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Team game - pairing probability issue (non-academic problem) Watch

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    So I'm playing a two team game.There are 20 players initially and we know that the makeup of team A is 16 players and team B is 4 players.

    Assuming random selection), given each player is 'attached' to another player into pairs, in a way not related to the teams A or B, what is the probability of an A-A pair, a A-B pair and a B-B pair please?
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    P(A) = 4/5
    P(B) = 1/5
    N = 20

    P(A-A) = 4/5 * 4/5 = 1/5
    P(A-B) = 4/5 * 1/5 = 4/25
    P(B-B) = 1/5 * 1/5 = 1/25

    So that's the probability for the first draws, I think. Where do I go from here?

    edit: or is it different because there is now 1 less element in the set and so it's not the same probability multiplied by itself?

    edit 2: yeah, this looks wrong. Please help! xD
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    I'd very much appreciate even a methodology here if an answer is too tedious
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    (Original post by BizzStrut)
    I'd very much appreciate even a methodology here if an answer is too tedious
    This is how I would approach the problem. After being selected a person is removed from the list of people that can be selected.
    Therefore the following probabilities occur for the first draw:
    P(AA)=\frac{16}{20}\times\frac{1  5}{19}=\frac{60}{95}
    P(AB)=2\times\frac{16}{20}\times  \frac{4}{19}=\frac{32}{95}
    P(BB)=\frac{4}{20}\times\frac{3}  {19}=\frac{3}{95}
    Note that there are two permuations for the AB pairing as it can be paired AB or as BA.
    Use this principle to now look at the question from a combination standpoint. Any further help needed just ask.

    Cryptokyo
 
 
 
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