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![Two points were chosen randomly on a number line [0,1]. The points](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "Two points were chosen randomly on a number line [0,1]. The points"){kind=icon}
08 May 2023, 01:30
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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
A. 1/2
B. 2/3
C. 1/4
D. 7/9
E. 5/9
08 May 2023, 11:56
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Bunuel
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 = \(\frac{1}{2} * \frac{1}{2} = \frac{1}{4}\)
17 May 2023, 07:49
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If we break the line at two random points, x and y , the three resulting pieces will have lengths x, (y−x), and (1−y) if x is to the left of y.
-----------------x ----------------y------------------1 ------(1)
{------(x)------}{----(y-x)-----}{------(1-y)------} length of each section
and y, (x−y), and (1−x) if x is to the right of y
-----------------y ----------------x------------------1 ------(2)
{------(y)------}{----(x-y)-----}{------(1-x)------} length of each section
Taking case (1)
The three pieces form a triangle if sum of any two sections is greater than third.
In other words,
if (x) + (y-x) > (1-y) => y> 0.5 , AND
if (y-x) + (1-y) > x => x< 0.5 , AND
if x + (1-y) > (y-x) => y< x+0.5
Similarly taking case (2) , we will get x > 0.5 AND y < 0.5 AND x < y+0.5
We can draw these inequalities in X-Y graph using X and Y axis , y=0.5 line , x=0.5 line , y=x+0.5 line and x=y+0.5 line and then get the area bounded by these conditions as shown in below pic .
If we plot all six of these inequalities we get the area that represents the proportion of triangles formed from these conditions.
The shaded regions represent the conditions that form a triangle each with area ½ * 0.5 * 0.5 =1/8 .
Both triangles together add up to 1/4 of the total area, or a 0.25 probability of forming a triangle.
pic.png [ 571.69 KiB | Viewed 3570 times ]
-----------------x ----------------y------------------1 ------(1)
{------(x)------}{----(y-x)-----}{------(1-y)------} length of each section
and y, (x−y), and (1−x) if x is to the right of y
-----------------y ----------------x------------------1 ------(2)
{------(y)------}{----(x-y)-----}{------(1-x)------} length of each section
Taking case (1)
The three pieces form a triangle if sum of any two sections is greater than third.
In other words,
if (x) + (y-x) > (1-y) => y> 0.5 , AND
if (y-x) + (1-y) > x => x< 0.5 , AND
if x + (1-y) > (y-x) => y< x+0.5
Similarly taking case (2) , we will get x > 0.5 AND y < 0.5 AND x < y+0.5
We can draw these inequalities in X-Y graph using X and Y axis , y=0.5 line , x=0.5 line , y=x+0.5 line and x=y+0.5 line and then get the area bounded by these conditions as shown in below pic .
If we plot all six of these inequalities we get the area that represents the proportion of triangles formed from these conditions.
The shaded regions represent the conditions that form a triangle each with area ½ * 0.5 * 0.5 =1/8 .
Both triangles together add up to 1/4 of the total area, or a 0.25 probability of forming a triangle.
Attachment:
pic.png [ 571.69 KiB | Viewed 3570 times ]
