<s>[INST] Combinatoric question - arrange balls in cells [Probability] We want to arrange $10$ (different) balls in $4$ cells, what is the chance that $5$ will be entered to cell number $1$, and the rest of them will enter an even cell ($2$ or $4$)?
My attempt:
Note that we have a symmetric space, in which we have $4^{10}$ options to arrange the balls.
Let us choose $5$ balls to enter cell $1$ - $\binom{10}{5}$.
Then we have to arrange the next $5$ balls in an even cell, so each one of them has $2$ options, to enter cell number $2$ or cell number $4$, which is: $\binom{5}{5} \cdot 2^5$.
So in total we have: $$\frac{\binom{10}{5} + \binom{5}{5} \cdot 2^5}{4^{10}}=0.027 ~\% $$
Is that correct?
Thanks!
 [/INST] We might view this problem as an example of the Multinomial Distribution. There are $10$ independent trials, and in each trial either (1) a ball lands in bin $1$, with probability $1/4$; (2) a ball lands in bin $2$ or bin $4$, with probability $2/4$; or (3) a ball lands in bin $3$, with probability $1/4$.  Then the probability of the outcome $(5,5,0)$ is
$$\binom{10}{5 \quad 5 \quad 0} \left( \frac{1}{4} \right)^5 \left( \frac{2}{4} \right)^5 = \frac{10!}{5! 5! 0!} \left( \frac{1}{4} \right)^5 \left( \frac{2}{4} \right)^5 = 0.00769043$$
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