Simple probability: $m$ balls are placed independently in container A with probability $a$, and B otherwise. Probability A has fewer than x balls?

Question

Each of $m$ balls is placed independently into a container A with probability $0 <a<1$, or in container B otherwise.

What is the probability that container A will have fewer than $x$ balls, where $0<x\leq m$?

Now immediately to me this looks like a binomial setup. So I would be tempted to say that the probability is

$$P = \sum_{i=0}^{x-1} {m \choose i} a^i (1-a)^{m-1}$$

This seems correct from everything I know about the binomial distribution. Nevertheless I am uncomfortable with it because it seems to be summing up the probabilities of each of the events "this specific set of $i<x$ balls is in A, and all other balls are in B$."

This makes me uncomfortable because the events are of course dependent in general. If you take one set of size $i>2$ containing ball number one, and another different set of size $i$ containing ball number one, then the events that the first set is the set of balls in A, is not independent from the event that the second set is the set in A.

If someone could help to clarify where I am not understanding that would be much appreciated.

Answer 1

1. The computation of the probability is correct.
2. For disjoint events $E_1, E_2$, we have $P(E_1 \cup E_2) = P(E_1) + P(E_2)$. This holds even if $E_1$ and $E_2$ are not independent (in fact, if they are disjoint, they are not independent except if one of the events have probability $0$ or $1$). Summing up the probability of disjoint events (to compute the probability of their union) is perfectly valid, regardless of dependency between the events. Where independence matters is when you are doing the computation $P(E_1 \cap E_2) = P(E_1) P(E_2)$.