Randomly Cutting a Rope into Two Segments
Date: 05/14/98 at 20:12:26 From: William Luh Subject: Probability of cutting ropes A rope 20m long is randomly cut into two segments, each of which is used to form the perimeter of a square. (a) Find the probability that the larger square has an area greater than 9m^2. (b) Find the probability that the total area of the two squares is greater than 20.5m^2. The reason why I am stumped is because the rope can be cut to any length. Essentially, I can have 0.01m and 19.99m. Here's what I've tried: a) Let piece 1 = p Let piece 2 = 20 - p restrictions 0 < p < 20 Then make squares with pieces 1 and 2: Area of square for piece 1: (p/4)^2 Area of square for piece 2: ((20-p)/4)^2 Let piece 1 be the larger of the two, so: p > 20 - p 2p > 20 p > 10 This makes sense. If p = 10, then both squares would have the same dimensions, and we would not be able to determine a LARGER square from the two because they would be equal. Now, for part a), we are told that the larger area has to be greater than 9m^2. So: (p/4)^2 > 9 p^2 > 144 p > 12 To have an area greater than 9 units square, the perimeter or length of the piece would have to be greater than 12 units. So the question asks for the probability, and I answer: P(Area > 9|larger square) = P(Area > 9 AND larger square)/P(larger square) = 7/9 I'm trying to decide how to interpret the inequalities. Should the probability be 7/9 or 8/10? I'm confused. ============================================================== Part (b) is also interesting, because I used some calculus. I will restate the question now: Find the probability that the total area of the two squares is greater than 20.5m^2. Let function A(p) represent the sum of the areas, so: A(p) = (p/4)^2 + ((20-p)/4)^2 using info from first part The graph of the function A(p) is a parabola concave up, so we will find a minimum area if we set A'(p) = 0: A'(p) = 2(1/4)(p/4) + 2(-1/4)((20-p)/4) Set A'(p) = 0 and solve for p. The result is p = 10, as predicted. When p = 10, A(10) = 12.5. The maximum would occur when one piece of the string approaches 0 while the other approaches 20m, yielding a maximum area of 25m^2. So: 12.5 <= A(p) <= 25 The question asks to find probability that A(p) > 20.5. If we work through the mechanics of A(p) > 20.5, we find that this occurs when: 0 < p < 2 (1) 18 < p < 20 (2) Now we must remember that the maximum area occurs when p -> 20 and minimum area occurs when p = 10, so there is a 10 unit range. Looking at restriction (2), there is a 2 unit range. Therefore, the probability is 2/10 = 0.2 Please help me, Dr. Math. I know there is something missing or something flawed, but I can't seem get my brain around this problem. If I'm totally on the wrong track, please show me how to do at least part (a) or (b). Thanks, William
Date: 05/15/98 at 10:58:56 From: Doctor Pat Subject: Re: Probability of cutting ropes William, Wow, you put a lot of good work into this, and I think you got the second part exactly right. In the first part you did some great analysis and walked right up to the answer, but didn't see it. I would suggest thinking of the problem as making a single cut, instead of making two pieces. One thing is easier to focus on than two, usually. Think of the rope as laying along the x-axis from zero to 20. If you use your value p as the coordinate of the cut, then one of the pieces is p units long, and the other is (20 - p), as you also figured. Now revisit the problem thinking of this question: If I picked a point at random along the segment from 0 to 20 to make the cut, where would it have to be to produce a square with an area greater than 9 sq. units? You have really found this answer already. Just figure what fraction of the time this would occur by chance, and I think you will see the solution. Good luck! -Doctor Pat, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 05/16/98 at 18:47:51 From: William Subject: Re: Probability of cutting ropes Dear Dr. Pat, Thank you for the encouragement, and finding my mistake. I have revised part (a). I think the probability is conditional because we are asked to examine the large square and not just any square. I took your advice and drew a number line: ------------------------------------------------------------------ 0 10 12 20 |----------------------------------| P(larger square) |---------------------------| P(area > 9 AND larger square) Therefore, IF this is indeed a conditional probability, then: P(area>9|larger square) = P(area>9 AND larger square)/P(larger square) = (8/20) / (10/20) = 4/5 Would this be correct now? Thank you, William
Date: 05/19/98 at 16:02:31 From: Doctor Anthony Subject: Re: Probability of cutting ropes For an area of 9m^2, the side must be greater than 3, so perimeter will be greater than 12. If you have a 20m length of rope, then a cut in the range 0 < x < 8 or in the range 12 < x < 20 will give you one piece with a perimeter greater than or equal to 12. The probability of a cut in one or the other of those regions (assuming the cut to be equally likely at any point) is 16/20 = 4/5. If the cut is at a distance x from one end, the two portions are of length x and 20 - x. The sides are of length x/4 and (5 - x/4), and the areas total to: A = x^2/16 + (5 - x/4)^2 > 20.5 x^2/16 + 25 - 5x/2 + x^2/16 > 20.5 x^2/8 - 5x/2 + 9/2 > 0 x^2 - 20x + 36 > 0 (x - 2)(x - 18) > 0 This will be satisfied if x < 2 or x > 18 So probability of being in either of those two ranges is 4/20 = 1/5. So probability that area is greater than 20.5 is 1/5. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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