There are 3 machines, what is the probability that the one I pick is working? There are 3 machines, machines 1,2 and 3.
The probability that they're working when turned on is shown below:
P(1) = 0.9
P(2) = 0.8
P(3) = 0.4

If I pick one machine at random from those 3, what is the probability that that one is working?
Clearly P for picking any machine in particular is 1/3
The solution in the book is .7, which is P(1)+P(2)+P(3)/3.
It 'makes sense' to me that the solution would look something like that but I really do not understand where that came from.
How do you arrive at that using Kolmogorovs axioms (along with derived theorems)?
 A: Let $X_i$ denote the event that you picked machine $i$ and let $A$ denote the event that the picked machine works. Then
$$
P(A) = \sum_{i=1}^{3}P(A ; X_i) = \sum_{i=1}^{3}P(A \mid X_i)P(X_i) = \frac{1}{3}\sum_{i=1}^{3}P(A \mid X_i).
$$
We used the law of total expectation here.
A: We can use the law of total probability. The law of total probability is the proposition that if $\left\{{B_n : n = 1, 2, 3, \ldots}\right\}$ is a finite or countably infinite partition of a sample space (in other words, a set of pairwise disjoint events whose union is the entire sample space) and each event $B_n$ is measurable, then for any event $A$ of the same probability space
$$
\Pr(A)=\sum_n \Pr(A\cap B_n)
$$
or, alternatively,
$$
\Pr(A)=\sum_n \Pr(A\mid B_n)\Pr(B_n),
$$
where, for any $n$, for which $\Pr(B_n) = 0$, these terms are simply omitted from the summation, because $\Pr(A\mid B_n)$, is finite.
In this particular case
$$
\Pr\{\text{the picked machine is working}\}=0.9\cdot\frac13+0.8\cdot\frac13+0.4\cdot\frac13=0.7.
$$
