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Date: 8 Mar 1995 17:04:44 -0500 From: cchen@walrus.mvhs.edu (Chianne Chen) Subject: question Hi, I have a problem: A wooden dowel is randomly broken in 2 places. What is the probability that the 3 resulting fragments can be used to form the sides of a triangle? Thanks, cchen@walrus.mvhs.edu
Date: 9 Mar 1995 17:55:17 -0500
From: steve@mathforum.org (Stephen Weimar)
Subject: Re: question
I'd say it makes a difference how it's broken and what you mean
by random. For instance, is the stick broken once and then one
of two pieces is selected at random to be broken again ("pick one
and break again")? Or do you pick two spots on the dowel at
random where the breaks will be made (two-at-once)?
Let's look at the second situation because the first seems similar
but a little harder to calculate.
Two-at-once is like "pick one and break again" except that with
the later we have a fifty percent chance of choosing the bigger
piece for the second break; whereas for the two-at-once, since
the location of the second break is made with reference to the
whole stick, you have a greater than fifty percent chance of
making the next break in the bigger piece since the bigger piece
is more than fifty percent of the whole stick. So the probability
of "two-at-once" resulting in a triangle should be better than
that of "pick one and break again".
Now what does it take for three sticks to be able to make a
triangle? Will any three sticks do? Once you express the
minimum conditions the broken pieces must satisfy in relation
to each other if they are to make a triangle, then you can make
3 statements, in this case, 3 inequalities to solve. Try to figure
this out before reading the "spoiler" below. When I set up my
equations, I let x be the distance from the left side of the dowel
to the first break. Let y be the distance from the left side to the
second break. Now we have two sets of inequalities. When y
is to the right of x, we have one set. When y is to the left of x,
we have another. But we can see that these two situations are
equally likely, in fact the probability in each case should be the
same, so once we have calculated one the total will be easy to
figure.
*********************
"y to the right of x"
*********************
("Semi-Spoiler")
The lengths of the pieces are x, 1-y, y-x. So we can set up
three inequalities since any two sides added together have to
be longer than the third.
What happens if you set up these inequalities and solve them?
I'll leave it to you to write out the inequalities.
When doing problems such as this I like to make a 1 by 1 square,
the area of which represents the total probability (1). Let's say
the length of the unbroken stick is one unit. The horizontal side
of our square could represent the first break which is anywhere
from 0 to 1. The vertical side of our square will be the second
break, y.
If you graph the inequalities you get a triangular area occupying
the lower right hand half of the upper left quadrant. I don't know
if this picture comes out on your screen, you may have to change
fonts. The area of that triangle is equal to the probability of "y
to the right of x".
| /|
| / |
| / |
|/_ _|
|
|
|_ _ _ _ _ _
0 1/2 1
Don't forget about "y to the left of x".
Once you have completed this version of the problem you might
want to go back and see if you can set up the inequalities for
"pick one and break again". The idea is the same but the math
can be harder.
-- steve "chief of staff"
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