<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] Probability for a ball to end up in any of the $4$ bins is uniform $(=0.25)$ so
$$P=\sum_{k=0}^5\underbrace{\frac{10!}{5!(5-k)!0!k!}(0.25)^{10}}_{\textrm{Probability of 5 balls in B1, 5-k in B2, 0 in B3, k in B4}}$$
so
$$\frac{10!\cdot 0.25^{10}\cdot 2}{5!}\bigg(\frac{1}{5!}+\frac{1}{4!}+\frac{1}{3!2!}\bigg)\approx 0.0077$$
</s>