A binary relation on $A$ is any subset of the Cartesian product $A\times A$. There are $n^2$ ordered pairs, and each can independently be included or omitted. Hence the number of binary relations is $2^{n^2}$. This also gives one relation when $n=0$, namely the empty set.

The set difference $\varnothing \setminus \varnothing$ is the empty set. Its power set contains exactly one element, the empty set itself. Therefore the cardinality is $1$, not zero.

The disjoint decomposition into sets $A=(A\cap B)\cup (A\setminus B)$ uses only events in $\mathcal{F}$. Finite additivity and nonnegativity giveThus $\mathbb{P}(A\cap B)\le \mathbb{P}(A)$ already follows from the two stated requirements. Equality is attained, for example, by choosing $B=A$, since then $A\cap B=A$. More generally it holds whenever $\mathbb{P}(A\setminus B)=0$. Neither normalization nor countable additivity is needed for this particular inequality.

The missing axiom is normalization: $\mathbb{P}(\Omega )=1$. The zero set function satisfies both displayed requirements but is not a probability measure, showing why this condition is essential.

The incomplete axiom needs countable additivity: for every sequence of pairwise disjoint events $A_1,A_2,\dots \in \mathcal{F}$,The stated two-event version implies finite additivity, but does not supply this requirement for a countably infinite collection. Together with nonnegativity and normalization, this gives the usual probability measure axioms on the sigma-algebra $\mathcal{F}$.

$\Omega$ is the sample space; its elements are elementary outcomes, also called sample points. An event is a measurable subset of $\Omega$, belonging to the sigma-algebra $\mathcal{F}$, rather than necessarily a single outcome.

This is the ordinary factorial of $X$, rather than a falling-factorial moment. Substituting the probability mass function of the Poisson distribution gives the nonnegative seriesThe geometric series converges exactly when $0<\lambda <1$. Therefore $\mathbb{E}[X!]=\frac{e^{-\lambda }}{1-\lambda }$ for $0<\lambda <1$. For $\lambda \ge 1$, its terms fail to tend to zero and the expectation is infinite. This is the factorial of a Poisson random variable.

By the preceding convolution, $Z=X_1+X_2$ has a Poisson distribution with parameter $\nu =\lambda +\mu$. Directly from its probability mass function, shifting the index in the absolutely convergent sums gives $\mathbb{E}[Z]=\nu$ and $\mathbb{E}[Z(Z-1)]={\nu }^2$. Since $Z^2=Z(Z-1)+Z$, $\mathbb{E}[(X_1+X_2{)}^2]=(\lambda +\mu {)}^2+(\lambda +\mu )$. This is also the identity $\mathbb{E}[Z^2]=\mathrm{Var}(Z)+(\mathbb{E}[Z]{)}^2$.

Let $X_1,X_2$ be the two independent Poisson random variables. For $k\ge 0$, their discrete convolution iswhere the last equality is the binomial theorem. Thus $X_1+X_2\sim \mathrm{Poisson}(\lambda +\mu )$.

Subtract the first three values of the probability mass function of the Poisson distribution from one:

For a Poisson distribution with parameter $\lambda >0$, the probability mass function isand is zero at all other values. Its normalization follows from the exponential series ${\sum }_{k\ge 0}{\lambda }^k/k!=e^{\lambda }$.

Let $M$ count the ringed birds among the two captures, and let $D$ be the event that the detector records at least one ring. The hypergeometric distribution gives the prior probabilitiesUnder the usual additional assumption of independent per-ring detection, with no false records, the likelihoods $\mathbb{P}(D\mid M=m)$ are $0$, $9/10$, and $1-(1/10{)}^2=99/100$. Therefore $\mathbb{P}(D)=12/25+33/250=153/250$. Applying Bayes' theorem yields $\mathbb{P}(M=0\mid D)=0,\mathbb{P}(M=1\mid D)=\frac{40}{51},\mathbb{P}(M=2\mid D)=\frac{11}{51}$. Storing only the last detected number means that a stored record does not identify how many rings were present. By label symmetry, the identity of that stored ring does not further alter this distribution.

