Date: 8 Mar 1995 17:04:44 -0500 From: firstname.lastname@example.org (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, email@example.com
Date: 9 Mar 1995 17:55:17 -0500 From: firstname.lastname@example.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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