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Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives $\mathrm{Var}(N)=np(1-p)=24(3/4)(1/4)=9/2$. Its positive square root is the standard deviation:

When only a tiny fraction of either species is removed, the trout fraction among the remaining fish stays nearly constant at $p=K/N$. Successive catches are then approximately independent trials with that fixed success probability. The count therefore approaches the binomial distributionThe approximation comes from negligible depletion, not from an exact independence property of sampling without replacement.

The number of tails has a binomial distribution. The complement of at least two tails consists of zero tails or exactly one tail. Therefore

Assume that wrong-note events are independent events and that each note has the same error probability $p$, with $0\le p\le 1$. Then their count $N$ has a binomial distribution. Choose the $n$ mistaken positions in $\left(\genfrac{}{}{0ex}{}{100}{n}\right)$ ways and multiply the probabilities of the errors and correct notes:The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, soThe single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

Let $N$ count rotten pieces. Each item supplies an independent Bernoulli trial with success probability $p$, so $N$ has a binomial distribution with parameters $10,p$. There are $\left(\genfrac{}{}{0ex}{}{10}{n}\right)$ choices of the rotten items, and each such choice has probability $p^n(1-p{)}^{10-n}$. HenceFor $p=0$ or $p=1$, interpret this as the corresponding distribution concentrated at zero or ten; no ambiguous $0^0$ evaluation is needed.

Let $N_A$ be the independent avocado count. There are three disjoint ways to have fewer than two total trees: no trees, only one avocado, or only one lemon. Their probabilities giveThe sum is generally a Poisson binomial distribution, rather than a single binomial distribution, because the two success probabilities may differ.

Let $N_L$ count successful lemon seeds. By independence it has a binomial distribution with five trials and success probability $y$. Therefore

There are three possible count pairs conditional on two trees. Their probabilities, from the two independent binomial distributions, areThe first two correspond to matching kinds, so the conditional probability isThis form remains valid at endpoints whenever the conditioning event has positive probability; if the denominator vanishes, the conditional probability is undefined. As a check, when $x=y\in (0,1)$ the result is $20/45=4/9$, the fraction of two-element subsets containing matching kinds.

The successful ordered outcomes are $TTH$, $THT$ and $HTT$. Each has probability $1/8$ by independence, and these events are mutually exclusive. Adding their probabilities givesEquivalently, the binomial distribution gives $\left(\genfrac{}{}{0ex}{}{3}{2}\right)(1/2{)}^2(1/2)=3/8$.

Independent increments give independent Poisson variables $N[0,s]$ and $N(s,t]$ with means $\lambda s$ and $\lambda (t-s)$. Dividing their joint probability by $\mathbb{P}(N[0,t]=n)$ yieldsThus the conditional distribution is binomial distribution with parameters $n,s/t$.

The event $T_i>s$ means that at most $i-1$ events have occurred by $s$. ConsequentlyFor $1\le i\le n$, the conditional event time lies in $[0,t]$. Integrate its survival function and substitute $u=s/t$:The index range $i\le n$ is implicit in the source's last assertion. For $i>n$, the $i$th event is after $t$, and independent future increments instead give expectation $t+(i-n)/\lambda$; the displayed $it/(n+1)$ formula does not apply.

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

Write $s=e^{-\lambda t}$ for individual survival probability. Conditional on a clump of size $n$, its death probability is $(1-s{)}^n$. Averaging with the geometric distribution givesThusHere $0\le \theta <1$ and $t>0$; different clumps and experimental repetitions are understood independent for the sampling model.

The live-clump counts have independent binomial distributions, all with the same $p$. Up to a constant, their combined likelihood function is $p^R(1-p{)}^{N-R}$, maximized at $\hat{p}=R/N$ when $0<R<N$. Solving $p=s/(1-\theta +\theta s)$ for $s$ gives $s=p(1-\theta )/(1-\theta p)$. ThereforeMonotonicity of $p$ in $\lambda$ makes this the maximum-likelihood estimator. At the sampling boundaries, $R=N$ gives $\hat{\lambda }=0$, while $R=0$ gives only a supremum as $\lambda \to \infty$, with no finite maximizer. These are the limiting interpretations of the formula.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

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.

