Why do we consider a critical region instead of individual values? Why does the alternative hypothesis determine which tail to consider for rejection? Setup:
Suppose a coin is tossed 8 times and I'm trying to determine whether the coin is biased in favour of landing on heads. Let $X$ be the number of heads in 8 tosses, so $X \sim B(8,p)$. Conducting a hypothesis test with a significance level of $5\%$, my null hypothesis would be $H_0\!: p = 0.5$ and my alternative hypothesis would be $H_1\!: p > 0.5$ .
The method of conducting this test that I have learned is that I would first need to consider $X \sim B(8,0.5)$. Since the alternative hypothesis is p greater than 0.5, I would need only consider the right tail of the distribution. Then I would need to find the minimum value of $a$ such that $P(X\geq a) < 0.05$, so $[a,8]$ is the critical region. Thus if we observe $X$ to be between $[a,8]$, it is statistically probable that the die is biased.
Question:

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*Why do we consider a critical region instead of individual values?

It makes sense to me why we would consider a region in the context of continuous distributions, but why should we do it in the case of a discrete distribution? Suppose from the setup, $P(X=8) = 0.03$ and $P(X=7) = 0.049$, then $X=7$ would not be part of the critical region even though individually its probability is still below the set threshold of $5\%$. I don't understand why we should disregard $X=7$.

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*Why does the alternative hypothesis determine which tail to consider for rejection?

The rule I have been taught is, if the alternative hypothesis contains a ">", consider the right tail and if it contains a "<", consider the left tail.  Suppose from the setup, $P(X\geq7) < 0.05$, then it's also true that $P(X\leq2) < 0.05$. Thus why are we only considering one of the tails even though both tails contain values which are below the $5\%$ significance level? As $p$ increases, both of these regions are going to change in values, so I don't understand why we're focusing on the right tail. How does the alternative hypothesis play into this?
Thanks for taking the time to read my question. Please don't respond with too much jargon-heavy language since I'm still in high school! I appreciate any and all help.
 A: You seem to have a good start figuring this out. Let me
say a little more toward finishing the problem.
(I see that @Heropup (+1) already has given a nice answer--but with no Acceptance. Maybe one answer or the other or both together will be helpful.)
Let's do consider all nine probabilities in the
distribution $\mathsf{Binom}(n=8, p=.5).$ rounded to four places.
I'll use R statistical software. But you could use the formula for
the PDF of this binomial distribution, use a statistical
calculator, or maybe find such a table at the back of your
textbook. [Ignore the line numbers in brackets in the first column.]
k = 0:8
PDF = round(dbinom(k, 8,.5), 4)
 cbind(k, PDF)
      k    PDF
 [1,] 0 0.0039
 [2,] 1 0.0312
 [3,] 2 0.1094
 [4,] 3 0.2187
 [5,] 4 0.2734
 [6,] 5 0.2187
 [7,] 6 0.1094
 [8,] 7 0.0312
 [9,] 8 0.0039 

If the coin is fair, you would get $8$ Heads with
probability $0.0038,$ or $7$ heads with probability $0.0312,$
or $7$ or more Heads with probability $0.0038 + 0.0312 =0.0350.$
[If you find the probability of $6$ or more Heads, it's greater
than $0.05.]$
Thus, if we're testing $H_0: p = 0.5$ against $H_a: p > 0.5,$
then we will reject $H_0$ in favor of $H_a,$ concluding
that the coin is biased in favor of Heads, if we get $7$ or more
Heads.
This is a test at the significance level
$0.035 = 4.5\%.$
Because the alternative hypothesis is $H_a: p > 0.5$ (bias
in favor of Heads), we look only at outcomes and probabilities is the upper tail of the distribution.
Because of the discreteness of this
binomial distribution, we can't find an 'upper tail' probability that is exactly 5%, so we settle for the largest
level that doesn't exceed 5%.
Terminology: You may find $7$ called the 'critical value' because any more than seven Heads leads to rejection.
You may find the probability $0.035$ called the P-value
for observing $7$ Heads because it is the probability of a result of observing $7$ or more Heads with a fair coin (null
hypothesis).
