Simple combinatorics problem - assigning people to groups $20$ people that A, B and C belong to are to be randomly seperated in groups of 4. I have a few questions about this problem:
Q1: What is the probability, that A, B and C place in the same group?
My intuition would be, that the person A has a $4/20=1/5$ chance to land in group 1 and then person B has a $3/19$ chance to land in the same group and finally person C has a $2/18$ chance to do the same. Because there are $5$ groups, this means that $\mathbb{P}(Q1)=5\frac{1}{5}\frac{3}{19}\frac{2}{18}=\frac{1}{57}$.
Q2: What is the probability that they all are placed in different groups?
Place person A in group $a$ (probability $1/5$). Then place person B in group $b\neq a$ - probability $4/19$. Finally place person C in group $c\notin\{a,b\}$ - probability $4/18$. There are $\binom{5}{3}$ ways to choose $a,b,c$, so $\mathbb{P}(Q2)=\binom{5}{3}\frac{1}{5}\frac{4}{19}\frac{4}{18}$
This somehow seems wrong to me, because it does not account for the other $17$ people.
Q3: What is the conditional probability, that A,B,C land in different groups if it is already known that two of them are in different groups?
This one I really struggle with. My attempt: Because there are three groups to choose from and $18$ people to match, I think that $\mathbb{P}(Q3)=3\frac{4}{18}$.
Please check my solutions and tell me what is wrong.
EDIT: Thank you to everyone for the discussion and solutions, especially to @chrslg for even throwing together a little simulation. Also, Q3 is for sure related to conditional probability, so $\mathbb{P}(Q3)=\frac{\mathbb{P}(Q1)}{1-\mathbb{P}(Q2)}$, as pointed out by you all.
 A: $\require{cancel}$
Your first answer is correct, but your second isn't.
What I consider the simplest approach is given below.
Consider $20$ slots in $5$ groups of $4$,
$\quad\boxed{o o o o}\quad\boxed{o o o o}\quad\boxed{o o o o}\quad\boxed{o o o o}\quad\boxed{o o o o}$

*

*(1) The first person can be anywhere, the remaining two can be placed in the same group with $Pr =\frac{3}{19}\frac{2}{18} = \frac1{57}$


*(2) The first person can be anywhere, the other two can be placed in different groups with $Pr = \frac{16}{19}\frac{12}{18} = \frac{32}{57}$


*(3) Two blocks are already occupied, $12$ "good" slots are available, $\cancel{Pr =\frac {12}{18}}$
Correction to part (3)
Since this is a conditional probability question,
$Pr = \frac{All \;in \;different\; groups}{All \;not\; in\; same\; group} = \frac{32}{57}\big/\frac{56}{57} = \frac47$
A: Your answer to the first question is correct.
Q2: Any particular outcome has probability $\frac{4!^5}{20!}$, where $n!=n\cdot(n-1)\cdots 1$ is the factorial. The reason is that no matter what person you put in which group, you will fill up the 4 spots in the 5 groups one by one.
Let's assume that A,B,C are the first three persons. Let us also fix groups $a,b,c$. Now, we can iterate through all possibilities where A ends up in $a$, B in $b$ and C in $c$. Since the probability of each outcome is the same, and equal to $\frac{4!^5}{20!}$, we only need to count the possible assignments of the remaining $17$ persons to groups, that is we have the letters $d$, $e$ four times, and the others three times, yielding $\binom{17}{3,3,3,4,4}=\frac{17!}{3!^34!^2}$, using the multinomial coefficient. Summing over these outcomes gives $\binom{17}{3,3,3,4,4}\frac{4!^5}{20!}=\frac{4^3}{20\cdot 19\cdot 18}$.
Next, we sum over the choice of groups.
Now, we have to choose one group for A, one for B and one for C, yielding $\frac{5\cdot 4\cdot 3\cdot 4^3}{20\cdot 19\cdot 18}=\frac{32}{57}$.
Q3: There are several ways to consider to think of the conditional probability, in case of doubt one approach is always $P(A|B)=P(A\cap B)/P(B)$.
In this case, the probability that all three land in different groups, and two of them land in different groups, well, is the probability that all three land in different groups (the former event is a subset of the latter, hence the intersection is the former). But this probability is exactly the probability from Q2. On the other hand, the complement of the event that there exist two that are in different groups, is the event that all three are in the same group, the event from Q1. Hence, the probability is $\mathbb P(Q2)/(1-\mathbb P(Q_1))=\frac{57}{56}\frac{32}{57}=\frac{4}{7}$.
