Urgent Probability Question?

Question

Take a stick of unit length and break it into three pieces, choosing the break points at random, independently and with uniform density. What is the probability that the three pieces can be used to form a triangle?

Detailed explanation is greatly appreciated. Thanks!!!!

Answer 1

1/4

To see why, draw the unit square and let x be the position of one break and y the position of the other (the stick will be of length 1). Then the probability that the three pieces form a valid triangle is just the area of the x,y break positions that form a valid triangle.

Now, first you can knock out both the upper right and lower left quarters, because they correspond to the leftmost and rightmost pieces being of length over 1/2. You can also knock out two triangles of size 1/8 -- one touching point (0,1) and the other touching point (1,0) (these triangles correspond to the case when the middle piece is of length over 1/2).

What is left is two triangles each of size 1/8. This is a bowtie shaped region bounded by the lines x=1/2,y=1/2,y-x<1/2,x-y<1/2. Thus the remaining area (which corresponds to the probability that the three pieces form a valid triangle) is 1/4.

You can also do this with a double integral, but it is a bit tricky to get right that way.

Answer 2

Call the lengths of the legs of the triangle 'a', 'b' and 'c' where 'c' is the longest leg. Notice that the sum of 'a' and 'b' must be greater than 'c' otherwise it won't work (do draw some pictures to see this). Therefore, piece 'c' must be less than half the length of the stick (otherwise, 'a' and 'b' won't be able to "reach" a vertex).

So now the problem's reduced to finding the probability that the two break points will not fall on the same half of the stick. The first break point can be anywhere. Whatever side of the midpoint it's on, the second break point must be on the other... that's a 50% chance.

Feel free to email me for any clarifications.

Answer 3

This means the stick has to be broken twice. If one of the pieces is longer than 1/2 you can't make a triangle.

If the first break is at:

Zero - the second break is on the interval (0,1/2) → 50%
1/4 - the second break is on the interval (0,3/4) → 75%
1/2 - the second break is on the interval (0,1) → 100%
The rest is by symmetry.

It averages to 75%. So the chance of being able to form a triangle is 75%.

Answer 4

It seems to me that there is a 100% probability that three pieces of straight whatever, of random size, could be used to form a triangle.
I cannot think of a condition where they couldn't.

Answer 5

peeps give some retard questions but this is how u do it 1)100 divide by 35% 2)unlikely 3)1/65 4)even chance 5)likely 6)1/3 please give me 10 points

Answer 6

any three sides can form a triangle....

Answer 7

80 - 99%. hope this helps