First establish the limiting law needed by the delta method. For fixed $t$, put $u_n=1/2+t/\sqrt{n}$ and let $N_n$ count how many of the $U_i$ are at most $u_n$. Then $N_n$ has a binomial distribution with expectation $n/2+t\sqrt{n}$ and variance $n/4-t^2$. The event $M_U\le u_n$ is exactly $N_n\ge (n+1)/2$. Its standardized threshold tends to $-2t$, so the central limit theorem for these binomial counts givesHere $\Phi$ is the cumulative distribution function of the standard normal distribution. Equivalently,Use the usual local regularity for the delta method: $f$ is continuous at the population median $\theta$ and $f(\theta )>0$. The inverse function theorem, applied at $F(\theta )=1/2$, givesThereforeThe asymptotic normality gives the requested leading variance approximationHere the variance parameter for the rescaled limiting law is $1/(4f(\theta {)}^2)$, while the source states the corresponding $1/n$ approximation for the unscaled estimator. Strict increase of $F$ alone does not force $f(\theta )>0$; that local condition is needed for this form of the asymptotic distribution of a sample median.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

Let $S={\sum }_i{X}_i$, which has a binomial distribution with parameters $(n,p)$. For $0<p_1<1/2$, the alternative-to-null likelihood ratio of the independent Bernoulli trials isSince $p_1/(1-p_1)<1$, this is strictly decreasing in $S$. A threshold on large $R$ is therefore a lower-tail likelihood-ratio test, and a threshold between $R(2)$ and $R(1)$ gives the stated critical region $S<2$. The Neyman-Pearson lemma also proves it is a most powerful test at its attained test size.

Using the binomial probability mass, its test size and statistical power are respectivelyThe associated power function is $\pi (p)=(1-p{)}^n+np(1-p{)}^{n-1}$. These formulas hold for positive integer $n$; when $n=1$ the test always rejects and has size one. If the boundary value $p_1=0$ is admitted, all alternative probability is at $S=0$; the same test still has statistical power one and can be represented using a threshold with ties in the likelihood ratio.

Set $S={\sum }_i{x}_i$ and $N=nm$. Conditional independence of the binomial distribution observations gives the likelihood functionThe prior distribution is the beta distribution with parameters $(1,\beta )$. By Bayes' theorem, multiplying the prior density by the likelihood function gives a posterior density proportional toRecognizing and normalizing this beta distribution givesfor $0<\theta <1$. This is Beta-binomial conjugacy; the original binomial coefficients cancel in the normalization because they are independent of $\theta$.

Let each step have length $a$, and let $K$ be the number of right steps among $N$ independent steps. Then $K$ has a binomial distribution and the physical displacement is $x=a[K-(N-K)]=a(2K-N)$. Therefore $K=(N+x/a)/2$, and the position probability mass isThis is zero unless $x/a$ is an integer, $|x/a|\le N$, and $N+x/a$ is even. The parity condition matters because the walk alternates between the two sublattices. If distance is measured in lattice spacings, replace $x/a$ by $x$.

Normalization follows from the binomial theorem. Its first two moments provide useful checks:These are exact discrete-time statements; a smooth continuum probability density requires coarse-graining over the lattice and its parity oscillation.

For this radial random walk on a regular tree, the drawing is an infinite regular tree of degree $z$, not a planar lattice with closed loops. The radial coordinate is the number of edges in the unique path to the root. A non-root vertex has one neighbour inward and $z-1$ neighbours outward, so equal probabilities for its $z$ edges giveLet $r$ be radial distance measured in edge units. For $r>1$, probability arriving in shell $r$ comes from shell $r-1$ by an outward step or from shell $r+1$ by an inward step:This has precisely the same bulk master equation as the one-dimensional biased random walk, withThe outward bias is geometric: beyond the first shell, the number of vertices is $M_r=z(z-1{)}^{r-1}$, so there are more outward than inward choices even though no edge direction is dynamically preferred. $P(r,N)$ is the probability summed over an entire shell, not the probability at one specified vertex; no additional shell-size factor belongs in the boxed recurrence.

The equality concerns the evolution equation away from the root, not equality to the unrestricted full-line binomial distribution at all positions. The root imposes a different boundary rule.