If $n$ biscuits are distributed at random among $N$ beggars, find the chance that a particular beggar recieves $r (Before the moderators close my question, I cant think of any starting approach to the question. 
Another question of the similar type I am having trouble with is: 12 balls are distributed at random among 3 boxes. What is the probability that the first box will contain 3 balls?
For the second question I can figure out the exhaustive number of outcomes will be 3 raised to the power 12 since each ball has 12 options.
 A: Hint: how many possible selections of $r$ biscuits out of $n$ are there? For each selection, what is the probability that each selected biscuit goes to the first beggar, but none of the other biscuits do? (Probabilities for different biscuits are independent, so you can just multiply.)
A: All questions can be answered by the binomial distribution, if we distinguish between the given beggar und the group of the remaining $N-1$ beggars.
$$\binom{n}{k}p^k(1-p)^{n-k}\tag{1}$$
For the question in the title we set in eq.$(1)$ $k=r,p=1/N$
$$\binom{n}{r}\frac{1}{N^r}\left(1-\frac{1}{N}\right)^{n-r}$$
This question can also be understood that a given beggar becomes any $r$ with $r<n$ biscuits. For this case we first calculate the probability that the beggar becomes all biscuits. In eq.$(1)$ we set $k=n,p=1/N$ and get
$$\binom{n}{n}\frac{1}{N^n}\left(1-\frac{1}{N}\right)^{n-n}=\frac{1}{N^n}$$
The probability that a beggar gets r biscuits and $r<n$ is then
$$1-\frac{1}{N^n}$$
For the $2^{\textrm{nd}}$ question we have $n=12,k=3,p=1/3$ and get a probability of $21.2\%$.
A: Question 1 : 
If B biscuits are distributed among M men, find the chance that a particular man receives r ( < B ) biscuits.
Solution :
X = The number of ways in which r biscuits can be given to any particular man 
Y = The number of ways in which remaining (B - r) biscuits can be distributed among the remaining (M - 1) men 
Z = The total number of ways in which B biscuits can be distributed at random among M men.
$P(\text{a particular man receives r Biscuits)} = \frac{X*Y}{Z}$
Total Men = M and Total Biscuits = B 
r biscuits can be given to any particular man in $\binom{B}{r}$ ways.
Remaining Men = (M - 1) and Remaining Biscuits = (B - r)
Now, Remaining Biscuits = (B - r) can be distributed among Remaining Men = (M - 1) in $(RemainingMen)^{RemainingBiscuits} = (M - 1 )^{B -r}$ ways. 
Then, the Number of Favorable Cases = $\binom{B}{r}(M - 1 )^{B - r}$
Again,
Total number of ways in which B biscuits can be distributed at random among M men = $(M)^{B}$
$P(\text{a particular man receives r Biscuits)} = \frac{\binom{B}{r}(M - 1 )^{B - r}}{(M)^{B}}$
Question 2 :
12 balls are distributed at random among 3 boxes. What is the probability that the first box will contain 3 balls?
Solution :
X = The number of ways in which r = 3 balls out of B = 12 can go to the first box which can be done in  $\binom{12}{3}$ ways.
Y = The number of ways in which remaining $(B - r) = (12 -3 )= 9$ balls can be placed in the remaining $(Total Boxes - 1) = (3-1) = 2$ Boxes.This can be done in $(RemainingBoxes)^{RemainingBalls} = (2 )^{B -r}= (2 )^{9}$ ways.  
Z = The total number of ways in which B=12 Balls can be placed in 3 Boxes. This can be done in $3^{12}$ ways.
$P(\text{The first ball will contain 3 balls i.e. a particular ball will contain 3 balls out of 12 balls)} =\frac{X*Y}{Z}= \frac{\binom{12}{3}*(2 )^{9}}{3^{12}} = 0.211$
There is another way of doing it. The number of balls in the first box is binomially distributed.
Therefore we can use $\binom{n}{r}(p)^{r}(1-p)^{n-r}$ formulae where p denotes success.
P(Any ball to be in the first box) = $\frac{1}{3}$
P(Any ball NOT to be in the first box) = $\frac{2}{3}$
Then, The Probability that we obtain exactly 3 balls in the First Box  = $\binom{12}{3}*(\frac{1}{3})^{3}*(\frac{2}{3})^{12 - 3} = 0.211$
