Conditional Rules Use a Restricted Sample Space
Once an event \(B\) is given, conditional probability treats \(B\) as the relevant sample space. The complement, union, and inclusion rules must therefore be interpreted within that restricted space. This viewpoint is useful whenever a question asks for a probability among cases already known to meet some condition, such as reviewed claims, selected applicants, or members of a specified risk group.
Let \(A\) and \(B\) be events in a probability model. Conditional probability given \(B\) is defined only when \(P(B)>0\). For any event \(E\), the definition is
The intersection \(E\cap B\) is the part of \(E\) that remains possible after restricting attention to \(B\). In particular, the complement of \(A\) inside the restricted sample space is \(A^c\cap B\), not all of \(A^c\) without regard to \(B\).
The Complement Rule
Within \(B\), the events \(A\cap B\) and \(A^c\cap B\) are disjoint and together make up all of \(B\). Consequently, their conditional probabilities add to one.
The condition \(P(B)>0\) is essential because both conditional probabilities use \(P(B)\) as a denominator. The identity concerns the complement of \(A\) given the same event \(B\). It does not assert that \(P(A^c\mid B)=1-P(A)\).
Worked Example: The Complement Among Reviewed Claims
Method: Treat the reviewed claims as the restricted sample space and subtract the conditional probability from one. In a hypothetical review, 100 claims are selected for enhanced review. Of these, 28 have incomplete documentation. Let \(B\) be the event that a claim is selected for enhanced review, and let \(A\) be the event that its documentation is incomplete. Find the probability that a claim has complete documentation given that it was selected for review.
Among claims in \(B\), 28 of 100 are in \(A\). Thus
Complete documentation is the complement of incomplete documentation, so
As a check, 100 reviewed claims minus 28 incomplete claims gives 72 complete claims, and \(72/100=0.72\). The common mistake to avoid is subtracting from one a probability calculated for a different population; both terms in the complement rule must use the same conditioning event \(B\).
The Conditional Union Rule
When either of two events may occur, adding their conditional probabilities counts cases in both events twice. Subtract the conditional probability of their overlap once. The formula is the usual two-event inclusion-exclusion identity, applied within the restricted sample space.
If \(A\) and \(C\) cannot both occur within \(B\), so that \(A\cap C\cap B\) is empty, the overlap term is zero. In that case the conditional probabilities add directly. More generally, the events need only be disjoint after restriction to \(B\); they need not be disjoint outside \(B\).
Worked Example: Either of Two Loss Types in a Risk Tier
Method: Use conditional inclusion-exclusion, subtracting the overlap once. A hypothetical insurer examines 200 policies in a specified risk tier \(B\). Among them, 60 have a cyber loss \(A\), 50 have an equipment loss \(C\), and 20 have both types of loss. Find the probability of at least one of these loss types given that a policy is in this tier, and the probability of neither type.
First calculate each conditional probability using the 200 policies in \(B\) as the denominator:
Apply the union rule:
The conditional complement rule then gives
Checking by counts, the union contains \(60+50-20=90\) policies, leaving \(200-90=110\) with neither loss type. Since \(90/200=0.45\) and \(110/200=0.55\), both results are verified. Adding 0.30 and 0.25 without subtracting 0.10 would count policies with both losses twice.
Why the Conditional Rules Hold
The rules follow directly from the definition and set relationships. This short derivation also shows why the condition \(P(B)>0\) appears throughout.
The relevant parts of \(A\), \(A^c\), and \(A\cup C\) are \(A\cap B\), \(A^c\cap B\), and \((A\cup C)\cap B\), respectively.
The sets \(A\cap B\) and \(A^c\cap B\) partition \(B\). Also, the two-event inclusion-exclusion identity gives \(P((A\cup C)\cap B)=P(A\cap B)+P(C\cap B)-P(A\cap C\cap B)\).
Because \(P(B)>0\), divide each identity by \(P(B)\). Each resulting ratio is a conditional probability by definition, giving the complement and union rules.
For the complement identity, the partition of \(B\) gives \(P(A\cap B)+P(A^c\cap B)=P(B)\). Dividing by \(P(B)\) yields \(P(A\mid B)+P(A^c\mid B)=1\). For the union identity, dividing the set inclusion-exclusion equation by \(P(B)\) gives precisely the stated conditional union formula. These arguments require no assumption that \(A\), \(C\), or \(B\) are independent.
