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Topic: Bullet Holes (Read 664 times) |
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ThudnBlunder
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Bullet Holes
« on: Feb 1st, 2005, 12:56pm » |
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A man fires a rifle from a long distance at a target which is 10 cm. square, each bullet making a circular hole 1 cm. in diameter. The holes are uniformly distributed in the target area with none extending beyond the boundary.
1) How many bullets must be fired before there is a 50% chance that two bullet holes overlap?
2) How many bullets must be fired before there is a 50% chance that the next bullet hole will overlap an existing hole?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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JocK
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Re: Bullet Holes
« Reply #1 on: Feb 1st, 2005, 2:22pm » |
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Obtaining estimate answers to both question is relatively easy:
:: given the Poisson nature of the process, firing N bullets will lead to a hole density of
[phi] = 1 - e-x
where x = [pi]N/400 is the hole density one would obtain if none of the holes in the target would overlap.
The N value for which [phi] [approx] x - [pi]/400 corresponds to the number of bullets at which the chance that two holes overlap is about 50%. This happens for N [approx] 16.
The N value for which [phi] [approx] 1/2 corresponds to the number of bullets for which there is an approximate chance of 50% that the next hole overlaps. This happens for N [approx] 88. ::
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| « Last Edit: Feb 1st, 2005, 2:27pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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JocK
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Re: Bullet Holes
« Reply #2 on: Feb 2nd, 2005, 2:25pm » |
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Did no one spot the deliberate mistakes..? 
:: The N value for which [phi] [approx] x - [pi]/800 corresponds to the number of bullets at which the chance that two holes overlap is about 50%. This happens for N [approx] 11.
The N value for which [partial][phi]/[partial]N [approx] [pi]/800 corresponds to the number of bullets for which there is an approximate chance of 50% that the next hole overlaps. This happens for N [approx] 88. ::
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| « Last Edit: Feb 2nd, 2005, 2:26pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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SWF
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Re: Bullet Holes
« Reply #3 on: Feb 2nd, 2005, 6:55pm » |
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I did not notice deliberate mistakes, but immediately thought your value for part b) was way too high. I'd say 10 for a) and 19 for b).
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SWF
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Re: Bullet Holes
« Reply #4 on: Feb 3rd, 2005, 5:43pm » |
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Nobody noticed my deliberate mistake either. Here is an adjustment along with description of the method I used to get an approximate answer:
The bullet holes have radius 0.5 cm. If the centers of two hole are within 1 cm of each other, they will overlap. Since they do not extend beyond the boundary, the center of each hole is within a 9 by 9 cm square.
How much area within the 9 by 9 square does an existing hole take away from from potential location for center of a non-overlapping hole? If it does not overlap the border of the 9 by 9 square, it blocks area [pi]. 64 of the 81 square cm fit this condition. If an existing hole overlaps the border (and ignoring the corners), it will block somewhere between [pi]/2 and [pi] square cm, call it 0.75*[pi]. So the average area blocked up by hole is roughly ([pi]*64 + 0.75*[pi]*17)/81. Divide by area of the 9 by 9 square again to get probability of the center of 2nd hole overlapping a given hole: 0.03675. Probability of missing a given existing hole is then 0.96325; call this number p. Because of the way I estimated, this value for p is probably a little bit too low.
For part 2): After n holes, the probability of the next one missing all the existing holes is roughly pn. Set this to 0.5, solve for n and round to the nearest integer to get n=19.
For part 1): Probability of no holes overlapping after n shots equals probability of each bullet missing all the previous ones multiplied together: p0*p1*p2*...*pn-1 = p(n-1)*n/2. Set this to 0.5, solve for n, and round to nearest integer to get n=7.
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