Question

An army tank 2.1 m wide needs to travel 22 m to cross a minefield. The enemy that laid the minefield is known to have a standard practice of randomly placing 180 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.

（a）What is the probability that the tank will make it to the other side of the minefield?

(b)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?

(c)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.

*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) The tank sweeps an area of 2.1m times 22m= 46.2 sqm of area to reach to the other side.

Given that the density of mines is uniform over an area of 10000/180 sqm.

Hence Probability that the tank reaches other side= probabilty that the tank encounters at most one mine in its path= probability of no mine+ probability of one mine

=(1-46.2*180/10000)+46.2*180/10000*(1-46.2*180/10000)=1-46.2^2*180^2/10000^2=0.30844

(b) probability that armour protected given tank reached=

P(no mine|at most one mine)=P(no mine)/P(at most one mine)=(1-46.2*180/10000)/((1-46.2*180/10000)(1+46.2*180/10000))=1/(1+46.2*180/10000)=0.54597

(c) Let X be the random variable denoting the distance travelled by the tank.

Then P(X<=x)=P(at least two mines till distance x)=(2.1*x*180/10000)^2

To find median distance, we need to solve P(X<=x)=0.5

which leads to