The text specifies each ring's marginal detection probability, but does not explicitly specify independence of the detection errors. To keep this modelling qualification visible, let $\ell =\mathbb{P}(D\mid M=2)$. The marginal detection rate only implies $9/10\le \ell \le 1$. Without the extra independence assumption, the full family of possible posterior distributions isIn particular $1/5\le \mathbb{P}(M=2\mid D)\le 5/23$. The displayed numerical answer corresponds to $\ell =99/100$; the general answer exposes exactly what the detector's marginal specification alone establishes.

Let $R_1,R_2$ be the events that the first and second birds are ringed. Under sampling without replacement,Thus the conditional probability is $\mathbb{P}(R_1\mid R_2)=\frac{2/15}{2/5}=\frac{1}{3}$. Alternatively, once a particular ringed second bird is specified, three of the nine possible first birds are ringed.

The two captures form a uniform ordered sample of distinct birds, that is, sampling without replacement. By exchangeability, each of the ten birds has the same chance of occupying the second position, so the probability is $4/10=2/5$. Equivalently, conditioning on the first capture gives $(4/10)(3/9)+(6/10)(4/9)=2/5$.

The finite-partition form of the law of total expectation statesBecause the events form a partition of a sample space of the sample space, ${\sum }_j{1}_{B_j}=1$ pointwise. Multiply by $X$ and use linearity of expectation to obtainAn event of zero probability contributes zero to the middle expression and is omitted from the final expression, so no undefined conditional expectation is multiplied by zero.

Using the indicator random variable, sum over the values of the discrete random variable $X$:Dividing by the positive probability $\mathbb{P}(B)$ proves $\mathbb{E}[X\mid B]=\mathbb{E}[X{1}_B]/\mathbb{P}(B)$. On $B^c$ the product vanishes, so no additional contribution is omitted.

For an integrable discrete random variable, the conditional expectation given an event $B$ of positive probability isThe sum ranges over the possible values of $X$; integrability ensures absolute convergence.

Set $D(s)=4-2s-s^2$, so $D(s)G(s)=s^2$. Differentiate once and evaluate at $s=1$, using $D(1)=1$, $D^{\prime }(1)=-4$, $G(1)=1$:A second differentiation gives $D^{\prime \prime }G+2D^{\prime }{G}^{\prime }+D{G}^{\prime \prime }=2$, henceThe derivatives of a probability generating function give $G^{\prime }(1)=\mathbb{E}[F]$ and $G^{\prime \prime }(1)=\mathbb{E}[F(F-1)]$. Consequently $\mathrm{Var}(F)=52+6-6^2=22$. The earlier geometric tail bound from a uniform escape probability ensures the needed moments are finite.

Write $G(s)=\mathbb{E}[s^F]$. The initial possibilities are $H$ with probability $1/2$, after which a fresh wait begins; $TH$ with probability $1/4$, after which a fresh wait begins; or $TT$ with probability $1/4$, which ends the experiment. Accounting for the flips already made in each case gives the probability generating function equationTherefore $G_F(s)=\frac{s^2}{4-2s-s^2}$ for $0\le s\le 1$. In particular, $G_F(1)=1$ and there is no mass below two flips, as required.

Let $E_0$ be the expected additional flips when there is no immediately preceding tail, and $E_1$ the expectation when the preceding flip is a tail. A one-step first-step analysis givesSolving gives $E_1=4$ and $\mathbb{E}[F]=E_0=6$. To justify finiteness before manipulating these expectations, split the flips into independent two-flip blocks. Each block has probability $1/4$ of being $TT$; hence $\mathbb{P}(F>2m)\le (3/4{)}^m$. This geometric tail bound from a uniform escape probability makes the expectation finite. The state $E_1$ is essential because successive possible pairs overlap: the waiting time is not itself a geometric distribution with parameter $1/4$.

Write $r$ for a positive integer moment index. The moment isThis improper integral converges exactly when $r-\theta -1<-1$, namely $r<\theta$. For those indices,At $r=\theta$ the divergence is logarithmic; for $r>\theta$ it is a power divergence. Thus, in the source's moment notation, the positive integers $k$ allowed are exactly $k<\theta$. If $\theta \le 1$, no positive integer moment is finite.