
Re: probability of a triangle (SPOILER)
Posted:
Jul 12, 2004 7:49 PM


I believe that the answer to part (b) should still be 1/4. 1. Heuristically or intuitively, although the 2 sequences of events as "described" in part (a) and part (b) may "appear" to be different, in the end they really amount both to the same event of breaking a unit length into 3 pieces, whether by making 2 "simultaneous" cuts (part a) or making 1 cut followed by a second one (part b). 2. Mathematically, suppose the first cut in (b) results in 2 pieces, and the piece picked to be further broken has length X, with the other piece of length c = (1X). Then supposed X is further divided in 2 pieces of length b = y and a = (Xy). Clearly, 0 < y < X In the cartesian plane (X, y), the domain of all possible points (X, y) is the right triangle intersection of the 3 areas y>0, y<X and X<1. The subdomain of all points (X, y) resulting in a, b and c forming a triangle must satisfy the 3 conditions a < 1/2, b < 1/2 and c < 1/2, or: y > X  1/2, y < 1/2 and X > 1/2 The area of this triangular subdomain is 1/4 the area of the entire domain, thus the probability remains 1/4, conforming with what intuition indicated.
On Jun 7 12:48:12 1996, Daniel A. Asimov wrote: >In article <9604298333.AA833373730@ccmail.odedodea.edu> Pat_Ballew@ccmail.odedodea.edu writes: >> b) If the length is broken at a random point, and then one of the two >> pieces is randomly selected and broken at a random point on its length >> what is the probability that the three pieces will form a triangle >> >> Pat Ballew >> Misawa, Japan >> Pat_Ballew@ccmail.odedodea.edu > > >Here's my answer to question b): > >After the first random break, we can (without loss of generality) skip the step >of randomly selecting one of the pieces, since the distribution of the left >and right pieces will be identical anyway. This simplifies the problem a bit. > >Say we always choose the leftmost piece; call its length x. > >At this stage we have 2 pieces: [0,x] and [x,1]. > >Now we want to break [0,x] "at a random point", which amounts to choosing >(independent of x) a number y at random in the interval [0,1], where y >represents the fraction of x where the break will occur. > >So, we end up with 3 pieces: [0,xy], [xy, x], and [x,1]. Their lengths are >clearly xy, xxy = x(1y), and 1x. > >The distribution on x and y is uniform on the unit square [0,1]x[0,1]. > >And the region where the 3 pieces form a triangle corresponds (as Mark Burkey, >and via email, Pat Ballew have pointed out) to precisely the condition that >all 3 sides are of length < 1/2. > >The subset of the square, then, where xy < 1/2, x(1y) < 1/2, and 1x < 1/2 >(i.e. x > 1/2) is the thornshaped region bounded on the left by the line >x = 1/2 and on the right by the 2 hyperbolas xy = 1/2 and x(1y) = 1/2. > >Calculus then gives the area of this region  which must be the probability we >are seeking  as Prob(triangle) = ln(2)  1/2 = .1931471805599453094172321.... > >Dan Asimov

