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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?

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4 Answers 4

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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:$$P(A\cap B)=1-P(A^{\complement}\cup B^{\complement})=1-P(A^{\complement})-P(B^{\complement})+P(A^{\complement}\cap B^{\complement})=$$$$1-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}-\frac{\binom30\binom{n-3}{m}}{\binom{n}{m}}+\frac{\binom60\binom{n-6}{m}}{\binom{n}{m}}=$$$$\frac{\binom{n}{m}-2\binom{n-3}{m}+\binom{n-6}{m}}{\binom{n}{m}}$$

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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.

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  • $\begingroup$ Why did you put the factor $\frac{1}{2}$ ? $\endgroup$
    – David
    Feb 20, 2018 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – nonuser
    Feb 20, 2018 at 14:38
  • $\begingroup$ @Aqua this assumes that groups $A$ and $B$ are indistinguishable which need not be the case. For example, suppose individuals in group $A$ manage a company's finances while individuals in group $B$ manage a company's internal affairs. From this perspective, the number of ways to assign individuals to group $A$ and group $B$ is ${n \choose 3}{n-3 \choose 3}$. $\endgroup$
    – user801306
    Nov 29, 2020 at 19:42
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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: $$1-\dfrac{{n-m \choose 6}\cdot {6 \choose 3}}{{n-m \choose 6}\cdot {6 \choose 3} + 2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3} + 2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3} + 2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}}$$

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I think our approach to this problem should ultimately depends on whether or not group $A$ and group $B$ are distinguishable, and they certainly could be. For example, if the individuals in group $A$ are assigned to control a company's finances while the individuals in group $B$ are assigned to control a company's internal affairs, then group $A$ and group $B$ are distinguishable.

First suppose $A$ and $B$ are distinguishable. How many ways can we construct two distinguishable groups of three people so that at least one group contains no malicious people? Well, there are $${n-m \choose 3}{n-m-3 \choose 3}+2{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}$$ ways to do this. Since there are ${n \choose 3}{n-3 \choose 3}$ different ways to select the two distinguishable groups of three people, the probability that both distinguishable groups contain at least one malicious individual equals $$1-\frac{{n-m \choose 3}{n-m-3 \choose 3}+2{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}}{{n \choose 3}{n-3 \choose 3}}$$

Now suppose group $A$ and group $B$ are indistinguishable. Let's count the number of ways we can choose the two indistinguishable groups of three people so that at least one group contains no malicious people: $$\frac{1}{2!}{n-m \choose 3}{n-m-3 \choose 3}+{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}$$ Because there are $\frac{1}{2!}{n \choose 3}{n-3 \choose 3}$ ways to choose the two indistinguishable groups, the desired probability in this situation equals $$1-\frac{\frac{1}{2!}{n-m \choose 3}{n-m-3 \choose 3}+{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}}{\frac{1}{2!}{n \choose 3}{n-3 \choose 3}}$$

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