Two opposite events fill whole probability event space, a process selects them 
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Probability of two opposite events 

Suppose there is string of eight bits, e.g.:

00100110

Bits are randomly chosen from the string. Location of bit (in the string) does not influence its selection probability.
Probability of choosing $0$:
$p_0 = \frac{5}{8} = 0.625$
Prob. of choosing $1$:
$p_1 = \frac{3}{8} = 0.375$
Suppose there is an ongoing process of selecting $0$ and 1. So at each moment, $0$ or 1 is selected, and represents current state C of the process.
Probability of choosing opposite state (to current state C), and then again opposite state – such complex event is called cycle – is given with:
$$p_\text{cycle} = p_0 \cdot p_1 \space\space\space (1)$$
Question: define $p_a = p_{cycle}$. Then, opposite event is $p_b = 1 - p_{cycle}$. We have again two opposite events. How will the $p_b$ look like? I.e. what sequences of $0$ and $1$ will belong to events of kind A and events of kind B. I have problem defining B-set.
 A: This is an answer to the updated question. Let $C$ be the current state, either $0$ or $1$. Let $\overline C$ be the opposite state, either $1$ or $0$, respectively. Let $E$ be the event not-cycle. Then $E$ occurs when the next choice is $C$, or when the next choice is $\overline C$ and the choice after that is also $\overline C$.

If $C=0$, $E$ occurs if the next choice is $0$, or if the next two choices are both $1$; the probability of this is $p_0+p_1^2$.
If $C=1$, $E$ occurs if the next choice is $1$, or if the next two choices are both $0$; the probability of this is $p_1+p_0^2$.

Now the probability of being in state $0$ at any given time is $p_0$, and the probability of being in state $1$ is $p_1$. Thus, the probability of being in state $0$ and having $E$ occur is $p_0(p_0+p_1^2)$, and the probability of being in state $1$ and having $E$ occur is $p_1(p_0+p_1^2)$. Combining the two, we find that the probability of $E$ is $$\begin{align*}p_0(p_0+p_1^2)+p_1(p_1+p_0^2)&=p_0^2+p_0p_1^2+p_1^2+p_1p_0^2\\
&=p_0^2+p_0p_1(p_0+p_1)+p_1^2\\
&=p_0^2+p_0p_1+p_1^2\;,
\end{align*}$$
since $p_0+p_1=1$. And this agrees with your calculation of $p_0p_1$ as the probability of a cycle, since $$(p_0^2+p_0p_1+p_1^2)+p_0p_1=p_0^2+2p_0p_1+p_1^2=(p_0+p_1)^2=1\;:$$ the probabilities of cycle and not-cycle must add up to $1$.
