What is the probability that some people are chosen in different groups?

Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What is the probability that there will be at least a malicious person in every group?

I'm not sure, but it might be something like this:

The number of ways we can select two groups with 3 people each is $${n\choose 3}\cdot{n-3\choose 3}\cdot{1\over 2}$$ and the number of ways that at least one group has no malicious people is (the use of principle of inclusion and exclusion) $$2\cdot {n-m\choose 3}\cdot {n-3\choose 3}-{n-m\choose 3}{n-m-3\choose 3}{1\over 2}$$ i.e. the number of ways A doesn't have malicious people+ the number of ways B doesn't have malicious people -the number of ways A and B doesn't have malicious people.

• Why did you put the factor $\frac{1}{2}$ ? – David Feb 20 '18 at 13:57
• Because you can choose some set $\{1,2,3\}$ at first selection and then a set $\{4,5,6\}$ at second selection, but this is the same as I firs choose $\{4,5,6\}$ at first selection and then $\{1,2,3\}$ at second selection. – Aqua Feb 20 '18 at 14:38

Not sure if this is the correct answer, but maybe my reasoning can help you nonetheless.

The total number of arrangements is: ${n \choose 6} \cdot { 6 \choose 3}$.

${n \choose 6}$ represents the total ways of choosing 6 people out of the n people

${ 6 \choose 3}$ represents the number of ways of splitting the people into two groups

Now I will (attempt) to find the number of event in which one of the groups has no malicious persons...which is the complement of what you require.

This requires 4 separate cases:

Case 1: None of the chosen 6 people are malicious

${n-m \choose 6}\cdot {6 \choose 3}$

Case 2: one person is malicious

$2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3}$

Case 3: Two persons are malicious and placed in the same group

$2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3}$

Case 4: Three persons are malicious and placed in the same group

$2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}$

So the final answer is: $$1-\dfrac{{n-m \choose 6}\cdot {6 \choose 3}}{{n-m \choose 6}\cdot {6 \choose 3} + 2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3} + 2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3} + 2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}}$$

You can also think of it like this: $n$ persons are split up in group $A$ containing $3$ persons, group $B$ containing $3$ persons and group $C$ containing $n-6$ persons. Now $m$ of the $n$ persons are chosen randomly and are labeled as "malicious". What is the probability that the groups $A$ and $B$ contain at least one labeled person? In that way of thinking my answer is best understood.

Let $A$ denote the event that a malicious person is in group $A$ and let $B$ denote the event that a malicious person is in group $B$. Then:$$P(A\cap B)=1-P(A^{\complement}\cup B^{\complement})=1-P(A^{\complement})-P(B^{\complement})+P(A^{\complement}\cap B^{\complement})=$$$$1-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}+\frac{\binom60\binom{n-6}{m}}{\binom{n}{m}}=$$$$\frac{\binom{n}{m}-2\binom{n-3}{m}+\binom{n-6}{m}}{\binom{n}{m}}$$