A stick of length 1 is randomely broken in 3 pieces, whats the probability pieces can form sides in a triangel

Question

This is a geometrical probability problem

Answer 1

I trust you mean broken in 2 place to form 3 pieces; is this why Aviophage is laughing?

Vlee is correct in using the Triangle Inequality to show that you need no piece of length 1/2 or more. But I don't buy his probability analysis.

Let's assume the 2 breakpoints are chosen independently and randomly along the stick. Call them x & y and let's draw the sample space of solutions (hard to do here). In the 1 by 1 box, the solution is a sort of bow tie, two triangles, one connecting (0, .5), (.5, .5) and (.5, 1), the other (.5, .5), (,5,0), and (1, .5), all with dashed lines since the boundary values are not solutions.

How do you get this? Look at enough values. For x = .1, what works are y values between .5 and .6 (.5 or less, the third piece is too long, .6 or more, the middle piece is too long). For x = .4, the y-values for pieces that can form a triangle are between .5 and .9. For x = .8, the y-values are between .3 and .5. Draw this and you'll see the rotated bow-tie.

Each of the two triangles is half of a 1/2 by 1/2 square, so together their area is 1/4. That means the probability of 2 random breaks giving pieces that can form a triangle is 1/4.

Answer 2

First of all the answer is NOT 100% Anyone who is assuming so is forgetting a very important fact from elementary mathematics: namely the triangle inequality, which reads as follows:

For any triangle, the measure of any side must be less than the sum of the measure of the other two sides.

Hence if one piece is of length 0.6 and the other two pieces are each of length 0.2 (for a total of 0.6 + 0.2 + 0.2 = 1 as expected) one can NOT make a triangle (if you don't believe me try it or try drawing a picture where this works).

Now there is an issue with this problem that may prevent us from solving it, namely HOW does one break a stick into three random pieces. There are probably several ways of doing this, but the most basic way I can think of is probably that one chooses two random points (say a and b, where a < b) on the stick and one piece is represented by the interval from 0 to a, another from the interval from a to b, and the last from the interval from b to 1.

In order to denote intervals I will use the notation (p, q) to denote an interval from the point (number) p to the point q. The notation (p, q) also implies that p < q in that case.

Now observe that the length of any piece is greater than the sum of the lengths of the other two pieces if and only if the length of one piece is greater 0.5. So how is this possible.

First assume we have two randomly chosen numbers from (0, 1). Call these numbers x and y. Here are the cases that allow for this to occur (I'm kind of doing this problem by finding the opposite answer, so these are the cases in which you CAN'T form a triangle):

CASE I

0.5 < x < y < 1

In this case the meausre of (0, x) > 0.5 > measure of (x, y) + the measure of (y, 1).

CASE II

0.5 < y < x < 1

Because: m(0, y) > 0.5 > m(y, x) + m(x, 1)

CASE III

0 < x < y < 0.5

Because: m(y, 1) > 0.5 > m(0, x) + m(x, y)

CASE IV

0 < y < x < 0.5

Because: m(x, 1) > 0.5 > m(0, y) + m(y, x)

CASE V

0 < y < 0.5 + x < 1

Because: m(y, x) > 0.5 > m(0, y) + m(x, 1)

CASE VI

0 < x + .5 < y < 1

Because: m(x, y) > 0.5 > (0, x) + m(x, y)

You can verify on your own that all these cases are mutuality exclusive (only one can be true), and that no other cases allow for one piece to be longer than the sum of the lengths of the other two pieces.

Now to find the probabilities. We want to find the probability that one of CASES I - VI hold. Later we'll find the probability that none of them hold.


Recall that x and y are independent. Therefore the probability that x > 0.5 is 1/2 and the probability that y > 0.5 is also 1/2. Therefore the probability that both x and y are greater than 1/2 is 1/4. When both x and y are greater than 0.5 then either CASE I holds or CASE II holds.

P(CASE I or CASE II) = 1/4.

Likewise the probability that both x and both y are less than 0.5 is also 1/4. If both are less than 0.5, then either CASE III or CASE IV holds.

P(CASE III or CASE IV) = 1/4.

CASE V and CASE VI are most easily demonstrated using calculus. The probability of either case is given by

∫ (1 - x) dx [from 0 to 1] = 1/8.

So

P(CASE V or CASE VI) = 1/8 + 1/8 = 1/4.

So the sum of the probability of all cases (and we can add them up since they're mutually exclusive) is 1/4 + 1/4 + 1/4 = 3/4.

Once again, this is the probability that a triangle CANNOT be formed.

The probability that a triangle can be formed therefore is the probably that a triangle cannot cannot be formed (double negative intended), which is 1 - 3/4 = 1/4.

Once again the sticks can be broken (still randomly) in different ways. So you can find different "correct" answers.


For more insight check one of the links provided in the sources.

Answer 3

Ok you have 3 pieces of length a, b, c where 0
c < a + b

Because a+b+c=1, a+b = 1-c and so we require that c<1-c => c<1/2. This implies that as long as the largest piece is less than 1/2, then you can form a triangle.

Cut the stick once. Remove the part closest to your left and set it aside. With the remaining part of the stick, what's the probability that I will cut it so it can form a triangle? Draw a graph with x-axis being the length of the part closest to your left and y-axis being the remaining length of the stick. The area under this graph is the region of possibility and this region is a triangle of width 1 and height 1. Shade in the area where the point needs to be cut such that the largest piece is at most 1/2. This region is bounded by y=1/2, x=1/2 and a line from (0,1/2) to (1/2,0). This is a triangle of width 1/2 and height 1/2. Therefore the probability is the area of the triangle divided by the region of possibility = 1/8 / 1/2 = 1/4 => 25%.

Answer 4

the pieces have to satisfy the triangular inequality
a + b >= c
a + b + c =1
a + b = 1 - c >= c
or 2c <= 1
c <= 1/2

so for the 3 sides to form a triangle, no one side can be >1/2
so
Prob the 3 sides can form a triangle
= prob all 3 sides < 1/2
= (1/2)(1/2) (1/2)
= 1/8

Answer 5

I think there is a 100% chance that they will form a triangle- unless one of the pieces has a length of "0". In which case you have only two pieces. Any three lines can form a triangle regardless of their lengths.

Answer 6

The probability is 1:1, 1, or 100%. Choose the one you wish. This will most likely not be an equal lateral triangle but nevertheless, a triangle.

Answer 7

any 3 lines can form a triangle....

so probability is 1

Answer 8

Get out and get some fresh air. Take a walk. Relax. Sober up, if necessary. Then come back and read the question you asked. Think it over. See why we are all laughing?

Answer 9

100%