Question

We plan to employ precision munitions to support the NEO. Please note the following features of the planned munitions:

These munitions have a range error with mean zero and standard deviation of 25 meters, and the range error is normally distributed. The munitions have a bursting radius of 35 meters.

Prior planning data has indicated that if one lands within 35 meters of a friendly person, that person is assumed to become a casualty. If the round lands more than 35 meters away, the person is assumed to be non-injured.

Current rules of engagement call for no rounds to be aimed at a point closer than 125 meters from friendly personnel.

We plan to fire as many as 150 rounds in support of the mission. There is intense interest in avoiding fratricide. Please calculate some worst-case probabilities.

1. What is the chance that a round injures a friendly person, given that it is fired at an aim point exactly 125 meters from a friendly person?
2. If you fired 150 rounds, each exactly 125 meters away from a friendly position, what is the chance that no friendlies were injured?
3. In your professional opinion, is the minimum safe distance of 125 meters adequate? Why or why not?

Answer 1

$a.\ when\ we\ aim\ at\ exactly\ 125\ m\ away\ from\ person,\ then \\ person\ will\ be\ injured\ if\ error\ is\ more\ than\ 125-35 = 90\ m. \\ Pr(injured) = Pr(error \ge 90) = Pr(Z \ge \frac{90}{25}) =Pr(Z\ge 3.6) \\ = 0.000159 \\ b. \ Now, \ n=150, \ Let\ 'success' \ be\ 'the\ friend\ is\ injured. ' \\ Then, \ success\ probty\ = p = 0.000159 , \ X:no.of\ friends\ injured. \\ X\sim \ Binomial(np,\sqrt{np(1-p)}) = Binomial(0.02385,0.1544) \\ Pr(X=0) = ^{n}C_{0}(p)^{0}(1-p)^{n} = 0.97643 \\ c. \ Yes,\ minimum\ safe\ distance\ of\ 125\ m \ is\ adequate\ because\ \\ probty\ of\ casualty\ with\ this\ distance\ is\ very\ low\ i.e.\ 0.000159 \\ meaning\ at\ most\ 1 \ case\ in\ 10000\ will\ be\ injuring\ the\ person.$