Why is the application of probability in Quantum Mechanics (QM) fundamentally different from its application in other areas? QM applies probability according to the same probability axioms as in other areas of physics, engineering, etc.
Why is there a difference?
Naively one would assume one of these possibilities:

It is not the same probability (theory?)

It is a matter of interpretation (of the formalism?)

Something else?


Many answers (which I am still studying) focus on the fact that the combined probability of two mutually exclusive events in QM is not equal to the sum of the probabilities of each event (which holds classically by definition). This fact (appears to) makes the formulation of another probability (a quantum one) a necessity.
Yet this again breaks down to assumed independent or assumed mutually exclusive, if this is not so, the "classical probability" is applicable (as indeed in other areas). This is one of the main points of the question.

The theory of probability used in QM is intrinsically different from the one commonly used for the following reason: The space of events is non-distributive (more properly non-Boolean) and this fact deeply affects the conditional probability theory. The probability that A happens if B happened is computed differently in classical probability theory and in quantum theory,  when A and B are quantum incompatible events. In both cases probability is a measure on a lattice, but, in the classical case, the lattice is a Boolean one (a $\sigma$-algebra), in the quantum case it is not.
To be clearer, classical probability is a map $\mu: \Sigma(X) \to [0,1]$ such that $\Sigma(X)$ is a class of subsets of the set $X$ including $\emptyset$, closed with respect to the complement and the countable union, and such that $\mu(X)=1$ and:
$$\mu(\cup_{n\in \mathbb N}E_n) = \sum_n \mu(E_n)\quad \mbox{if $E_k \in \Sigma(X)$ with $E_p\cap E_q= \emptyset$ for $p\neq q$.}$$
The elements of $\Sigma(X)$ are the events whose probability is $\mu$. In this view, for instance, if $E,F \in \Sigma(X)$, $E\cap F$ is logically interpreted as the event "$E$ AND $F$".
Similarly $E\cup F$ corresponds to "$E$ OR $F$" and $X\setminus F$ has the meaning of "NOT $F$" and so on.
The probability of $P$ when $Q$ is given verifies $$\mu(P|Q) = \frac{\mu(P \cap Q)}{\mu(Q)}\:.\tag{1}$$
If you instead consider a quantum system, there are "events", i.e. elementary "yes/no" propositions experimentally testable, that cannot by joined by logical operators AND and OR.
An example is $P=$"the $x$ component of the spin of this electron is $1/2$" and $Q=$"the $y$ component is $1/2$". There is no experimental device able to assign a truth value to $P$ and $Q$ simultaneously, so that elementary propositions as "$P$ and $Q$" make no sense. Pairs of propositions like $P$ and $Q$ above are physically incompatible.
In quantum theories (the most elementary version due to von Neumann), the events of a physical system are represented by the orthogonal projectors of a separable Hilbert space $H$. The set ${\cal P}(H)$ of those operators replaces the classical $\Sigma(X)$.
In general, the meaning of $P\in {\cal P}(H)$ is something like
"the value of the observable $Z$ belongs to the subset $I \subset \mathbb R$" for some observable $Z$ and some set $I$. There is a procedure to integrate such a class of projectors labelled on real subsets to construct a self-adjoint operator $\hat{Z}$ associated to the observable $Z$, and this is nothing but the physical meaning of the spectral decomposition theorem.
If $P, Q \in {\cal P}(H)$, there are two possibilities: $P$ and $Q$ commute or they do not.
Von Neumann's fundamental axiom states that commutativity  is the mathematically corresponding of physical compatibility.
When $P$ and $Q$ commutes, $PQ$ and $P+Q-PQ$ still are orthogonal projectors, that is elements of ${\cal P}(H)$.
In this situation, $PQ$ corresponds to "$P$ AND $Q$", whereas  $P+Q-PQ$ corresponds to "$P$ OR $Q$" and so on, in particular "NOT $P$" is always interpreted as the orthogonal projector onto $P(H)^\perp$ (the orthogonal subspace of $P(H)$), and all classical formalism holds true this way.
As a matter of fact, a maximal set of pairwise commuting projectors has formal properties identical to those of classical logic: is a Boolean $\sigma$-algebra.
In this picture, a quantum state is a map assigning the probability $\mu(P)$ that $P$ is experimentally verified to every $P\in {\cal P}(H)$.
It has to satisfy: $\mu(I)=1$ and
$$\mu\left(\sum_{n\in \mathbb N}P_n\right) = \sum_n \mu(P_n)\quad \mbox{if $P_k \in {\cal P}(H)$ with $P_p P_q= P_qP_p =0$ for $p\neq q$.}$$
Celebrated Gleason's Theorem, establishes that, if $\text{dim}(H)\neq 2$, the measures $\mu$ are all of the form $\mu(P)= \text{tr}(\rho_\mu P)$ for some mixed state $\rho_\mu$ (a positive trace-class operator with unit trace), biunivocally determined by $\mu$.
In the convex set of states, the extremal elements are the standard pure states. They are determined, up to a phase, by unit vectors  $\psi \in H$, so that, with some trivial computation (completing $\psi_\mu$ to an orthonormal basis of $H$ and using that basis to compute the trace),
$$\mu(P) = \langle \psi_\mu | P \psi_\mu \rangle = ||P \psi_\mu||^2\:.$$
(Nowadays, there is a generalized version of this picture, where the set ${\cal P}(H)$ is replaced by the class of bounded positive operators in $H$ (the so-called "effects") and Gleason's theorem is replaced by Busch's theorem with a very similar statement.)
