 Solution

With a fixed budget of $n$ independent proposal trials, the accepted count $K_n$ has binomial distribution with success probability $1/M$. HenceFor $M>1$, the central limit theorem gives the one-sided approximationUse $z=2.33$, since now only the lower tail of the accepted count matters. Requiring $n-MN\ge z\sqrt{n(M-1)}$ and solving the resulting quadratic equation in $t=\sqrt{n}$ gives the one-sided normal approximation for a rejection proposal budgetEquivalently, before rounding, this is $MN+z^2(M-1)/2+(z/2)\sqrt{(M-1)[z^2(M-1)+4MN]}$. For the value in (a)(iii),When $M=1$, every proposal is accepted and $\boxed{n_0=N=66,564}$ exactly. Otherwise the positive safety margin above $MN$ allows for random fluctuations in the number accepted. The average cost in (b)(ii) describes stopping at the $N$th acceptance; this answer chooses a deterministic budget that reaches that acceptance with approximately $99\%$ probability.

The approximation requires a sufficiently accurate normal distribution approximation to the binomial distribution. An exact $99\%$ budget can instead be found from its tail probabilities. Also, a $99\%$ chance of enough accepted draws and approximate $99\%$ precision conditional on obtaining them are separate claims: their intersection is not automatically a $99\%$ event.