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3138
Own Kudos [?]: 4449 [4]
Given Kudos: 1856
Location: India
Manager
Joined: 28 Dec 2020
Posts: 148
Own Kudos [?]: 135 [1]
Given Kudos: 522
Manager
Joined: 28 Dec 2020
Posts: 148
Own Kudos [?]: 135 [1]
Given Kudos: 522
Re: Two points were chosen randomly on a number line [0,1]. The points [#permalink]
1
Kudos
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 ?
Intern
Joined: 15 Dec 2023
Posts: 4
Own Kudos [?]: 1 [0]
Given Kudos: 31
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?
Manager
Joined: 25 Oct 2017
Posts: 116
Own Kudos [?]: 46 [0]
Given Kudos: 683
GMAT Focus 1:
655 Q87 V80 DI80
GMAT 1: 690 Q49 V35
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.

Manager
Joined: 25 Oct 2017
Posts: 116
Own Kudos [?]: 46 [0]
Given Kudos: 683
GMAT Focus 1:
655 Q87 V80 DI80
GMAT 1: 690 Q49 V35
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 $$= 1/8 + 1/8 = 1/4$$­
Two points were chosen randomly on a number line [0,1]. The points [#permalink]
Moderator:
Math Expert
93373 posts