Monotonicity Under Conditioning
If one event is contained in another, its conditional probability cannot be larger, provided the same positive-probability event is used for conditioning. The relevant containment is \(A\subseteq C\), which implies \(A\cap B\subseteq C\cap B\).
Indeed, \(C\cap B\) consists of all the outcomes in \(A\cap B\) and possibly additional outcomes. Probability is monotone under set inclusion, so \(P(A\cap B)\le P(C\cap B)\). Dividing both sides by the positive number \(P(B)\) proves the conditional inequality. It is enough that \(A\cap B\subseteq C\cap B\); the stronger global inclusion \(A\subseteq C\) is a convenient sufficient condition.
Worked Example: A Subset of Coding Exceptions
Method: Verify event inclusion, then compare conditional probabilities using the same denominator. In a hypothetical audit, 120 claims belong to group \(B\). Nine claims in this group have a deductible coding error, event \(A\), and 30 have any coding exception, event \(C\). Every deductible coding error is classified as a coding exception. Find both conditional probabilities and check the monotonicity inequality.
The given classification states \(A\subseteq C\). Within the 120 audited claims,
Thus \(0.075\le 0.25\), as required. The difference is \(0.25-0.075=0.175\), corresponding to \(30-9=21\) claims in \(C\) but not \(A\), or \(21/120=0.175\). This count check confirms the comparison. The denominator must represent the same conditioning group for both probabilities.
Common Traps and Practical Checks
Conditional identities are compact, but small changes in the event being conditioned on can change the answer. Before calculating, write down the event after the vertical bar and preserve it throughout the calculation.
- Using an unconditional complement. Suppose \(P(A)=0.30\), \(P(B)=0.50\), and \(P(A\cap B)=0.20\). Then \(P(A\mid B)=0.20/0.50=0.40\). The incorrect calculation \(P(A^c\mid B)=1-P(A)=0.70\) mixes unconditional and conditional probabilities. The correct calculation is \(1-P(A\mid B)=1-0.40=0.60\).
- Ignoring overlap in a union. If \(P(A\mid B)=0.40\), \(P(C\mid B)=0.35\), and \(P(A\cap C\mid B)=0.15\), adding the first two gives 0.75 and counts the overlap twice. The correct union probability is \(0.40+0.35-0.15=0.60\).
- Assuming disjointness without checking. The simplified sum \(P(A\mid B)+P(C\mid B)\) applies only when \(A\cap C\cap B\) has probability zero. Events can overlap outside \(B\) and still be disjoint within \(B\), or overlap within \(B\) even if most of their outcomes lie elsewhere.
- Applying conditional probability when the condition has probability zero. If \(P(B)=0\), the ratio defining \(P(A\mid B)\) has a zero denominator. The formulas in this tutorial do not define conditional probabilities for that event.
These rules are aligned with the SOA VEE Mathematical Statistics syllabus. They provide reliable checks when translating a verbal condition into an event: identify the restricted group, calculate probabilities relative to that group, and account explicitly for overlap.
Check Your Understanding
For each problem, keep the stated conditioning event fixed throughout the calculation.
- Among 80 hypothetical applications selected for a second review, 22 contain a missing signature. Let \(B\) denote selection for second review and \(A\) denote a missing signature. Find the conditional probability of no missing signature. (Answer: 0.7250)
- Among 160 hypothetical policies in group \(B\), 64 have a water-damage claim, 48 have a fire-damage claim, and 16 have both. Find the conditional probability of at least one of these claim types and of neither type. (Answer: 0.6000 and 0.4000)
- In a hypothetical audit group \(B\) of 240 files, event \(A\) is a reserve error and event \(C\) is any valuation error. Every reserve error is a valuation error; 18 files have \(A\) and 72 have \(C\). Find \(P(A\mid B)\), \(P(C\mid B)\), and verify the inclusion inequality. (Answer: 0.0750, 0.3000, and \(0.0750\le 0.3000\))
- Suppose \(P(B)=0.40\), \(P(A\cap B)=0.12\), \(P(C\cap B)=0.16\), and \(P(A\cap C\cap B)=0.08\). Find the conditional probability of neither \(A\) nor \(C\), given \(B\). (Answer: 0.5000)
- Suppose \(P(B)=0.25\), \(P(A)=0.30\), and \(P(A\cap B)=0.10\). Find \(P(A^c\mid B)\), and explain why \(1-P(A)\) is not the answer. (Answer: 0.6000; \(1-P(A)=0.7000\) is unconditional)