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sateesp
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Probability for forming a triangle
« on: Aug 23rd, 2010, 9:34am » |
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Given a stick and randomly made two cuts on this stick, which makes stick into three pieces. What is the probability that with these three pieces we can form a triangle?
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towr
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Re: Probability for forming a triangle
« Reply #1 on: Aug 23rd, 2010, 9:39am » |
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It's equal to the probability that no piece is greater than half the stick.
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mmgc
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Re: Probability for forming a triangle
« Reply #2 on: Aug 24th, 2010, 5:14am » |
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on Aug 23rd, 2010, 9:39am, towr wrote:| It's equal to the probability that no piece is greater than half the stick. |
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yes ...I am little curious, can we calculate this..
say, L is the length of stick.
P ( we get a triangle ) = 1 - P ( one piece > L / 2)
P ( one piece > L / 2 ) = P ( Both the cuts were made in one half )
considering both events ( cutting ) are independent
P ( one piece > L / 2 ) = 1/2 * 1/2 = 1/4
thus, P ( we get a triangle ) = 3/4
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towr
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Re: Probability for forming a triangle
« Reply #3 on: Aug 24th, 2010, 5:56am » |
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You forget that even if the cuts are not in the same half, there may be half a stick distance between them.
Suppose we have cuts at a and b. Then we can't have a triangle if
a and b > 1/2; the probability of this is 1/4
a and b < 1/2; the probability of this is 1/4 (it's just the mirror of the previous case)
a - b > 1/2; perhaps the simplest way is to draw this in a square, with a,b on the axes. Then you'll see you get a small triangle with base 1/2 and height 1/2. So it's surface area (equal to it's probability) is 1/8
b - a > 1/2; same as the previous case, but on the opposite side of the square.
So, the cases where you don't have a triangle add up to a probability of 3/4, meaning that there's a 1/4 chance you do have a triangle.
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| « Last Edit: Aug 24th, 2010, 5:57am by towr » |
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sateesp
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Re: Probability for forming a triangle
« Reply #4 on: Aug 25th, 2010, 11:20am » |
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Following is my answer to this. Correct me if I am wrong.
Assume name A is given to the first cut happened from the left side of the stick and name B is given to the second cut happened.
If you consider A values (0 to 1) in x axis and B values (0 to 1) in y-axis, this is the complete possible area where A and B values to be considered.
To form a triangle no piece should be greater than 0.5.
1.
To form a triangle, A length should be less than 0.5. All the area where A value is greater than 0.5 is wont form a triangle. Now area will be reduced to 1/2 (Possible area to form a triangle)
2.
B-A length should be less than 0.5, which in turn reduces the reaming half area to 1/4
3.
1-B length should be less than 0.5, which in turn reduces the area to 1/8
Finally, 1/8 area of values for A, B we can form a triangle.
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towr
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Re: Probability for forming a triangle
« Reply #5 on: Aug 25th, 2010, 11:48am » |
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You have to multiply it by two, because the first cut you make may either end up being the first cut from the left or the second cut from the left.
And that way you get the correct answer of 1/4.
Actually, I'm not sure about your reasoning, even with that amendment.
If you draw the area you should have something like:
-------
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-------
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-------
where the red triangular pieces mark the values for a,b that give a triangle.
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| « Last Edit: Aug 25th, 2010, 11:54am by towr » |
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Aditya Ahuja
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Re: Probability for forming a triangle
« Reply #6 on: Dec 25th, 2010, 12:03pm » |
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A Brute Force Solution Sketch:
Let the cut lengths be x1, x2, x3.
Find the solution to the following systems using methods of linear programming problems:
First System - Unrestricted Cuts:
x1 + x2 + x3 = L
x1, x2, x3 > 0
Second System - Triangle Forming Cuts:
x1 + x2 + x3 = L
0 < x1, x2, x3 <= L/2
Find the ratio of the volumes of the convex polytopes obtained. This can be done easily by plotting the solutions in R3 where the results will be restricted to the first octant.
I don't know how helpful this is.
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| « Last Edit: Dec 25th, 2010, 12:09pm by Aditya Ahuja » |
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birbal
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Re: Probability for forming a triangle
« Reply #7 on: Jan 26th, 2011, 10:35am » |
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on Aug 25th, 2010, 11:20am, sateesp wrote:Following is my answer to this. Correct me if I am wrong.
Assume name A is given to the first cut happened from the left side of the stick and name B is given to the second cut happened.
If you consider A values (0 to 1) in x axis and B values (0 to 1) in y-axis, this is the complete possible area where A and B values to be considered.
To form a triangle no piece should be greater than 0.5.
1.
To form a triangle, A length should be less than 0.5. All the area where A value is greater than 0.5 is wont form a triangle. Now area will be reduced to 1/2 (Possible area to form a triangle)
2.
B-A length should be less than 0.5, which in turn reduces the reaming half area to 1/4
3.
1-B length should be less than 0.5, which in turn reduces the area to 1/8
Finally, 1/8 area of values for A, B we can form a triangle.
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i did not understand 1st step. Why 'A' should be < 0.5?
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