# How to equate hit probabilties on 2 different surfaces

I'm looking for help in determining how to equate the following:

1. We have a surface of 13 cm X 10 cm.
2. We have a 95% probability of hit on a 16 cm diameter surface.

I would like to equate $B$, probability of hit to the 13cm X 10 cm surface. 95% probability of hit on a circle of 16 cm represent what probability of hit on the rectangle of 13cm X 10cm.

What is the mathematical steps to arrive to the solution? It is a job related problem not a school one.

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What's a "hit"? Are you making an analogy to throwing darts at a dartboard? If so, you need additional information (or equivalent strong assumptions) to get an answer: you need to know the statistical distribution of "hits" around a target's center. –  whuber May 1 '12 at 19:34
Hi Whuber, a better analogy would be to go to the firing range and shoot for fun. So 95% probability of hit represent 1.96 sigma S.D. And the bullet is able to fly and hit the paper target with a precision or dispersion that would put all of the bullet in a 45 mm diameter. But the shooter with this quality of gun/bullet system need to be able to hit the target 95% of the time. –  Claude May 2 '12 at 16:12
OK. But precisely how is the 13 $\times$ 10 surface related to the target? At what point is the shooter aiming? And what distinction are you trying to make between darts and bullets: aren't they essentially the same physical analogy? –  whuber May 2 '12 at 19:13
As for the darts or the bullets analogy it is ok... Yes they are the same but I thought that it would be clearer or more easy to explain. The Point of Aim is the centre of the target The 13*10 (130 cm^2) surface is the new target. In one case, I have a 95% probability of hit on a circular surface of 16 cm diametre (201 cm^2). How does this translate in probability of hitting the 130 cm^2 target? p.s. thanks for the help in any cases –  Claude May 2 '12 at 20:16

If we assume the distribution of hits is binormal, then the distribution of the distance to the center of the target is a scaled Chi distribution with two degrees of freedom. From this (applying its inverse CDF), we find that its 95th percentile coincides with the radius of 16/2 = 8 centimeters when the scaling factor equals 1.33523. This factor is also the standard deviation of the components of the binormal distribution. Integrating the PDF of that binormal distribution over a 13 by 10 cm rectangle centered at the point of aim gives 0.83311, the desired value of $B$.

Here's a picture showing a shaded contour plot of the PDF restricted to that rectangle, with the circular target behind it for reference:

The value of 0.83311 was found with Mathematica:

With[{s = 8 / InverseCDF[ChiDistribution[2], 0.95]},
NIntegrate[PDF[BinormalDistribution[{0,0},{s,s},0],{x,y}], {x,-13/2,13/2}, {y,-10/2,10/2}]
]


It was checked by simulating 100,000 independent shots and reporting the proportions that (a) fell within the 16 cm circular target and (b) fell within the rectangular target:

With[{s = 8 / InverseCDF[g, 0.95], n = 100000},
data = RandomReal[BinormalDistribution[{0, 0}, {s, s}, 0], n];
old = Length[Select[data, Norm[#] <= 8 &]] / n;
new = Length[Select[data, Abs[#[[1]]] <= 13/2 && Abs[#[[2]]] <= 10/2 &]] / n;
{old, new} // N
]


The output of {0.94919, 0.83331} is close enough to the intended values of {0.95, 0.83311} to confirm the correctness of the calculations.

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Whuber, Thank you very much for this. I cannot clainm to understand everthing. But I'll make special diligence to understand this. –  Claude May 3 '12 at 18:20
WHuber, Together with a co-worker, we have understood your approach and solution... thanks –  Claude May 3 '12 at 18:39