I'm revising permutations, combinations and probability for an upcoming exam and would really appreciate if someone could take a look at my procedure to solve this problem and let me know if it's correct. I have no answers to refer to. Hence, I have no way of knowing if I'm solving this correctly.
Thank you in advance for your time! It is much much appreciated!
A basketball club consist of 5 girls and 11 boys. For the next game 5 persons are randomly chosen to take the following positions: point guard, shooting guard, small forward, power forward and center. Specify a probability space for the experiment and calculate the probabilities of the following events:
A: All players are girls
B: All players are boys
C: Small forward is the only girl
D: Small forward and point guard are girls
So we start with specifying the sample space.
$B=[1,2,3...16]$
$1,2,3,4,5$ - girls;
$6,7,8,9,10,11,12,13,14,15,16$ - boys
$\Omega=[c1...c5] ⊂B$, where ci≠cj for i≠j;
$n=16$ $k=5$
A: All players are girls
$\binom{5}{5}:\binom{16}{5}=1:4368=0.00022$
B: All players are boys
$\binom{11}{5}:\binom{16}{5}=462:4368=0.1057$
C: Small forward is the only girl
$\binom{5}{1}*\binom{11}{4}:\binom{16}{5}=(5*330):4368=0.377$
D: Small forward and point guard are girls
$\binom{5}{2}*\binom{14}{3}:\binom{16}{5}=3640:4368 = 0.83$
In order to model all sub-problems the same way, it is important to take into account which player plays on which position. The standard way to do this is to define an order among the positions, say
small forward: 1, point guard: 2, shooting guard: 3, power forward: 4, center: 5.
That order is arbitrary, I've chosen it such that we need to look at position 1 only in sub-problem (C) and positions 1 and 2 only in sub-problem (D).
Your result space is now not a 5-element subset of the 16 players, but an ordered 5-tuple (non-repeating) from the 16 players. This ordered tuple makes it possible to say that the player at index 3 of the tuple is playing shooting guard, for example, something that would be impossible to do with a subset of players.
Combinatorially, that means we move from combinations (selection without order, resulting in subsets) to variations (selection with order, resulting in tuples).
The number of possible selections $S_0$ is the same in all 4 sub-problems, it is the number of teams that one can be select, taking into account the played positions: $$ S_0=16*15*14*14*13={16 \choose 5}5!$$
Thats consistent with the number given in the posting; the differing factor is $5!$, which accounts for the $5!=120$ possibilities a selected set of 5 players can become a team with different positions.
(A): First position can be any of the 5 girls, second position any of the 4 remaining girls, etc., leading to
$$p_A=\frac{S_A}{S_0} = \frac{5*4*3*2*1}{{16 \choose 5}5!}=\frac1{16 \choose 5}$$
This is the same result as the OP got in their posting.
(B): This works essentially the same as (A), the first position can be any of the 11 boys, the second position any of the 10 remaining boys, etc.:
$$p_B=\frac{S_B}{S_0} = \frac{11*10*9*8*7}{{16 \choose 5}5!}=\frac{11*10*9*8*7}{5!}\frac1{16 \choose 5}=\frac{11 \choose 5}{16 \choose 5}$$
Again, this is the same result as the OP got in their posting.
(C): Position 1 (small forward) is a girl, there are 5 possibilities. The remaining positions are all boys, the position 2 is one of 11 boys, position 3 is one of 10 remaining boys, etc.
$$p_C=\frac{S_C}{S_0} = \frac{5*11*10*9*8}{{16 \choose 5}5!}=\frac{5*11*10*9*8}{5!}\frac1{16 \choose 5}=\frac{11*10*9*8}{4!}\frac1{16 \choose 5} = \frac{11 \choose 4}{16 \choose 5}$$
This is the first discrepancy to the OP's result, the correct one is only one fifth of their value. The reason is that this sub-problem, which takes one position of the team into account, cannot be modeled where "all possibilities" are still modeled as 5-element subsets of the players. This model cannot described that a certain player is small forward.
(D): Position 1 is a girl: 5 possibilities, Position 2 is also (another) girl, so 4 possibilities. The remaining positions can be boys or girls, so position 3 is one of the remaining 14 players, position 4 one of the remaining 13 and position 5 one of the remaining 12:
$$p_D=\frac{S_D}{S_0} = \frac{5*4*14*13*12}{{16 \choose 5}5!}=\frac{5*4*14*13*12}{5!}\frac1{16 \choose 5}=\frac{14*13*12}{3!}\frac1{16 \choose 5} = \frac{14 \choose 3}{16 \choose 5}$$
Again we see that the real answer is smaller than OP's answer, this time by a factor of ${5 \choose 2} = 10$. Again the reason is similar as in (C): The modeling of "all possibilities" as subset is incorrect.
It should also be easy to see that OP's value is too big: The probability that the small forward is a girl is $5/16$. Adding another requirement (point guard also a girl) cannot increase that probability. I'd guess if the problem would have continued with a condition for 3 positions, the resulting probabilities, using the OP's method, would have gotten bigger than 1.