What is the probability that in an endless series of independent Bernoulli trials, with probability $p$ for getting $1$ and probability $q=1-p$ for getting $0$, the pattern of $101$ will appear infinite times?
I don't really know how calculate this, but i assume that the probability is $1$.
I mean, if it's not, then it means that in some point, we won't get any $101$ at all. I don't know, it sounds impossible.
I tried to see it as: $Y \sim G(pqp)$ = Geometric random variable, where we say that a success is a getting a sequence of $101$, that's with the probability of $pqp$.
Then, $P(Y=k)=(1-pqp)^{k-1}pqp$ implies that the probability that after $k$ times we get the pattern of $101$.
So when $k \to \infty$, $P(Y=k)=0$ meaning that there is not chance that we won't get any $101$ in the future. So, $\lim_{n \to \infty}{\big(1-P(Y=k)\big)}=1-0=1$
What's your thoughts about this manner? Do you think my final answer is correct at all?
Let $X_n$ denote the outcome of the $n$-th Bernoulli trial and define, for $i=1,2,3, \ldots$, the events $$A_i= \{ X_{3i-2} = 1, X_{3i-1} = 0, X_{3i} =1 \}$$ Notice that the events $A_i$ are independent and $P(A_i) = p^2 (1-p)$. Furthermore, $$\sum_{i=1}^\infty P(A_i) = \sum_{i=1}^\infty p^2 (1-p) = \infty$$ Thus, by the second Borel-Cantelli lemma, $P(A_i \text{ infinitely often}) = 1$. Since the $A_i$'s happen infinitely often with probability $1$, so does the sequence $101$, as you conjectured.