$20$ 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.
Best Answer
$\require{cancel}$
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,
$Pr = \frac{All \;in \;different\; groups}{All \;not\; in\; same\; group} = \frac{32}{57}\big/\frac{56}{57} = \frac47$