Chop a stick randomly into three pieces, Probablity

[Ahmed Abdullah]

Chop a stick randomly into three pieces (by hitting twice with an axe). What is the probablity that these three pieces of stick will form a triangle?

when two sides of a triangle are given the third should be such as to satisfy the following inequallity: a-b<c<a+b

Thx for your response.

 

[Ahmed Abdullah]

Is the following argument correct?
"Without loss of generality, assume the stick is of unit length and let X and Y be the distance along from one end of the stick (denoted 0) that the cuts are made. Clearly, X and Y have independent, continuous uniform distributions on the interval from 0 to 1. Suppose X < Y; the lengths of the three pieces are then given by X, (Y - X) and (1 - Y), and so the triangle inequality yields:

X =< (Y - X) + (1 - Y) = 1 - X, and so X =< 1/2;

(Y - X) =< X + (1 - Y), and so Y =< X + 1/2;

(1 - Y) =< X + (Y - X), and so 1/2 =< Y.

Hence, we can find the probability that the pieces form a triangle when X < Y by integrating the joint pdf of X and Y over the region given by

0 =< X =< 1/2 and 1/2 =< Y =< X + 1/2,

which is a right triangle in the xy-plane with vertices (0, 1/2), (1/2, 1/2) and (1/2, 1). Since the joint pdf is 1 in this region, the integral is the area of this triangle, and so is equal to 1/8.

When Y < X, a symmetric argument shows that the probability that the pieces form a triangle is also 1/8. Since Y = X has probability 0, we can thus conclude that the total probability the pieces form a triangle is

1/8 + 1/8 = 1/4. "

 

[AlephZero]

Looks pretty good to me.

FWIW I did a simulation of this in Excel. Ten repetitions of 1000 trials gave these numbers of triangles

237 240 245 246 249 256 256 258 269 277

Which suggests you are probably right