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Use conditional independence at the common age $Y=y$. A system with a red giant and a second evolved component is either a pair of red giants, with conditional probability $g(y{)}^2$, or a red giant and a white dwarf, with conditional probability $2g(y)w(y)$. The factor two counts which component is the white dwarf; a pair of red giants is counted only once. Since $w=9g$, the unconditional probability of this event isThis event is contained in the event of at least one red giant. Therefore the requested conditional probability, among systems selected for containing a red giant, isIn particular the denominator is a fraction of systems, not the individual-star fraction $1/180$.

Use discrete dates $j=0,\dots ,N$, a bank account $B_j=R^j$ with $R=1+r>0$, and a positive stock with each one-period gross return either $u$ or $d$, where $0<d<u$. On the natural tree assume both moves have positive physical probability. The Cox--Ross--Rubinstein model is arbitrage-free precisely whenOtherwise a stock/bank position gives nonnegative profit in both states and positive profit in at least one. Under the condition, the unique equivalent martingale measure has independent upward moves with conditional probabilityIndeed $qu+(1-q)d=R$, making the discounted stock a martingale. Both branches have positive probability under this measure, ensuring equivalence. The uniqueness/completeness claim is for the stock-generated binomial tree; an enlarged filtration carrying additional untraded risk would change that claim.

The Kolmogorov axioms for a probability measure $\mathbb{P}$ are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity on pairwise disjoint events. For $\mathbb{P}(B)>0$, conditional probability is defined byBecause the events $A\cap B_i$ are pairwise disjoint and their union is $A$, countable additivity gives the law of total probabilityFor any $j$ with $\mathbb{P}(A)>0$, the definition of conditional probability and this identity give Bayes theorem:

A cipher has perfect secrecy when the conditional probability of every plaintext given a ciphertext equals its prior probability: $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$ whenever $\mathrm{Pr}(C=c)>0$. Equivalently, the plaintext and ciphertext are independent random variables.

Fix a ciphertext $c$ that can occur. Perfect secrecy requires every possible plaintext $m$ to remain possible after observing $c$. For a deterministic decryption rule, two different plaintexts producing $c$ require two different keys, so the key space has cardinality at least that of the message space.

In a one-time pad, the message and key belong to the same finite additive group, the key is uniform and independent of the message, and encryption is $C=M+K$. For every $m,c$ there is exactly one key $k=c-m$, soindependent of $m$. Bayes theorem then gives $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$, proving perfect secrecy.

For $\mathbb{P}(B)>0$, the conditional probability isHenceBecause the $B_k$ form a partition, the law of total probability givesSubstitution proves Bayes theorem in the required form:

Reveal the field in the order $x_1,\dots ,x_K$ and evaluate its joint density at zero. By the Spatial Markov property of the Gaussian free field, after the values onhave been fixed to zero, the remaining field is the GFF killed on $R_i$. Part 3 therefore givesFactoring the joint density into conditional probability densities now yieldsThe left side is intrinsic and does not depend on the order used to factor the density. Thereforeis independent of the ordering.

Conditional on label $1$ still having frequency one, it is avoided at step $i+1$ with probability $(i+1/2)/(i+1)$. ConsequentlyThe last observation is a singleton exactly when it introduces a new label, whose conditional probability is $(1+M_{n-1}/2)/n$. Taking expectations gives

For nontrivial events, strict attraction under conditional probability is equivalent to a positive covariance of their indicator variables. This condition is symmetric in the two events. Complementing exactly one of the events reverses the strict inequality. This two-event condition should be distinguished from association of an entire family of random variables.

For an edge with endpoints already connected without that edge, including any boundary wiring, the conditional open probability is $p$. Otherwise opening the edge merges two components, so the conditional probability is $p/(p+q(1-p))$. For $q\ge 1$, this conditional probability increases when more other edges are open or more boundary vertices are identified. This is the local mechanism of boundary monotonicity of the random-cluster measure.