 Solution

Applying the optimal single-copy filter independently gives a random number $M_n$ of exact successful targets with binomial distributionThe weak law of large numbers gives $M_n/n\to s/t$ in probability. In contrast, collective entanglement concentration and entanglement dilution yield asymptotically $n{h}_2(s)/h_2(t)$ targets with vanishing output error. The improvement is strict, not just nonnegative. Differentiating binary entropy givesSince $s<t$, it follows thatThis is collective advantage over independent two-qubit filtering. Independent filtering discards the entanglement in product-state failure branches; the collective reversible asymptotic procedure avoids that extensive loss. The separately filtered targets are exact conditional on success, whereas the collective statement uses the vanishing-error convention from the previous part. Here the counted outputs are target states $|\varphi ⟩$; that is also the notation in the original printed question.