I know the quantity of each carton made as well. For example: x = (0, 0, 1000, 0, 488, 0), not just 1488.
Then \mathrm{Count} \sim \mathrm{Pois}(\Phi \mathrm{diag}(r), \mathrm{shape}=(n_{obs}, n_{types}))
but ϕ is not independent of r_n
For the posterior this will certainly be the case (slower cartons should take more time on average) – though I’m not sure this should be encoded in the prior necessarily. I would see how this works before doing something more complicated.