Problem involving entanglement swapping

I’m trying to solve a problem. I have my own approach to it, as the one in the textbook seems overly complicated. I tried using my own approach, but I’m not sure if what I found was a coincidence or not.

Here’s the problem (it’s in French, I did my best to translate):

Alice and Bob are separated by a distance of $2L$. Suppose that Charlie is in the middle, i.e. he’s at a distance $L$ from Alice and at a distance $L$ from Bob. Charlie produces two entangled pairs in the state $left| B_{00} right>_{12}$ and $left| B_{00} right>_{34}$. Therefore, there are 4 bits called $1,2,3,4$. Charlie keeps in his lab the particles $2$ and $3$ and sends particles $1$ and $4$ to Alice and Bob. Now, Charlie makes a measure in the Bell basis on particles $2$ and $3$ that he kept in his lab. Which are the possible results (with probabilities) of the measure in Charlie’s Lab?

Here’s my approach

Before the measure, we know that the initial state is : $$left| psi right> = left| B_{00} right>_{12} otimes left| B_{00} right>_{34}$$ $$=frac{1}{2} left( left| 0 right>_{1} otimes left| 00 right>_{23} otimes left| 0 right>_{4} + left| 0 right>_{1} otimes left| 01 right>_{23} otimes left| 1 right>_{4} + left| 1 right>_{1} otimes left| 10 right>_{23} otimes left| 0 right>_{4} + left| 1 right>_{1} otimes left| 11 right>_{23} otimes left| 1 right>_{4} right)$$

Now, given that I know that $$left| 00 right>_{23} = frac{1}{sqrt{2}}(left| B_{00} right>_{23} + left| B_{10} right>_{23})$$, $$left| 01 right>_{23} = frac{1}{sqrt{2}}(left| B_{01} right>_{23} + left| B_{11} right>_{23})$$, and so on for the rest. I was able to rewrite the initial state as follows (although I’m not sure I can do this, given that I’m not sure I can do this):$$ left| psi right> =frac{1}{2sqrt{2}} left| B_{00} right>_{23} otimes (left| 00 right>_{14} + left| 11 right>_{14}) + frac{1}{2sqrt{2}} left| B_{10} right>_{23} otimes (left| 00 right>_{14} – left| 11 right>_{14}) + frac{1}{2sqrt{2}} left| B_{01} right>_{23} otimes (left| 01 right>_{14} + left| 10 right>_{14}) + frac{1}{2sqrt{2}} left| B_{11} right>_{23} otimes (left| 01 right>_{14} – left| 10 right>_{14})$$ which reduces to: $$left| psi right> =frac{1}{2} left| B_{00} right>_{23} otimes left| B_{00} right>_{14} + frac{1}{2} left| B_{01} right>_{23} otimes left| B_{01} right>_{14} + frac{1}{2} left| B_{10} right>_{23} otimes left| B_{10} right>_{14} + frac{1}{2} left| B_{11} right>_{23} otimes left| B_{11} right>_{14}$$

So, each state in each term of the sum is a possible state in Charlie’s Lab after the measurement. In other words, measurements states are: $left| B_{00} right>_{23} otimes left| B_{00} right>_{14}$ $left| B_{01} right>_{23} otimes left| B_{01} right>_{14}$ $left| B_{10} right>_{23} otimes left| B_{10} right>_{14}$ $frac{1}{2} left| B_{11} right>_{23} otimes left| B_{11} right>_{14}$

Therefore, the probability of measuring each state is $p=frac{1}{4}$. I found this without using Born’s rule. I used the fact that in another state, such as $left| phi right>=alphaleft| 0 right> + betaleft| 1 right>$, the probability of measuring $left| 0 right>$ is $|alpha|^2$(and the idea is the same for $left| 1 right>$). I used this fact, but I’m not sure why I would be allowed to do this. Now that I’m writing this question, I realize that perhaps, given that the states after the measure form a basis for our space, writing the initial state as a sum of basis vectors, I am allowed to do this. Would you say that I’m correct?