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A discrete-time stopping time for a filtration $({\mathcal{F}}_n)$ is a random variable $\tau$ taking values in $\{0,1,\dots \}\cup \{\infty \}$ such that $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. It means that whether stopping has occurred by time $n$ can be determined from the information available then.

For the first elimination time,Each event in this finite union belongs to ${\mathcal{F}}_j\subseteq {\mathcal{F}}_n$, by the definition of the natural filtration. Therefore $\tau$ is a stopping time. This verification does not require first knowing whether $\tau$ is finite almost surely.

For a Bernoulli distribution with fixed $0<p<1$, the mean and variance are $p$ and $p(1-p)$. The odds transformation $g(x)=x/(1-x)$ has derivative $g^{\prime }(x)=(1-x{)}^{-2}$. Applying the delta method from part (ii),More precisely, $\sqrt{n}\{\overline{X}/(1-\overline{X})-p/(1-p)\}$ converges to a normal distribution of mean zero and variance $p/(1-p{)}^3$, after defining the statistic arbitrarily on $\overline{X}=1$. That exceptional event has probability $p^n\to 0$, so this extension does not change the limit. The unextended ratio is undefined there and has no finite exact variance if interpreted as infinite. At $p=0$ it is zero almost surely; at $p=1$ the denominator is always zero. This is the delta method for Bernoulli odds.

The partition theorem is the law of total probability. If $B_1,B_2,\dots$ form a finite or countable partition of a sample space into pairwise disjoint events with positive probabilities, then for any event $A$,Indeed, $A$ is the disjoint union of the events $A\cap B_j$, so countable additivity gives the formula after applying the definition of conditional probability. Zero-probability members can be omitted.

For events $A,B$ with $\mathbb{P}(B)>0$, the conditional probability isIt measures the probability of $A$ after restricting the sample space to $B$ and renormalizing. This quotient does not define conditional probability given an event of zero probability.

Let $W$ be the event that $H$ wins, and $F$ the event that his first toss is a tail. For a nontrivial starting state $1\le i\le n-1$, a first tail has probability $1/2$ and changes his holding to $i-1$. By the chain rule for probabilities,Since $\mathbb{P}(W)=i/n>0$, the definition of conditional probability givesIn particular, when $i=1$, losing the first coin would end the game in defeat, so the conditional answer is zero. At $i=0$ conditioning on victory is undefined, while at $i=n$ the game has already ended and no first toss occurs.

The events $A\cap B$ and $A\cap B^c$ are disjoint and their union is $A$. Additivity of probability givesBoth conditioning events have positive probability because $0<\mathbb{P}(B)<1$. Apply the definition of conditional probability to each term to obtain the law of total probability:

First restrict to integers in the marginal supports: $2\le i\le 12$ and $-5\le j\le 5$. Their marginal probabilities are positive. By part (i), a feasible joint pair identifies a unique ordered pair of rolls, so its joint probability is $1/36$; an infeasible pair has joint probability zero.

Put $a_i=6-|i-7|$ and $b_j=6-|j|$. If the two events were independent, a feasible pair would requireSince $1\le a_i,b_j\le 6$, equality forces $a_i=b_j=6$, hence $(i,j)=(7,0)$. But that pair is infeasible. An infeasible pair cannot give independence either, since its joint probability is zero while the product of its positive marginal probabilities is positive. ThusIf all real values of $i,j$ are allowed literally, an impossible event is independent of every event. The complete answer without a support restriction is therefore

Complete the square in the transformation:Since $0<X<2$ almost surely, $0<Y\le 1$. For $0\le y<1$, put $r=\sqrt{1-y}$. The event $Y>y$ is equivalent to $1-r<X<1+r$. Integrating the given probability density function over both branches at once gives the survival functionThus, including the tails,The cumulative distribution function on $(0,1)$ is $1-\sqrt{1-y}$. Differentiation givesBoth inverse branches matter: $x=1\pm \sqrt{1-y}$. The endpoint divergence is integrable, and $\mathbb{P}(Y=1)=\mathbb{P}(X=1)=0$, so no endpoint atom is needed.

Let $(B_i)$ be a finite or countable partition of a sample space into events of positive probability. The events $A\cap B_i$ are disjoint and their union is $A$. Countable additivity and the definition of conditional probability therefore give the law of total probabilityCells of zero probability contribute zero through $\mathbb{P}(A\cap B_i)$ and can be omitted rather than assigning an undefined conditional probability.

A partition of a sample space is a finite or countable family of nonempty events such thatThus exactly one event in the family occurs in each outcome. To use ordinary event-conditioned conditional expectations, one usually assumes $\mathbb{P}(B_i)>0$; any zero-probability blocks can instead be omitted from the expectation formula.

Each $X_i$ has the same probability distribution as $-X_i$. By independence, the entire vector and its negative have the same joint law, so $S_m$ has a symmetric distribution about zero. The two disjoint tail events consequently have equal probabilities:Apply Chebyshev's inequality with the mean and variance from (i):Using only containment in the two-sided event would give $8/(3m)$; the symmetry supplies the necessary extra factor $1/2$.

Since the $B_i$ form a partition of a set of the sample space, the events $A\cap B_i$ are pairwise disjoint and their set union is $A$. By countable additivity of probability, followed by the definition of conditional probability,Every ratio is legitimate because each $\mathbb{P}(B_i)>0$. This proves the countable law of total probability.

For events $A,B$ with $\mathbb{P}(B)>0$, the conditional probability isThis renormalizes the probability of $A$ within $B$. The elementary ratio definition does not assign a value when $\mathbb{P}(B)=0$.

A partition of a set $\Omega$ is a collection of nonempty subsets $B_i$ such that $B_i\cap B_j=\varnothing$ for $i\ne j$ and ${\bigcup }_{i\in I}{B}_i=\Omega$. Thus every outcome lies in exactly one cell. For a probability-space partition the cells must also be measurable events. Nonempty cells may nevertheless have probability zero.

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.

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.

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 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 $\mathcal{F}$.