Probability of Making Triangle with Three Stick PiecesDate: 02/24/2007 at 03:26:40 From: Kiran Subject: Find probability that 3 parts of a stick form a triangle. Take a stick of unit length and break it into two pieces, choosing the break point at random. Now break the longer of the two pieces at a random point. What is the probability that the three pieces can be used to form a triangle? I'm not sure what to do after taking the initial condition. Let a be the point of breakage and its distance from one end is x. Then x > 1/2. The second probability gives y > 1/4. How to proceed from there and get the probability? Date: 02/24/2007 at 09:37:07 From: Doctor Greenie Subject: Re: Find probability that 3 parts of a stick form a triangle. Hello, Kiran - I made a rather extensive search of the Internet and the Dr. Math archives and found no satisfactory discussion of this problem. There are several sites I found which discuss the similar problem where the stick is broken randomly in two places--unlike your problem, where the first break is made and then the second break is made in the longer piece. Even in that other case, I found disagreement about the correct answer. And all of the sites I found used methods of solving the problem which seem more difficult than mine. So here is my solution to your problem.... To make a triangle out of three pieces of a stick of unit length, the longest piece must have a length which is less than half a unit. Let's make our first break and let x be the length of the shorter piece. x can vary between 0 and 0.5. Let's pick values in this range and consider the probability that we can make a triangle for that value of x. Suppose x = 0.2; then the long stick has length 0.8. The break on this stick must be made so that the longest piece has length less than 0.5. This means the break must be made between 0.3 and 0.5. So the probability of being able to make a triangle with the three pieces if the first break is at 0.2 is 0.2/0.8. Now let's pick one more value for where we make the first break. Suppose x = 0.4; then the long stick has length 0.6. Again, the break on this stick must be made so that the longest piece has length less than 0.5. This means the break must be made between 0.1 and 0.5. So the probability of being able to make a triangle with the three pieces if the first break is at 0.4 is 0.4/0.6. Using our two specific examples, we can see that, in the general case, the probability of being able to make a triangle out of the three pieces, as a function of the length x of the short piece we get on our first break, is x/(1 - x) Our variable x can range from 0 to 0.5, so the overall probability that we can make a triangle with the three pieces of the stick, given your statement of the problem, is integral from 0 to 0.5 of (x/(1 - x)) --------------------------------------- 0.5 - 0 This turns out to be 2(-.5 - ln(.5)) = 2(.19314718....) = .38629436... I hope this helps. Please write back if you have any further questions about any of this. - Doctors Greenie and Peterson, The Math Forum http://mathforum.org/dr.math/ |
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