Question

An army tank 3.9 m wide needs to travel 47 m to cross a minefield. The enemy that laid the minefield is known to have a standard practice of randomly
placing 160 mines per hectare (10 000 m2). The tank will set off any mine that it passes over. If the tank sets off a single mine then the armour protecting
the engine will crack, but the engine will keep working. However, setting off a second mine will damage the engine and stop the tank in its tracks. What is
the probability that the tank will make it to the other side of the minefield? (Answer to 2 decimal places)
____________
Assume that the tank makes it to the other side. What is the probability that the armour protecting the engine will not have been cracked? (Answer to 2
decimal places)
____________
Assume that there is another minefield with the same density of mines that is wide enough that it is unlikely that a tank will be able to get to the other side
without being stopped. A large number of tanks line up side by side and start crossing this minefield. What do you expect the median distance travelled by
the tanks to be. (Answer to 2 decimal places)*
____________
*Note: Less tanks would be blown up if they followed one another, as the first tank would explode the mines and the following tanks would have a clear
path. This is not happening in this question because the commander is worried that it is easier for the enemy to shoot tanks when they all follow the same
line.

Answer 1

(a) Area of the field which the tank needs to cover = 3.9 x 47

                                                                        = 183.3 m²

For simplicity, we can assume that the area of the minefield is 10000m². It will not make any difference in the final result.

So the tank's path is 0.01833 of the field.

A tank can make it to the other side of the minefield if there are either 0 or 1 mines in the tank's path.

Consider the situation:

A single mine can be in the tank's path with probability 0.018 and will not be in the tank's path with probability 0.98. This is true for all 160 mines.

So, the number of mines in the tank's path is a binomial random variable with p = 0.018 and n = 160.

Probability that there are 0 mines in the tank's path = (1 - 0.01833)¹⁶⁰

                                                                           = 0.052

Probability that there is only 1 mine in the tank's path = (1 - 0.01833)¹⁵⁹ (0.01833)¹ x 160

                                                                              = 0.15481

So, probability that the tank will cross the field = 0.052 + 0.158

                                                                    = 0.21

(b) The armor protecting the tank's engine will not be cracked if there were no mines in the tank's path. We are given that the tank has made it to the other side of the field. So we know that there were either 0 or 1 mines in the tank's path. We need to find the probability that there were 0 mines in its path.

Probability that the tank's armor will not be cracked = 0.052/0.21

                                                                           = 0.25

(c) This question is a bit ambiguous. We need more information to solve it. ('large number' of tanks??)