How $p(\theta\mid x)$ and $ P(\text{good bus tomorrow}\mid\theta, x)$ are independent? In a previous answer (click here to see), I have been told following -

$P(\text{good bus tomorrow}\mid x)$ is the probability of getting on a
good bus tomorrow given the data $x$ you've already observed, which in
this example is that 2 out of 5 buses were good.
$P(\text{good bus tomorrow}\mid\theta, x)$ is the probability of getting
on a good bus tomorrow given that you already know exactly what
$\theta$ and $x$ are. Here we treat $\theta$ and $x$ as fixed and
known, so if you already know $\theta$, then the probability of
catching a good bus tomorrow is simply $\theta$. Hence, $P(\text{good bus tomorrow} \mid\theta, x) = \theta$.

Now look at the following summation -

Mathematically, we can obtain the posterior probability of catching
the good bus tomorrow using the sum rule: $$P(\text{good bus tomorrow}\mid x) = \sum_{\theta} p(\theta\mid x) \times P(\text{good bus tomorrow}\mid \theta, x) $$$$=  \sum_\theta p(\theta\mid x) \times \theta$$

Here, $P(\theta|x)$ is the posterior probability. It describes $\textbf{how certain or confident we are that hypothesis $\theta$ is true, given that}$ we have observed data $x$.
$P(\theta)$ is the prior probability, which describes $\textbf{how sure we were that}$ $\theta$ was true, before we observed the data $x$.
$P(x|\theta)$ is the likelihood. $\textbf{If you were to assume that $\theta$ is true, this is the probability}$ that you would have observed data $x$.
$P(x)$ is the marginal likelihood. This is the probability that you would have observed data $x$, whether $\theta$ is true or not.
So, $$P (\theta\mid x) = \frac{P (\theta) P(x\mid\theta)}{P (x)}$$
For a detail description or context, click here.
QUESTION
It looks like,   $p(\theta|x)$ and $ P(\text{good bus tomorrow}\mid\theta, x)$ are considered independent of each other in the $\sum_{\theta} p(\theta\mid x) \times P(\text{good bus tomorrow}\mid\theta, x)$ as they are multiplied using Product Rule, but I don't see how  $p(\theta\mid x)$ and $ P(\text{good bus tomorrow} \mid \theta, x)$ are independent of each other, can anyone plz explain?
 A: As @jlewk already stated, they do not use independence. What they use here is the so-called total law of probability (https://en.wikipedia.org/wiki/Law_of_total_probability). This law states that if e.g. $B_1,B_2,B_3$ are disjoint events which together form the whole sample space, then:
$$
\mathbb{P}(A) = \mathbb{P}(A \cap B_1) + \mathbb{P}(A \cap B_2) + \mathbb{P}(A \cap B_3).
$$
(See also the example provided on Wikipedia)
Now assume that we are looking at your problem, i.e., computing
$$
\mathbb{P}(\text{good bus tomorrow} | x) =: \mathbb{P}(A | x)
$$
if we let $A$ be the event that we take the good bus tomorrow.
We cannot compute this probability directly, but we know what the probability should
be if $\theta$ is known. Thus we want to rewrite our asked probability in terms of $\theta$, as in that case we know what the probability is. To introduce $\theta$ in our probability, we use the law of total probability. Consider e.g. the case in which $\theta$ can attain the values $\theta_1,\theta_2,\theta_3,\dots$etc. and define the corresponding events $B_1,B_2,B_3,\dots,$etc. with
$$
B_i := \{\theta = \theta_i\}
$$
Then $B_1,B_2,B_3,\dots$ are disjoint and together form the whole sample space. Thus we can use the law of total probability, which now gives:
\begin{align*}
\mathbb{P}(A | x) = \sum_{i = 1}^\infty \mathbb{P}(A \cap B_i | x)
= \sum_{i = 1}^\infty \mathbb{P}(A | B_i, x)\mathbb{P}(B_i | x)
\end{align*}
with the second step following from the definition of conditional probability.
Inserting the known $\mathbb{P}(A|B_i,x) = \theta_i$ yields
\begin{align*}
\mathbb{P}(A | x) = \sum_{i = 1}^\infty \theta_i \mathbb{P}(B_i|x).
\end{align*}
