permutations and conditioned probabilty

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A soccer trainer is making a team, he can choose 11 out of 22 players. The trainer has to take into consideration the players preferred position:

2 players want to be keeper. 8 players want to play defense. 8 players wants to play middle field. 4 players want to play attack.

The team shall consist of 1 keeper, 4 defensive players, 4 middle field players and 2 attackers. How many different teams can the coach create?


I found: 2C1 * 8C4 * 8C4 * 4C2 = 58800.


Now the follow up question is what confuses me:

John and Peter wants to play middle-field.

Given that the coach takes into consideration of the player preferences (like above), what are the odds that:

John makes the team as middle-field player

Peter makes the team as middle-field player

Peter making the team as a middle-field player if we know that John made the team as a middle-field player.


I understand that I have to apply the conditional probability theorem, but I am unsure as to how.

Any help is much appreciated.

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The number of ways to not choose Peter is $7\choose 4$. The total number of ways to choose a team altogether is $8\choose 4$. Therefore the answer is the ratio netween these two, which happens to be the much expected $1/2$.

As for Question 3, given that we have one player already, there are only $3$ to choose from $7$ now, and the updated figures are $$\frac{6\choose 3}{7\choose 3}=\frac{4}{7}$$

therefore the probability that Peter makes it is $3/7$.