Are these two events independent (Comey and Mccabe tax returns)? How would we tell? The NYT reports:

The former F.B.I. director and his deputy, both of whom former
President Donald J. Trump wanted prosecuted, were selected for a rare
audit program that the tax agency says is random.
The odds of being selected for that audit in any given year are tiny —
out of nearly 153 million individual returns filed for 2017, for
example, the I.R.S. targeted about 5,000, or roughly one out of
30,600.

I'm trying to understand how we can know if these are independent events -- just in terms of the statistics, without knowing anything more about the case.
If we say independence means: P(XY) - (P(X) + P(Y)) = 0
Then we are saying: the probability of the joint occurrence is equivalent to the linear addition of the individual probabilities.
So is this true?
Let's say the process of selection is not random but is instead corrupt and the two have not yet been selected for tax audit. It's 2017, and the probability of joint occurrence (P(XY)) is high, say 95%.
On the face of it, the probabilities do not seem independent. However, maybe they are. We could say that the probability that one gets audited is 1/30,600 (P(X)) and the other is also 1/30,600 (P(Y)).
That implies dependence by our definition, even though this doesn't intuitively make sense to me. Is it correct to think about the joint probabilities this way?
It seems intuitively that once you know that one is selected, the odds the other is selected goes up massively. So there's some dependence.
But should we simply consider the individual probabilities so low -- or should they also be much higher because each person has a high chance of being audited, jointly or not? I'm confused how we rate individual probabilities in this instance and how it connects to independence.

If two events $X$ and $Y$ are independent then
$$
P(X \cap Y) = P(X)P(Y),
$$
where $X\cap Y$ is the event that both $X$ and $Y$ do happen.
IIUC, you want to know how likely it is to have both being selected for the audit provided the selection is completely random (i.e. in particular independent) and $P(X) = P(Y) = 1/30,600$. According to the above formula, that probability is:
$$
P(X\cap Y) = P(X)P(Y) = \frac{1}{30,600}\cdot \frac{1}{30,600} = \frac{1}{936,360,000} \approx 0.000000001.
$$
By the way, if the selection of both is independent, then knowing that one is selected will not change the probability that the other one is selected, that is the very meaning of independence.
