Probability of getting 3 balls in 1st box if 12 balls are distributed randomly among 3 boxes

$$12$$ balls are distributed at random among $$3$$ boxes.The probability that the 1st box will contain $$3$$ balls is_______?

My Approach:
$$\quad \quad \quad \quad \quad \quad \text{Method}1$$[Considering all balls different]:
Tatal no. of ways= $$3^{12}$$
No. of ways in which 1st box will contain $$3$$ balls:$$\binom{12}{3} * 2^9$$ Therefore,required probability= $$\frac{\binom{12}{3} * 2^9}{3^{12}}$$ $$=0.2119$$
$$\quad \quad \quad \quad \quad \quad \text{Method}2$$[Considering all balls identical]:
Tatal no. of ways: $$x_1 + x_2 + x_3=12$$ ,where $$\,\, 0 \leq x_1 \leq 12,\,\, 0 \leq x_2 \leq 12,\,\, 0 \leq x_3 \leq 12$$ $$= \binom{3+12-1}{12}$$ $$=\binom{14}{12}$$ No. of ways in which $$1st$$ box will contain $$3$$ balls $$=$$ No. of ways in which $$2nd$$ and $$3rd$$ boxes will contain $$9$$ balls (as the remaining $$3$$ balls will be given to $$1st$$ box)
so, $$x_2 + x_3=9$$ ,where $$\,\, \,\, 0 \leq x_2 \leq 9,\,\, 0 \leq x_3 \leq 9$$ $$=\binom{2+9-1}{9} =\binom{10}{9}$$ Therefore,required probability= $$\frac{\binom{10}{9} }{\binom{14}{12}}$$ $$=0.11$$
so, which method is the correct approach for the given problem and why the other is wrong,please explain...

The first calculation is correct and the second is not. There are indeed $${14 \choose 2} (= {14 \choose 12})$$ ways to partition $$12$$ into an ordered sum of three numbers, but not all of them are equally likely. To take just one example, there is only one way for all $$12$$ balls to end up in the third box, but there are $$12$$ ways for $$11$$ balls to end up in the third box and $$1$$ ball to end up in the first box. Your second calculation assumes each of the possible partitions is equally likely and that's not so.

The number of balls in the first box is binomially distributed. Each ball is in this box with probability $${1\over3}$$ and not in this box with probability $${2\over3}$$. The probability that we obtain exactly $$3$$ balls in the first box then is $${12\choose 3}\cdot\left({1\over3}\right)^3\cdot\left({2\over3}\right)^{12-3}\ ,$$ as you obtained in your first solution.

In your second solution you have counted the number of ways to make a first, second, and third heap with totally $$12$$ undistinguishable balls, and then you counted how many of these partitions have three balls on the first heap. Looking at the way the experiment is performed in reality there are $$3^{12}$$ equiprobable allocations of the "secretly numbered" balls to the three boxes, and your first solution counts all allocations having three balls in the first box, whatever their secret number. But the partitions you count as "one" in your second solution are not equiprobable. E.g., all $$12$$ balls in the third box counts as "one", and so does the partition having four balls in each box. But the latter is much more probable in the real experiment.

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 = $$\binom{12}{3}*(\frac{1}{3})^{3}*(\frac{2}{3})^{12 - 3} = 0.211$$