ags@pucc-i (Seaman) (02/22/84)
> You get a call to bring two lengths of chain with you to a job. Where to > you cut the chain to maximize your chances of both pieces being long enough? Let the two required lengths be x and y. Without loss of generality, x >= y. Therefore the needed lengths can be represented as the coordinates of a point in the first octant of the plane: the area in the first quadrant below the line y=x. In order to answer the question, a probability distribution is needed. There must be a way to assign the relative probability of different points within the designated region. Since the area of the region is infinite and the probability distribution must have an integral of 1 over the region, it follows that some areas are more probable than others -- there is no such thing as a uniform probability distribution over the designated region. The question as stated cannot be answered. If you knew an upper bound to the lengths, you could assign a uniform probability -- but no upper bound is stated. -- Dave Seaman ..!pur-ee!pucc-i:ags "Against people who give vent to their loquacity by extraneous bombastic circumlocution."
trough@ihuxa.UUCP (Chris Scussel) (02/28/84)
You are in your shop with one length of chain and your big, immovable chain-cutter.
You get a call to bring two lengths of chain with you to a job. Where to you cut
the chain to maximize your chances of both pieces being long enough?
Trick answer: in the shop!
Of course, what's being asked for is the non-trick answer. And, if you get that,
what if you need three pieces? n pieces? Enjoy!
Chris Scussel
Bell Labs
Naperville, Ill
{BTL}!ihnp4!ihuxa!trough