Past exam of the quantum information course of the University of Cambridge / 2013 / Paper 1 / 1 / d / Solution

Applying both final splitters givesThe charged-particle coincidence probabilities are $9/16,1/16,1/16,1/16$ in the displayed order, and the annihilation probability remains $1/4$. In particularConditioning on surviving pairs would change this number to $(1/16)/(3/4)=1/12$; it would not change the amplitudes of the unconditioned all-trials state above.