Quantum probability is therefore given by the map, for a given generally mixed state $\rho$,
$${\cal P}(H) \ni P \mapsto \mu(P) =\text{tr}(\rho_\mu P) $$
It is clear that, as soon as one deals with physically incompatible
propositions,  (1) cannot hold just because  there is nothing like $P \cap Q$ in the set of physically sensible quantum propositions.
All that is due to the fact that the space of events ${\cal P}(H)$ is now a non-commutative set of projectors, giving rise to a non-Boolean lattice.
The formula replacing (1) is now:
$$\mu(P|Q) = \frac{\text{tr}(\rho_\mu QPQ)}{\text{tr}(\rho_\mu Q)}\tag{2}\:.$$
Therein, $QPQ$ is an orthogonal projector and can be interpreted as "$P$ AND $Q$" (i.e., $P\cap Q$) when $P$ and $Q$ are compatible. In this case (1) holds true again. (2) gives rise to all "strange things" showing up in quantum experiments (as in the double slit one). In particular the fact that, in QM, probabilities are computed by combining complex probability amplitudes arises from (2).
(2) just relies upon the von Neumann-Luders reduction postulate stating that, if the outcome of the measurement of $P\in {\cal P}(H)$ is YES when the state was $\mu$ (i.e., $\rho_\mu$), the the state immediately after the measurement is $\mu'$ associated to $\rho_{\mu'}$ with
$$\rho_{\mu'} := \frac{P\rho_\mu P}{\text{tr}(\rho_\mu P)}\:.$$
ADDENDUM. Actually, it is possible to extend the notion of logical operators AND and OR  for all pairs of elements in ${\cal P}(H)$ and that was the program of von Neumann and Birkhoff (the quantum logic). In fact just the lattice structure of ${\cal P}(H)$ permits it, or better is it.  With this extended notion of AND and OR, "$P$ AND $Q$" is the orthogonal projector onto $P(H)\cap Q(H)$ whereas "$P$ OR $Q$" is the orthogonal projector onto the closure of the space $P(H)+Q(H)$. When $P$ and $Q$ commute these notions of AND and OR reduce to the standard ones. However, with the extended definitions, ${\cal P}(H)$ becomes a lattice in the proper mathematical sense, where the partial order relation is given by the standard inclusion of closed subspaces ($P \geq Q$ means
$P(H) \supset Q(H)$).
The point is that the physical interpretation of this extension of AND and OR is not clear. The resulting  lattice is however non-Boolean. In other words, for instance, these extended AND and OR are not distributive as the standard AND and OR are (this reveals their quantum nature). However, also keeping the definition of "NOT $P$"  as the orthogonal projector onto $P(H)^\perp$, the found structure of ${\cal P}(H)$ is well known: A $\sigma$-complete, bounded, orthomodular, separable, atomic, irreducible and verifying  the covering property, lattice. Around 1995 it was definitely proved, by Solér, a conjecture due to von Neumann stating that there are only three possibilities for practically realizing such lattices: The lattice of orthogonal projectors in a separable complex Hilbert space,   the lattice of orthogonal projectors in a separable real Hilbert space, the lattice of orthogonal projectors in a separable quaternionic Hilbert space.
Gleason's theorem is valid in the three cases.  The extension to the quaternionc case was obtained by Varadarajan in his famous book 1 on the geometry of quantum theory, however a gap in his proof has been  fixed in this published paper I have co-authored 2
Assuming Poincaré symmetry, at least for elementary systems (elementary particles), the case of  real and quaternionic Hilbert spaces can be ruled out (here is a pair of published works I have co-authored on the subject: 3 and 4).
ADDENDUM2. After a discussion with  Harry Johnston, I think that an interpretative  remark is worth to be mentioned about the probabilistic content of the state $\mu$ within the picture I illustrated above. In QM $\mu(P)$ is the probability that, if I performed a certain experiment (in order to check $P$), $P$ would turn out to be true.
It seems that there is here a difference with respect to the classical notion of probability applied to classical systems. There, probability mainly refers to something already existent (and to our incomplete knowledge of it). In the formulation of QM I presented above, probability instead refers to that which will happen if...
ADDENDUM3. For $n=1$ the theorem of Gleason is valid and trivial. For $n=2$ there is  known counterexample. $\mu_\nu(P)= \frac{1}{2}(1+ (v \cdot n_P)^3)$ where $v$ is a unit vector in $\mathbb R^3$ and $n_P$ is the  unit vector in $\mathbb R^3$ associated to the orthogonal projector $P: \mathbb C^2 \to \mathbb C^2$ in the Bloch sphere: $P= \frac{1}{2} \left(I+\sum_{j=1}^3 n_j \sigma_j \right)$.
ADDENDUM4. From the perspective of quantum probability, the von Neumann-Luders reduction postulate has a very natural interpretation. Suppose that  $\mu$ is a probability measure over the quantum lattice ${\cal P}(H)$ representing a quantum state and assume that the measurement of $P \in {\cal P}(H)$, on that state,  has outcome $1$. The post measurement state is therefore represented by $\mu_P(\cdot) = \mu(P \cdot P)$, just in view of the aforementioned postulate.
It is easy to prove that  $\mu_P : {\cal P}(H) \to  [0,1]$ is the only probability measure such that
$$\mu_P(Q) = \frac{\mu(Q)}{\mu(P)} \quad \mbox{if $Q \leq P$}\:.$$
