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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 a finite or countable partition of a set into events $B_i$, the law of total probability statesThe events $A\cap B_i$ are pairwise disjoint and have union $A$, so countable additivity proves the formula. Terms with $\mathbb{P}(B_i)=0$ contribute zero and are omitted to avoid undefined conditional probabilities.

Starting from $X$ at time $n$, the markov property and the time-independent transition rules give $\mathbb{P}(M_{2n}=X\mid M_n=X)=p_n$. ThereforeWrite $p_n=\frac{3}{4}(\frac{3}{4}{)}^n\left(1+\frac{1}{3}(-\frac{1}{3}{)}^n\right)$. The conditional probability becomesThe conditioning selects paths which have avoided the absorbing state up to time $2n$; it is therefore quite different from the unconditioned probability $p_n$, which tends to zero.

For $\mathbb{P}(X=x)>0$, the conditional probability mass function isThe denominator is the sum of the numerator over all possible $y$, so these conditional probabilities sum to one.

Summing the joint probability mass function over $y$ with the geometric series gives the marginal probability mass functionDivide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:It is the geometric distribution counting failures before the first success, with success probability $1-q^{10}$. For the other nonnegative integers $x\ge 10$, $\mathbb{P}(X=x)=0$; the elementary conditional probability given $X=x$ is undefined and supplies no additional case.

After a red first draw, one red remains among six counters. The multiplication rule for conditional probability gives

The definition of conditional probability givesConditioning on a red in the second position leaves one red among the six counters that could occupy the first position, giving the same result directly.

Let $D$ denote having the disease and $+$ a positive result. Assume I am sampled uniformly from the stated population, so $\mathbb{P}(D)=n/N$. By the law of total probability, $\mathbb{P}(+)=an/N+b(N-n)/N$. Thus Bayes' theorem givesThe denominator must be positive; otherwise the observed event has zero probability in this model and the conditional probability is undefined. The answer depends on disease prevalence as well as the test's two error rates.

The survival function of this exponential distribution is $\mathbb{P}(X\ge t)=e^{-kt}$ for $t\ge 0$. Since $\{X\ge 3\}\subseteq \{X\ge 1\}$, the conditional probability isThis is the memoryless property: after surviving the first unit, surviving two additional units has the same probability as it did initially.

The conditioning event selects the exponential component: $P(Y\ge 0)=1/2$ and $P(Y\ge 1)=e^{-1}/2$. Thus conditional probability givesThe atom at $-1$ is excluded by the conditioning event.

For $s,\tau \ge 0$, use conditional probability and the survival function:It is independent of $s$ and equals $P(T>\tau )$. This is the memoryless property of the exponential distribution, equivalent here to a constant hazard function $\alpha$.

By Bayes' theorem, the required conditional probability isThis is about $1.94\%$; it is a different conditioning direction from the test's $99\%$ positive probability among carriers.

Let $D$, $C$, and $V_0$ denote the three disjoint categories, and $+$ or $-$ the test result. The law of total probability givesTo calculate the requested conditional probability, first combine the two categories with the virus. Their joint negative-test probability is $0.1(0.05)+0.2(0.1)=0.025$, whereas $\mathbb{P}(-)=1-0.31=0.69$. Bayes' theorem therefore givesThese numbers use the specified population proportions and test probabilities; the negative result reduces the initial combined proportion $0.3$ to about $3.62\%$ within this model.

For attendance probability $9/10$, absence on both independent observations has conditional probability

Here “absent” denotes absence at both observed lectures. The assumed independence makes its conditional probability

The conditional probability product rule gives $P(A\mid B)P(B)=P(A\cap B)$. Applying it once more givesThe conditionals in the original expression require their conditioning events to have positive probability.