Three machines produce the same car parts and have the following production shares and defect rates:
A part from the total production is selected at random. a) Set up a suitable model $(Ω, F, P)$ with $Ω = \{1, 2, 3\} \times \{0, 1\}$ and $F = 2^Ω$.
In addition to $P$, also define the events
$D$: ”Part is defective”
$M_k$ : ”Part was produced by machine $k$”.
Use $D$ and $M_1$ to $M_3$ to formulate which six probabilities are given in the table above.
We have that $Ω = \{1, 2, 3\} \times \{0, 1\}= \{M_1,M_2,M_3\}\times \{D,D^c\}$
What I have:
So I have always problems finding $P$ So we must find a function that $P(m,d): 2^{\Omega} \rightarrow [0,1]$ where $m \in \{M_1,M_2,M_3\}, d \in \{D,D^c\}$
Now I thought on taking $P(m,d)=P(m)P(d|m)$, so far everything right?
My second problem is always to understand if the problem is talking about $A|B$ or about $A \cap B$ in this case I'm pretty sure that for example $0,14$ refers to $P(D|M_1)$, right? However there are other exercises where I have a lot of trouble figuring out which of the two they mean. Do you have any advice on how to distinguish the two?
One think to note is that $D$ and $M_k$ are events, i.e. sets of outcomes, not outcomes themselves. The event $D$ would be $\{1,2,3\} \times \{0\} = \{(1,0),(2,0),(3,0)\}$, while $M_1$ would be $\{1\} \times \{0,1\} = \{(1,0),(1,1)\}$, and similarly for $M_2$ and $M_3$.
As you said, we need to find $P: 2^{\Omega} \rightarrow [0,1]$. Note that $2^\Omega$ is a set of sets, so the goal is to find $P(A)$ for all $A \subset \{1,2,3\} \times \{0,1\}$. We can do this by finding $P(\{(m,d)\})$ for all $m \in \{1,2,3\}$ and $d \in \{0,1\}$, then obtaining $P(A)$ for more general subsets by adding. The table gives $P(M_k)$ for all $k$ and $P(D|M_k)$, so as you said one can find $P(\{(k,1)\}) = P(D \cap M_k) = P(D|M_k)P(M_k)$ for all $k$, from which we can also find $P(\{(k,0)\})$ for all $k$ and be done.
As you said, the table does indeed refer to $P(D|M_k)$, not $P(D \cap M_k)$. Distinguishing the two can be tricky and largely comes from experience and labels in the table. The phrase "of which" in the row label here is a big indicator that this is a conditional probability rather than the probability of an intersection.