Monday, March 30, 2009

Probability Problem


Lee found this nice problem in the Minitab manual that accompanies "Elementary Statistics" by Mario F. Triola. The problem posed is:

Two points along a straight stick are randomly selected. The stick is then broken at these two points. Find the probability that the three pieces can be arranged to form a triangle.

pic by flowers & machinery

6 comments:

HM said...
This comment has been removed by the author.
HM said...

SPOILER WARNING: Don't click the following link until you solved the problem. Lee has nice geometric solution over at Prime Puzzle.

Max said...

Hello everyone. This comes to you
from "Chip." I'm Lee's younger brother - and also referred to as 'Max' on blogs. I've been enjoying what I've seen of yofx and just felt like stopping by to say hello -- add a comment. In looking at this probability problemI tried to rub two sticks together (in my head) to get a fire going (mathematically) but must confess I resorted to what Lee knows as one of Chip's "gut feeling approaches" to what the answer might be. For my admittedly very dangerous logic, you can visit this picture, but only if you are certain you know the correct solution. Again, this could be entirely wrong, but Lee said it is better to try (and be wrong) than not to try at all.

HM said...

Chip,

Thanks for the visit! Like the picture, but didn't see a probability anywhere? I'm not sure what your answer is?

Tinyc Tim said...

I think what may be going on is Chip is simply giving an example of a line that has two points marked on it and with his particular marks, a triangle can be formed. But you're right - this does not answer the probability question. There are other points that could be put on the stick that would correspond to three line segments that would *not* make a triangle.

A "simulation" approach to this problem might be useful: Think of the line as being marked, starting at the left and ending at the right, with digits 0 1 2 ... 20. Be sure the digits are equally spaced apart. Then put 19 pieces of paper, labeled 1 thru 19, in a hat. Pull two of them out of the hat and cut your stick at these places. Try to make a triangle with your pieces. Glue your stick back together. Put your pieces of paper back in the hat. Shake the hat. Pull two more pieces of paper out. Cut your stick. etc. Do this routine 20 times. (If necessary, go to the drug store to restock your glue supply). You of course are keeping track of the number of times you were able to make a triangle. Call the final number of times you were able to make a triangle t.

An estimate for the probability of getting three pieces that will make a triangle should be provided by the value of t / 20.

(Of course the above is more of a "thought experiment" and, unless you're really into wasting glue, paper, sticks etc., I'm only about 1/4 serious you do this (and thus 3/4 serious you do something less ridiculous)).

It might be worth doing this a few times, in your head. I think you'll begin to understand better what is required of the pieces when they are being asked to make a triangle. For example, what if you drew the numbers 2 and 4 out of the hat. Could you make a triangle with the pieces that would be associated with these numbers?

What is the key requirement of the longest piece of the three pieces in order for the three to form a triangle?

I wrote a computer program that uses the above approach. SPOILER WARNING: Don't click this if you want to solve this problem by writing your own computer simulation.

Max said...

Well, it wasn't until just now that I see my link was not the link I had intended to use. Sorry about that. I was not aware that my link was pointing to a .jpg file -- when it fact I really meant for it to point to an html file. As I post the hopefully corrected link here, I am also (as I mentioned to Lee in a separate email) not pleased with my reasoning (in the now visible text) and I must confess again that it is lacking mathematically in many respects, but it was just a hunch in my own thinking -- so my answer of sorts is now here for the reading. I'll take another look at your good comment Lee and see it I can "put it together" in a better way. Thanks for posting my initial comment. This is all very interesting! Chip