Is there a basic solution to this probability problem? There are 16 army squads going to protect 8 different cities (labeled from A to H). Due to weather conditions, it's impossible the communication between the squadrons. Each squadron must deicide which of the 8 cities it's going to protect. Suppose that each army squad decides in an equiprobable way which one of the cities to protect.
Which is the probability that 4 squads end protecting city A?
What is the probability that exactly two cities end up without protection?
What is the probability that 9 squads protect the same city?
What is the probability that 2 cities are protected by 4 squads each one, 2 cities by 2 squads each one and the remaining 4 cities by a squad each one?
I don't really know how to attack this problem, i know it's solution may be by combinations or permutations, but i don't really know how to determine the adequate expression that are useful in order to get a solution, can you explain me please?
My probability knowledge is not more than basic probability axioms, conditional probability and combinatorial tools.
 A: We look at the last two problems. The other two have been asked recently.
We want the probability that $9$ squads end up at one city. Which city? It can be chosen in $\dbinom{8}{1}$ ways. So we find the probability that $9$ squads end up protecting city A, and the rest are elsewhere.
The probability that a particular squad ends up at city A is $p$, where $p=\frac{1}{8}$. So the probability that $9$ of them end up at A, and the rest elsewhere, is $\dbinom{16}{9}p^9(1-p)^7$.
For the required probability, multiply by $\dbinom{8}{1}$.
For the problem about $2$ cities with $4$ squads, $2$ by $2$ each, and the remaining $4$ by $1$ each, the analysis is complicated, thogh the components are relatively straightforward.
There are $2$ cities with $4$ squads each. These cities can be chosen in $\dbinom{8}{2}$ ways. For each of these ways, line up the $2$ chosen cities in alphabetical order. Call them P, Q.
Calculate first the probability that P has $4$ squads. This is $\dbinom{16}{4}(1/8)^4(7/8)^{12}$.
Given that P has $4$ squads, each of the other $12$ is equally likely to be at any of the remaining cities. So the conditional probability that Q has $4$ squads is $\dbinom{12}{1}(1/7)^4(6/7)^{8}$.
Given that $2$ cities have $4$ squads each, the $2$ cities which will have $2$ squads each can be chosen in $\dbinom{8}{2}$ ways. Line up the chosen cities in alphabetical order, say P and Q. The probability P has $2$ squads is $\dbinom{8}{2}(1/6)^2(5/6)^6$. Continue.
There are fancier ways to proceed, if you are familiar with the multinomial distribution.
