Suppose we have binary random variables $X_1,X_2,\dots,X_n$ on $\{0,1\}$ that are not necessarily independent or identically distributed. You can think them as a sequence of possibly biased coin flips that could also depend on each other.
Suppose also we can compute the probability $P(X_1=0,X_2=0,\dots,X_{n-1}=0,X_n=1)$ and we would like to compute $P(X_n=1|X_1=0,X_2=0,\dots,X_{n-1}=0)$. In other words we know the geometric distribution analogue and we would like to compute the probability in the Bernoulli distribution analogue at the $n$th trial, if that makes sense.
Is there any obvious reason why
$$P(X_n=1|X_1=0,X_2=0,\dots,X_{n-1}=0)=\frac{P(X_1=0,X_2=0,\dots,X_{n-1}=0,X_n=1)}{1-\sum_{i=1}^{n-1} P(X_1=0,X_2=0,\dots,X_{i-1}=0,X_i=1)}$$
is true?
I know by conditional probability for this to be true we require ${1-\sum_{i=1}^{n-1} P(X_1=0,X_2=0,\dots,X_{i-1}=0,X_i=1)}=P(X_1=0,X_2=0,\dots,X_{n-1}=0)$, which is clearly true if you check the LHS inductively. So $1-P(X_1=1)=P(X_1=0), P(X_1=0)-P(X_1=0, X_2=1)=P(X_1=0,X_2=0)$ and so on. Alternatively you could have started on the RHS with $$P(X_1=0,X_2=0,\dots,X_{n-1}=0)=P(X_1=0,X_2=0,\dots,X_{n-2}=0)-P(X_1=0,X_2=0,\dots,X_{n-1}=1)$$ and proceed inductively. However, is there some other probabilistic argument that makes the above expression immediately obvious?
It would have to be the complement rule. That is, $\sum_{i=1}^{n-1}P(X_1=0,X_2=0,...,X_{i-1}=0,X_i=1)$ is the probability that the X's are not all 0. Then the complement rule says that subtracting this from one gives the probability that the X's are all 0. Why is that the probability that all X's are not all 0? It is the probability that there is at least 1 1. $P(X_1=1)$ covers the case where the first flip is 1 and the rest can be 0's or 1's. $P(X_1=0, X_2=1)$ covers the case where the first flip is 0, the second is 1, and the rest flips can be whatever. This is an exhaustive disjoint covering of the possibilities that there is at least 1 1 in the $n-1$ flips. So there we have it that $1-\sum_{i=1}^{n-1}P(X_1=0,X_2=0,...,X_{i-1}=0,X_i=1)=P(X_1=0,X_2=0,...,X_{n-1}=0)$.