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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?

I have already asked help with this question but I am not sure about the correctness of the answers

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

I modified one of the proposed answers in this way.

So I need to calculate the complement of "Events where at least a group has no bad people"

so the number of ways in which I can choose 6 people and divide them in groups of 3 is: $$ {n\choose 6} {6\choose 3}$$

The events where a group has no bad people:

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

events where only a single person is malicious:

$$ {m\choose 1} {n-m\choose 5} {6\choose 3}$$

events where 2 people are malicious and in the same group

$$ {m\choose 2} {n-m\choose 4} {4\choose 3} 4$$

events where 3 people are bad and in the same group

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

so the total probability of my question is:

$$ P = 1 - \frac{{n-m\choose 6}{6\choose 3}+{m\choose 1} {n-m\choose 5} {6\choose 3}+{m\choose 2} {n-m\choose 4} {4\choose 3} 4+{m\choose 3}{n-m\choose 3}{3\choose 3}}{{n\choose 6} {6\choose 3}} $$

Is my method correct?

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  • $\begingroup$ It may be easier to find the probability that all of the malicious people are in a single group. $\endgroup$ – awkward Feb 20 '18 at 15:14

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