probability of forming a quadrilateral with 4 pieces of wire

[circuit]

A wire of 1m length cut into 4 pieces. what is the probability that i can form a quadrilateral with them ?

any suggestions ?

 

[flatulent]

Re: probability of forming a quadrilateral with 4 pieces of

How about the longest wire segment has to be less than the sum of the remaining three. This means that it has to be less than 50% of the original length.

 

[steve10]

Re: probability of forming a quadrilateral with 4 pieces of

(1) Formulation.
Cutting a segment of wire of 1m length into 4 pieces is equalent to sampling three uniformly distributed random numbers in (0,1). Assume X1, X2 and X3 are the three random variables. We can construct three ordered statistics X1:3, X2:3 and X3:3, where X1:3 is the minimum of {X1,X2,X3}, X3:3 is the maxmum of {X1,X2,X3} while X2:3 is something in between. Therefore, we have X1:3 < X2:3 < X3:3. Sometimes (very often), we denote them by Y1, Y2, Y3, and the joint pdf is
g(Y1,Y2,Y3)=6, when 0<Y1<Y2<Y3<1, and 0, otherwise.

(2) Transform.
The four sides are Y1, Y2-Y1, Y3-Y2, 1-Y3.
The necessary and sufficient condition that the four sticks form a quadrilateral is that non of them is equal to or larger than 0.5. Therefore, you have four events:
U1=(Y1<0.5), U2=(Y2-Y1<0.5), U3=(Y3-Y2<0.5), U4=(1-Y3<0.5)

Thus, all you have to do is to calculate the integral
P(U1ΠU2ΠU3ΠU4)=∫∫∫g(Y1,Y2,Y3)dY1dY2dY3 and notice that the region is U1ΠU2ΠU3ΠU4.
Forgive me for not carrying it through. That is just a simple but tedious multiple integral.