Independence of two r.v. Let $\{ X_n, n\ge \}$ be iid with $P[X_1=1]=p=1-P[X_1=0]$. What is the probability that the pattern $1,0,1$ appears infinitely often? 
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Hint: Let 
\begin{align*}
A_k=[X_k=1,X_{k+1}=0, X_{k+2}=1]
\end{align*}
and consider $A_1,A_4,A_7,.....$.
What I did.
I was thinking I can use Borel zero-one law. Since, $P[A_k]=p(1-p)p$ then $\sum P[A_k]=\sum p(1-p)p$ diverges. 
So, by Borel zero-one law $P[A_k  \text{ i.o. }]=1$.
Is this my reasoning correct? Thanks. 
 A: Hint: Use Borel-Cantelli Lemma 2 for sets like $A_{3k}$ or similar set with non overlapping string pieces!
Solution:
Define $B_k = A_{3k}$. One can simply see that $\{B_k\}_{k=1}^\infty$ are independent and $\sum {\mathbb P}(B_k)=\infty$, so from second Borel-Cantelli lemma one has that ${\mathbb P}(\{B_k,i.o.\})=1$. As $\{B_k,i.o.\}\subset \{A_k,i.o.\}$, the result is straightforward.
A: Assuming that you have a typo in the statement of the hint, this is basically the solution.  To employ this version of Borel-Cantelli we need the events to be independent.  But if you only look at the events $\{ A_{1 + 3k} \}_{k \geq 0}$, you'll have mutually independent events.  This is important (for example, if $A_1$ occurs $A_2$ is precluded from happening).  Now $$ \sum_{k=0}^{\infty} P(A_{1+3k}) = \sum p (1-p) p $$ which diverges if $p \neq 0,1$.
A: No, this is not correct, bacause the event are not independent.
But if you let some time pass between the $A_k$, so that the events become independent,
 you can apply the theorem (consider for instance $\sum P(A_{5k}) = \infty$).
