Suppose an event takes place, and there is a parameter $\theta$ with $\mathbb E[\theta] = \mu$. Suppose I take a sample to estimate the parameter by taking the average of sample $\bar X$ (i.e. $\bar X$ is a statistic). $I$ is an interval carefully chosen from our sample, and $H_0$ is a null hypothesis that $\mu \in I$, while $H_1$ is $\mu \not\in I$. Note that either $\mu \in I$ or $\mu \not\in I$, so $p$ is either $1$ or $0$. Please refer to the diagram below:
In Statistic I, this is what I have learned: if $\bar X \not\in I$, then I reject $H_0$. This methodology is straightforward, but I don't understand the justification that I need to reject $H_0$ if $\bar X \not\in I$ (if $p$ is either 1 or 0).
If I observe that $\bar X \not\in I$, then I reject $H_0$. This probability is given as $$\mathbb P(H_0 = 0 \mid \bar X \not\in I) = \frac{\mathbb P(H_1, \bar X \not\in I)}{\mathbb P(\bar X \not\in I)} = \frac{(1-\beta)(1-p)}{\alpha p + (1-\beta)(1-p)}. $$
Now, $p \in \{0, 1\}$. Very trivially, \begin{equation*}\mathbb P(H_1 = 1 \mid \bar X \not\in I) = \begin{cases} 0 \quad\quad \text{if $p = 1$} \\ 1 \quad\quad \text{if $p = 0$} \end{cases} \end{equation*}
Also,
\begin{equation*}\mathbb P(H_0 = 1 \mid \bar X \not\in I) = \begin{cases} 1 \quad\quad \text{if $p = 1$} \\ 0 \quad\quad \text{if $p = 0$} \end{cases} \end{equation*}
In other words, since $p \in \{0, 1\}$, the thing called "hypothesis testing" becomes trivial. Hence, is it really true that $p$ is either 1 or 0? In hypothesis testing, we are making an educated guess whether we embrace $H_0$ (or fail to reject it) or reject $H_0$. We are comparing the likelihood of $H_0$ being false or not, and if it is more likely that $H_0$ is false, then we reject it. Hence, we are assigning some probability value on $H_0$ being true or not, meaning $p$ should not be either 0 or 1. Can someone clarify my confusion?
