Why the two probability results are the same? Suppose there are 3 red balls and 2 white balls in a bag. We want to pick out 2 balls without replacement. What's the probability of the 1st and 2nd balls are both red?

Solution 1: Use the conditional probability
Let $E_1=$ 1st ball red  $E_2=$ 2nd ball red
$P(E_1)=3/5$  $P(E_2|E_1)=2/4$
then $P(E_1E_2)=P(E_2|E_1)\times P(E_1)=3/5 \times 2/4 = 3/10$
Solution 2: calculate the probability directly
Let $E=$ 1st and 2nd balls are red
There are $5!$ permutation of ball sequence. Of which $3\times 2 \times 3!$ are for $E$.
So $P(E)= 3 \times 2\times 3!/5!= 3/10$

The 2 results are identical just as expected. But why? Should we treat this coincidence as merely an evidence that the mathematical theory of probability is fortunately consistent with our practical experience? So that we can be more confident to apply the rule such as the conditional probability as long as no contradiction arise. Or is there any deep reason that ensures the results will inevitably be identical?
Here is a related question about the justification of Mathematical Theory of Probability.

A "deep reason" is probably how the conditional probability formula
$$
P(E_2|E_1) = \frac{ P(E_1 \cap E_2)}{P(E_1)},
$$
which you used in Method 1, arrives when using the second method.
Using the second approach, the LHS is
$$
P(E_2|E_1) = \frac{\text{number of ways to arrange the remaining 4 balls such that the first of them is red, given that the first ball was red}}{\text{number of ways to arrange the remaining 4 balls given that the first ball was red}}
$$
but hey, this becomes
$$
P(E_2|E_1) = \frac{\text{number of ways to arrange the 5 balls such that the first two are red}}{\text{number of ways to arrange the 5 balls such that the first ball is red}},
$$
dividing both the numerator and denominator by "number of ways to arrange the 5 balls", we get $P(E_1 \cap E_2)$ and $P(E_1)$, respectively. Thus,
$$
P(E_2|E_1) = \frac{ P(E_1 \cap E_2)}{P(E_1)}
$$
