Assuming that $i > 0$ and $p_1 = 5$, let $p_i$ denote an $i$-th prime. Then we can assume that the value of $b_i$ is $0$ if $p_i = 6n-1$ and the value of $b_i$ is $1$ if $p_i = 6n+1$ (where $n$ denote natural numbers).
Consider a real number $r$ such that $0 \leq r \leq 1$ and an $i$-th bit of the binary representation of the fractional part of $r$ is equal to $b_i$: $$r = 0.010101001\ldots$$
Is it possible to prove that $r$ is irrational? If yes, then how?
Yes, $r$ is irrational.
Let $\pi_1(x)$ denote the number of primes less than $x$ of the form $1 \pmod 6$, and $\pi_{-1}(x)$ the number if primes less than $x$ of the form $-1 \mod 6$.
If $r$ is rational, then the binary expansion is eventually repeating. The "repeating part" will then have a fixed number of $0$s and $1$s, say $a$ and $b$ respectively. It then follows that the function
$$b \pi_{-1}(x) - a \pi_{1}(x)$$
will be bounded uniformly for all $x$. This, however, is inconsistent with known properties of these functions. First, by the proof of Dirichlet's theorem (or by the Cebotarev density theorem), one has
$$\pi_1(x) \sim \frac{x}{2 \ln(x)}, \quad \pi_{-1}(x) \sim \frac{x}{2 \log(x)},$$
which implies that $a = b$ is the only possibility. However, a more refined analysis by Littlewood (who considered the very similar case of $1 \pmod 4$ and $-1 \pmod 4$) shows that
$$\pi_1(x) - \pi_{-1}(x)$$
is unbounded and achieves positive and negative values of order at least $x^{1/2 - \epsilon}$.
Relevant: https://dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf
Actually, something weaker than Littlewood's theorem is required here. Assume that $r$ is rational and that $a = b$. Let $\chi(n)$ be the Dirichlet character of conductor $3$ (so it is $+1$ on primes $1 \pmod 6$ and $-1$ on primes $-1 \pmod 6$). Assuming that $r$ is rational and $a = b$, then
$$\sum_p \frac{\chi(p)}{p^s}$$
would be convergent for all $s > 0$ by an application of the alternating series test. But that would mean that
$$\sum_{k=1}^{\infty} \sum_p \frac{\chi^k(p)}{k p^{sk}}$$
converges for $s > 1/2$ and has a pole at $s = 1/2$ coming from the "second term" $k = 2$ which is $\sum 1/p^{2s}$ (since $\chi^2 = 1$). But this expression is none other than
$$ \log L(\chi,s) = \log \left( \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}\right) = \log \left( \prod_{p} \left(1 - \frac{\chi(p)}{p^s} \right)^{-1} \right)$$ $$ = \sum_p - \log \left(1 - \frac{\chi(p)}{p^s} \right) = \sum_{p} \sum_k \frac{\chi^k(p)}{k p^{ks}} $$
where $L(\chi,s)$ is the Dirichlet series of conductor $3$. However, one can prove directly that $\log L(\chi,s)$ does not have a singularity at $s = 1/2$.
