Probability of obtaining one infinitely often in a sequence of Bernoulli trials Consider a sequence of Bernoulli trials $X_{1}, X_{2}, X_{3}, \dots$, where $X_{n}=1$ or $0$. Assume:
\begin{equation}
P\{X_{n}=1 \mid X_{1}, X_{2}, X_{3}, \dots, X_{n-1}\}\geq \alpha>0, \; n=1,2,\dots
\end{equation}
Prove that:
a) $P\{X_{n}=1 \; \text{for some} \; n\} = 1$
b) $P\{X_{n}=1 \; \text{infinitely often}\} = 1$
Part a) can be solved by conditioning on all possible realizations of $X_{n-1},\dots, X_{1}$ and using and inductive argument.
For part b) it seem easier to prove that the complement of the desired event occurs with probability 0. I have two questions regarding this approach:
1) What is the complement of the event described in b)?, to me it would be the event of $X$ not occurring infinitely often, i.e., $\{\forall n\geq k, \;X_{n}=0 \}$. My confusion arises since it has been suggested to me that the complement is $\{X_{n}=1$ except for a finite number of indices $\}$, which is equivalent to $\{X_{n}=1, \forall n \geq k\}$ 
2) In any case, the solution of b) revolves around finding the probability of a realization of the sequence $X_{k}, X_{k+1}, \dots$ given the sequence $X_{k-1}, X_{k-2}, \dots, X_{1}$. I need some suggestions about how to proceed here.
Any help would be great.
 A: The probability of any particular answer in a Bernoulli trial of equal probability distribution (flipping a coin, rolling a fair die, drawing a numbered card from a randomized deck with replacement) is $\dfrac{1}{x}$. For our purposes, $x=2$ and so the chance of obtaining $1$ from your trial is $\dfrac{1}{2}$.
Given the above, the probability of getting that same answer twice in a row is $\dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}$. Three times, $\dfrac{1}{8}$. This generalizes to the statement that the probability $P(n)$ of obtaining 1 every time in $n$ trials is $P(n)=\dfrac{1}{2^n}$. By the same token, the probability of any of the trials being 1 is the probability that they are not all zero (an event which would have the same probability as all trials resulting in 1), therefore $P(X_a=1\ for\ any\ a<n) = 1-\dfrac{1}{2^n}$.
Now we extend this to the infinite. We can't say $n=\infty$; that's nonsensical. What we can say is that these equations converge as $n\to \infty$; $\lim_{n\to\infty} \dfrac{1}{2^n} = 0$, and so $\lim_{n\to\infty} 1-\dfrac{1}{2^n} = 1$. In English, we will almost surely see at least one 1 in the series of infinite trials, and we will almost surely not see an infinite sequence of 1s. It's not impossible for us to never see a 1, or for us to never see a 0. These events simply occur with probability zero.
