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This is exactly the definition of conditional probability, provided $P(A)>0$:

There are eighteen equally likely first choices. Counting the relevant subsets givesFor conditional probabilities, restrict the denominator to the conditioning subset:The five counts come from the same finite sample space, so no independence assumption is involved.

Let $C$ denote a green pear on the first draw and $D$ a red apple on the second. The two categories are disjoint, so after $C$ there are still three red apples among seventeen remaining fruits. Thus $P(C\cap D)=(8/18)(3/17)$. Using $P(D)=1/6$ from (c), conditional probability givesEquivalently, conditioning on the second fruit being one particular category removes a red apple and leaves the other seventeen fruits symmetrically eligible for the first position.

The implication means that $A\subseteq B$, so $A\cap B=A$. By the definition of conditional probability,This implication gives $P(B\mid A)=1$; it does not make the two events independent.

The law of total probability givesFor $\alpha >0$, Bayes' theorem therefore yieldsAt the allowed endpoint $\alpha =0$, no bulb is defective: $\mathbb{P}(D)=0$. This conditional probability is then undefined, rather than $2/5$. The value $2/5$ is its limit as $\alpha \to 0^{+}$, but does not define conditioning on the impossible event within the stated discrete model.

The numerator event is contained in the conditioning event. The definition of conditional probability therefore givesThe denominator is positive for all $\alpha >0$, and endpoint strictness is irrelevant for a continuous probability distribution.

Let $C$ mean control-group membership and $E$ the adverse event. Equal allocation gives $P(C)=P(C^c)=1/2$, so the law of total probability gives $P(E)=(p_0+p_1)/2$. Bayes' theorem then givesThis is defined when $p_0+p_1>0$. If both are zero, the conditioning event has zero probability and the stated conditional probability is undefined.

Any two distinct draw positions have the same pair law by exchangeability. The marginal and pair probabilities in sampling without replacement therefore giveDivide this joint probability by the relevant marginal probability to get the two conditional probabilities:Conditioning on the third draw can change the probability of the second draw; conditional probability does not require the conditioning event to occur earlier in time.

Distinguish the $N$ balls by labels. For a fixed position $n$, each label appears there in exactly $(N-1)!$ of the $N!$ equally likely permutations. Since $N_B$ labels are black,This is the same marginal law at every position, even the last one. The argument averages over earlier draw outcomes; it does not assert that a conditional probability after a specified history remains $N_B/N$. The original PDF uses $B_n$, correcting the transcription's uppercase subscript $B_N$.

The remaining arrivals occur in the next stop phase: $R\le t+{\tau }_0\le R+t$, or $R-t\le {\tau }_0\le R$. B waits until $R+r$, while A departed at $r$, giving ${\tau }_1=R\ge {\tau }_0$. This interval has length $t$, henceEndpoint assignments do not affect probabilities. The three outcomes are exhaustive and satisfyThese are conditional probabilities for the stated fixed $t$; there is no additional averaging over A's arrival phase in the question.

The sole green ball comes from the first bag exactly on the event $(X,Y)=(1,0)$. Divide this joint probability by the conditioning event's probability:This is the conditional probability form of Bayes' theorem.

By Bayes' theorem, provided $p+q>0$,If $p=q=0$, singing has probability zero and the stated observation cannot occur in this model; that conditional probability is undefined.

There are $99$ possible stopping points, so the equal defective-ball probabilities are $d_k=1/99$. With $g_0>0$, the desired conditional probability givesThus, for a fixed good-ball stopping law,Since $g_0\le 1$, the smallest possible production probability over admissible stopping laws isIf strict positivity of every $g_k$ is additionally imposed in this general case, then $g_0<1$ and one only has the infimum $1/2$, approached as $g_0\to 1$. The paper states that strict-positivity assumption explicitly for part (a), not for the general case in part (b). Without choosing or bounding $g_0$, the exact fixed-law answer remains $1/(1+g_0)$.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

By the definition of conditional probability, or Bayes' theorem,Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for $0<r<1$, because replacement makes a repeated colour more likely.

Write $U=A\cup B\cup C$. Since $D\subseteq U$, their intersection is $U\cap D=D$. Therefore the definition of conditional probability givesThe positive-probability condition is necessary: if $P(D)=0$, this elementary definition of conditional probability does not assign a value.

Each position has apple probability $1/2$. Applying Bayes' theorem to the preceding conditional probability givesThis also agrees with exchangeability of the second and third positions.

Condition on an apple occupying position two in a uniformly random ordering. The nine other positions then contain four apples and five bananas, each equally likely to occupy position three. The conditional probability is thereforeThis uses exchangeability of the unconditioned positions; there is no need to assume the first lunch contained a particular fruit.

Of the ten outcomes in $F$, all except $(1,1,1)$ have exactly one repeated pair. Hence $P(T\mid F)=9/10$. Applying Bayes' theorem in the requested direction,Equivalently there are nine favorable outcomes among the ninety outcomes in $T$. This direct count confirms the conditional probability calculation.

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.