Given a wave function $\Psi(\vec r_1, \vec r_2)$, where $\vec r_1$ and $\vec r_2$ are the positions of particle 1 and 2, respectively, the probability of finding particle 1 at position $\vec r$ (irrespective of particle 2) is the marginalised probability distribution $$P_1(\vec r) = \int |\Psi(\vec r,\vec r_2)|^2 d^3\vec r_2. \tag 1$$
Similarly, the probability of finding particle 2 at position $\vec r$ is the marginalised distribution $$P_1(\vec r) = \int |\Psi(\vec r_1,\vec r)|^2 d^3\vec r_1. \tag 2$$
Question: How can I do this marginalisation in bra-ket notation? Given a ket $|\Psi\rangle \in \mathcal{H}_1 \otimes \mathcal{H}_2$, where $\mathcal{H}_1$ and $\mathcal{H}_2$ are the Hilbert spaces of particle 1 and 2, respectively, how do I define the probability of finding particle 1 at position $\vec r$ (irrespective of particle 2)?
Attempt: My guess to translate $(1)$ into bra-ket notation would be something like $$P_1(\vec r) = |\langle \vec r_1 | \Psi \rangle |^2,$$
where $\langle \vec r_1|$ only acts on the part of $|\Psi\rangle$ which is in $\mathcal{H}_1$. However, this is a notational mess because $\langle \vec r_1|$ (which is actually $\langle \vec r|$, an eigenstate of the position operator) does not have the same symbol as the argument of the function $P_1$.
Furthermore, how do I denote that nothing happens to the part of $|\Psi\rangle$ which is in $\mathcal{H}_2$? Shouldn't one write something like $$P_1(\vec r)=|\left(\langle \vec r_1 | \otimes I \right) |\Psi \rangle|^2,$$
where $I$ is the identity operator on $\mathcal{H}_2$. But this is wrong because $\langle \vec r_1 | \otimes I $ is meaningless, as it is the tensor product of a vector and an operator.
