Probability of Forming a TriangleDate: 08/09/99 at 09:38:24 From: Sharon Lawson Subject: Probability and Triangles Hi, I wonder if anybody can help me solve the following problem: Three numbers are chosen from the first n natural numbers. What is the probability that they can be sides of a triangle? The numbers are picked "without replacement," so we cannot pick the same number twice. I know that for this to be possible, the sum of any two sides must be greater than the third side, and obviously the sum of the three numbers cannot exceed 3*n - 3, but I don't know how to do it from there. Thanks in advance for any help. Sharon Date: 08/09/99 at 18:47:36 From: Doctor Anthony Subject: Re: Probability and Triangles If the numbers range from 1 to n we can find a pattern for numbers that will form a triangle. In the pattern that I derive below, the case where the two shorter sides sum to give the longest side is included as a collapsed triangle. I give the two shorter sides first, then the longest side. 1+2 - 3 2+3 - 4,5 3+4 - 5,6,7 1+3 - 4 2+4 - 5,6 3+5 - 6,7,8 1+4 - 5 2+5 - 6,7 3+6 - 7,8,9 : : : : : : 1+(n-1) - n 2+(n-2) - (n-1),n 3+(n-3) - (n-2),(n-1),n ------------ 2+(n-1) - n 3+(n-2) - (n-1),n Total = n-2 ------------------ 3+(n-1) - n Total = 2n-7 ------------------ Total = 3n-15 Continuing in this way we get the following series: (n-2), (2n-7), (3n-15), (4n-26), (5n-40), (6n-57), ... The numbers in the brackets 2, 7, 15, 26, 40, 57, ... are given by: r(3r+1)/2 starting at r = 1 The end of the sequence is more complicated and we have different results depending on whether n is odd or even. The top limit for r must be restricted to (n-1)/2 when n is odd and (n-2)/2 when n is even. Thereafter we have the sum of the triangular numbers up to the (n-3)/2 such number (for n odd) and the (n-2)/2 such number (for n even). The triangular numbers are 1, 3, 6, 10, 15, 21, ... and the nth number is given by: n(n+1)/2 The sum of n terms of this triangular number series is given by: (1/2)SUM[n^2 + n] = (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2] = (1/4)n(n+1)[(2n+1)/3 + 1] = (1/12)n(n+1)[2n+1+3] = (1/12)n(n+1)[2n+4] = (1/6)n(n+1)(n+2) So the formula for the number of triangles will be, for n odd: Let k = (n-1)/2 k SUM[r.n - r(3r+1)/2] + (1/6)(k-1)k(k+1) r=1 k SUM[r.n - r(3r+1)/2] + (1/6)k(k^2-1) r=1 And the formula for the number of triangles with n even: Let k = (n-2)/2 k SUM[r.n - r(3r+1)/2] + (1/6)k(k+1)(k+2) r=1 Now you could choose 3 numbers from n numbers in C(n,3) ways, so the probability of forming a triangle, for n odd, is given by: k SUM[r.n - r(3r+1)/2] + (1/6)k(k^2-1) r=1 -------------------------------------- where k = (n-1)/2 C(n,3) and for n even the probability is: k SUM[r.n - r(3r+1)/2] + (1/6)k(k+1)(k+2) r=1 ----------------------------------------- where k = (n-2)/2 C(n,3) To avoid tedious algebra we shall work out this probability in the case where n = 20. k = (n-2)/2 = 9 The top line is: 9 SUM[20r - r(3r+1)/2] + (1/6)k(k+1)(k+2) r=1 = 20.SUM[r) - (1/2)SUM[3r^2] - (1/2)SUM[r] +(1/6)(9)(10)(11) for r = 1 to 9 = 20(9)(10)/2 - (1/2)(3)(9)(10)(19)/6 - (1/2)(9)(10)/2 + 165 = 900 - 427.5 - 22.5 + 165 = 615 and C(20,3) = 1140 and so the required probability is 615/1140 = 41/76 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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