Conditional laws and joint independence

Try

A bag contains two red and four blue balls. Drawing a red changes what remains when sampling without replacement. The probability of a second red given a first red is 1/5, whereas the first-red probability was 1/3. Explain the dependence using the physical mechanism before multiplying fractions.

Predict

Predict the probability of two reds without replacement and compare it with independent draws with replacement. Then ask whether three events can be pairwise independent without being jointly independent. Pairwise checks involve fewer intersections than a statement about the entire collection.

Experiment

Without replacement, two reds has probability (2/6)(1/5)=1/15. With replacement and independent remixing, it is (1/3)²=1/9. Make a two-stage tree and label the second probabilities conditional on the first outcome.

For a separate example, list two independent fair bits: 00,01,10,11. Define A as first bit 1, B as second bit 1, and C as equal bits. Each has probability 1/2; each pair intersection has probability 1/4. But A∩B∩C is the single outcome 11, probability 1/4 rather than 1/8. Mark the memberships in a four-row table to inspect every claim.

Understand

For P(B)>0, P(A|B)=P(A∩B)/P(B). Multiplying through gives the general multiplication rule P(A∩B)=P(B)P(A|B). This rule does not assume independence. Independence adds the identity P(A∩B)=P(A)P(B); when conditioning probabilities are positive, it means the conditional probability equals the corresponding marginal probability.

Disjoint events have empty intersection. If both have positive probability, they cannot be independent: the intersection probability is zero while the product is positive. Independence means learning one event does not change the probability of the other; disjointness often changes it dramatically.

Mutual independence of a finite family requires product identities for every nonempty subcollection. For three events, the three pairwise identities are insufficient without the triple identity. For an infinite family, the usual definition requires independence of every finite subcollection. The bit example demonstrates why compressing these requirements into a vague claim that events are unrelated is unsafe.

Conditioning can create dependence. Given that two fair independent bits sum to one, only 01 and 10 remain. Each bit is still equally likely zero or one, but knowing one determines the other. Independence in an unconditional model need not persist inside a selected subgroup. This mechanism matters when selection depends on multiple variables.

The elementary ratio does not define conditioning on an event of probability zero. Continuous conditional distributions later use density ratios or more general conditional-expectation constructions. Do not assign a result by canceling two zero point probabilities.

Independence is a modeling assumption or a mathematical property of a specified joint law. A finite dataset with matching proportions does not prove population independence, and a story involving separate people does not guarantee it if they share a common environment.

The way an observation is produced changes the conditional law

Choose a uniform unordered hand of three distinct objects from six, two of which are marked. Let X count marked objects in the hand. There are20 hands: four with X=0, twelve with X=1 and four with X=2. Learning that at least one marked object is present leaves16 equally likely hands, giving P(X=2 | present)=4/16=1/4.

Now instead choose one member of the hand uniformly and observe that the selected member is marked. A hand with X marked members has X favorable reveal choices, not one. The twelve X=1 hands supply12 favorable(hand,member) outcomes and the four X=2 hands supply8, giving P(X=2 | marked reveal)=8/20=2/5. The words “a marked object was seen” are insufficient unless the observation rule is stated.

Present in the hand, or selected from the hand?

Predict the chance of at least two marked objects after positive evidence. The prior hand is always a uniform unordered size-k subset; only the observation mechanism changes.

Population N 468 Marked M 0123456 Hand size k 1234 Observed evidence At least one marked is presentUniformly selected hand member is marked Reset

Positive evidence occurs in 16/20 equally weighted elementary outcomes. P(X≥2 | evidence) = 4/16 = 0.25; conditional mean = 20/16 = 1.25.

Marked count X	Prior hand count	Positive-evidence weight
0	4	0
1	12	12
2	4	4
3	0	0

Inspect every hand and its observation multiplicity

Hand (first M labels marked)	X	Evidence multiplicity
1, 2, 3	2	1
1, 2, 4	2	1
1, 2, 5	2	1
1, 2, 6	2	1
1, 3, 4	1	1
1, 3, 5	1	1
1, 3, 6	1	1
1, 4, 5	1	1
1, 4, 6	1	1
1, 5, 6	1	1
2, 3, 4	1	1
2, 3, 5	1	1
2, 3, 6	1	1
2, 4, 5	1	1
2, 4, 6	1	1
2, 5, 6	1	1
3, 4, 5	0	0
3, 4, 6	0	0
3, 5, 6	0	0
4, 5, 6	0	0

Presence gives one favorable record per eligible hand. A uniform reveal gives X favorable marked-member records per hand, among k possible reveals. At most70 hands/280 reveal outcomes; temporary exact enumeration, not an observed dataset or automatic mastery.

Derive both laws for arbitrary population and hand size

Let N≥1 objects include M marked ones, and choose a uniform k-subset, where1≤k≤N and0≤M≤N. The prior count law is P(X=x)=C(M,x)C(N−M,k−x)/C(N,k), with zero binomial coefficients outside feasible ranges. Under presence evidence E=(X≥1), its probability is1−C(N−M,k)/C(N,k). For positive P(E), discard X=0 and divide the remaining masses by P(E); the conditional mean is E[X]/P(E)=(kM/N)/P(E). This uses X=0 outside E, not a blanket rule for arbitrary conditional means.

For uniform member reveal, each elementary pair(hand H, member j in H) has probability1/[k C(N,k)]. Count positive pairs in two ways: summing X over hands gives Σx·C(M,x)C(N−M,k−x), while choosing the revealed marked object first gives M·C(N−1,k−1). Their equality proves both the normalization and P(marked reveal)=M/N. Thus the posterior count masses are proportional to x times the prior masses, with normalization E[X]=kM/N. Equal prior hand probabilities do not imply equal posterior hand probabilities under this evidence.

Conditional on a specific revealed marked object, the other k−1 hand members form a uniform subset of the remaining N−1 objects, which include M−1 marked ones. The same count law holds when only its marked status is revealed, by symmetry across marked labels. Hence X is one plus a hypergeometric count with parameters N−1,M−1,k−1, and its conditional mean is1+(k−1)(M−1)/(N−1) when N>1. At N=1 and positive evidence, M=k=1 and X=1 deterministically. At M=0 positive evidence is impossible; no posterior normalization is defined. At k=N the hand is fixed even though the reveal may remain random.

A deliberate reveal is another protocol

A host who always chooses a marked object whenever one exists produces positive evidence exactly when X≥1. If only the marked/unmarked status is reported, its posterior is the presence law, not the uniform-member-reveal law. If the identity or other details are reported, additional selection probabilities may matter. The distinction is about likelihoods under a specified experiment, not an assumption that all ways of learning a true statement are interchangeable.

Apply

Construct a counterexample to the claim that pairwise independence always permits multiplying three probabilities. Verify all required pairs rather than only one. Then describe a real collection mechanism in which conditioning on inclusion could induce dependence, stating the model rather than asserting a causal conclusion.

Continue

Next use a partition to collect all routes to an observed event. Total probability and Bayes’ rule keep the relevant denominators explicit.