Allow me to give you a basic probability education (I'm assuming you're not being facetious; if you're trolling, then well played sir):
If this is a typical game, there will be 4-5 scum. Let's say 4, for the sake of argument. Probability is nothing more, nothing less than a count of how many possible outcomes out of the full set of all possible outcomes match a certain criteria. It is expressed as a fraction: the number of possible outcomes that match the criteria divided by the total number of possible outcomes. Very often, this is converted to a decimal. A probability can never be greater than 1, since you can't have more outcomes mathcing your criteria than the total number of possible outcomes in any given context. Since we are counting things, a pronbability can never be less than 0 either. All of the possible outcomes collectively are referred to as the "sample space".
In our context, we are lynching one player from a pool of 22. For the sake of simplicity, we'll ignore no lynch or any power role shenanigans. Any one of the 22 players could be lynched, giving us 22 possible outcomes, or a sample space of 22. We've already assumed there are 4 scum, which is the criteria for the outcomes we are interested in. So, we have 4 outcoems that match our criteria, and 22 possible outcomes. This gives us a probability of 4/22, or ~0.18, of catching a scum in one lynch.
To figure out the probability for two lynches is a fair bit more complicated, as the outcome for the second lynch is dependent on the outcome of the first. You can see how this situation is handled in my previous post, but allow me to explain a little more fully. Since by the time we get to the second lynch, any one of 22 players will have already been lynched, we have to do the same analysis as we did for the first lynch 22 times over (one for each possible outcome for the first lynch). Luckily there are shortcuts, but let's continue.
Let us assume we got lucky the first time and nabbed a scum. This means there are only 3 left alive, out of 21 players. So the probability of nabbing another scum is 3/21, or ~0.14. But assuming we got a scum first time out is a dangerous assumption, when we already know there was only an 18% chance of that happening. So how do we handle not making assumptions about the first lynch? Beyond that, how do we figure out the probabilities associated with the outcome of both lynches, taken together?
Recall above that I used the word count. It's helpful to reiterate this: probability is first and foremost a counting process. We are counting how many ways an uncertain event or sequence of related or unrelated events can occur (in most cases, the uncertainity is because the events haven't happened yet, but it can also be the case that we are dealing past events where we don't know what exactly has happened). In our case, there are 22 possible outcomes on the first lynch, and for each of these outcomes, another 21 possible outcomes on the second lynch. Now we know that if we count 21 objects 22 times, we are dealing with multiplication, so our sample space for what we want to know (the chance of catching a scum in 2 lynches) is 22 x 21 = 462. Put another way, there are 462 possible ways of lynching 2 people out of 22. What we want to know is how many of these 462 net us what we want - at least one scum (we want two, but we'll be content with one). We could reduce this number by realising that lynching Jason Maher and then Bad Ash is ultimately the same outcome as lynching Bad Ash and then Jason Maher, but as it would make our calculations more difficult, we won't do this.
Often in probability, as in life, there is more than one way to skin a cat (and they're all fun! :badteeth

. We could, in theory, count up all the possible outcomes where at least one scum is lynched, and then divide this by 462. Or, realising the front door is locked, and not desiring to go to the trouble of fishing out a key, we could nick 'round the back door instead, which we know to be unlocked. Which is to say, sometimes it easier to count the events we're not interested in, and figure out the result we want by a simple subtraction (since the probabilities of all possible outcomes always sum to 1 (or 462/462 in our case), or in other words, once you've counted every outcome that meets the criteria you're interested in and then counted every outcome that doesn't, you've counted everything there is to count).
So then, let's count all the outcomes we don't want - every outcome where we lynch two townies. There are 18 townies in the first lynch, and if we nab one, then there are 17 left for the second. So for our worst case scenario, using the same method as above when counting all possible outcome, we have 18 cases of lynching one townie, and for each one of these, 17 cases of lynching another one. So we're counting 17 18 times over, which gives us 17 x 18 = 306 cases (out of 462) where we nab 2 townies. It thus follows that the other 156 cases net us at least one scum (4 x 3 = 12 of them net us 2 scum, by the same analysis).
TL;DR version:
The chance of lynching one of 4 scum with 22 players is 4/22 = ~0.18
The chance of lynching at least one of 4 scum in 2 lynches with 22 players is 156/462 = ~0.34
The chance of getting lucky and catching 2 scum is 12/462 = ~0.026
Numbers different to my previous post due to differing assumptions.
Note that all of this analysis assume each outcome is equally likely. Not true in general in a Mafia game, but close enough on day 1. Assessing a situation where outcome is not equally likely requires weighting multipliers, greatly complicating the analysis. Also, it's often very difficult, if not impossible, to accurately estimate what the weightings should be.