I want to check my solutioons for this problem:
The captain of a soccer team is in good form $70%$ of the games, in fair form $20%$ and in poor form $10%$. In these cases, the team's chances of winning are $70%$, $40%$ and $20%$ respectively. a) Set up a suitable probability model $(Ω, F, P)$.
So let
$F_1=$ Good Form
$F_2=$ Fair Form
$F_3=$ Poor Form
$W=$ Win
$W^c=$ Lost Game
So for $\Omega$ I have $\Omega= \{F_1,F_2,F_3 \} \times \{ W,W^c\}$
$F=2^{\Omega}$ and $P=\left\{\begin{matrix} 0,7 & F_1\\ 0,2& F_2\\ 0,1& F_3\\ ?& W\\ ?& W^c \end{matrix}\right.$
Since we have $P(F_1|G)=0,7$ $P(F_2|G)=0,4$ $P(F_3|G)=0,2$ we can calculate $P(W)=\sum_{i=1}^3 P(F_i)P(W|F_i)$
And therefore $P(W^c)=1-P(W)$
I am not sure with the definition of $\Omega$. Is everything right?
Your sample space $\Omega$ and sigma-field $F$ look fine to me, but your probability measure seems off given how you've defined $F$.
The probability measure $P$ should map elements of $F$ to a value in $[0,1]$ and are $\sigma-$subadditive over unions of events in $F$ (strictly additive if the sets are disjoint).
In this case, $F$ consists of subsets of $\Omega$ which is the Cartesian product of the possible states of the coach and outcome of the game.
So you need $P$ to be able to handle subsets whose elements are $(f,w): \;f \in \{F_1,F_2,F_3\},\; w \in\{W,W^c\}$
Since this is a discrete probability space, that means that $P$ needs to assign a probability to each element of $\Omega$, each of which is disjoint with the other (i.e., the coach cannot be in two states at once):
$$P(f,w) = P(f)P(w|f)$$
And
$$\text{For event } E \subset F \implies P(E) = \sum_{i \in E} P(i)$$