# Probability of a Hit Results Problem

Probability theory and statistics

### Probability of a Hit Results Problem

Hello everyone,

I am currently involved in calculating the probability of a hit on a given target using the following bivariate normal distribution equation:

P_H_equation_2.png (3.53 KiB) Viewed 100 times

where,

Sigma-x = Standard Deviation of the distribution of all the horizontal errors
Sigma-y = Standard Deviation of the distribution of all the vertical errors
x_m = mean horizontal error / horizontal distance of aiming point and mean impact point
y_m = mean vertical error / vertical distance of aiming point and mean impact point
A = Area of the target

P_H = probability of a hit (should be a value between 0 and 1).

using the following values:

Sigma-x = 1.6 m
Sigma-y = 1.6 m
x_m = 0.0 m
y_m = 0.0 m
A = -6 < x < 6 and -6 < y < 6 The target is 12 m x 12 m in size. The aim point is seen as the origin (0,0).

=> P_H = Double Integral of (1/(2*pi*1.6*1.6))*(e^(-(((1/1.6)*(x-0.0))^2 +((1/1.6)*(y-0.0))^2))); for the given domain of A in terms of x and y.

These values give me a P_H of 0.5.

I do not understand this result. Given the target size (12 x 12 m) and dispersion (sigma-x and sigma-y?) should not the probability of a hit be 1.0 (100%)

I tried different values for all values and unfortunately i have never gotten a value that goes beyond 0.5. -> The maximum calculable value seems to be 0.5 (50%).

I have checked the equation several times and could not find an error.

Could i just add a "correctional factor" of 2.0 to P_H and consider the results to be valid ? In a way, I consider this "cheating" but I do not know what else to do at this point in time.

I would appreciate any help or feedback to solve this problem. Thank you all in advance !
Stoff

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