Why Factor an Intersection?
Many probability problems describe a result that requires several conditions to hold together. A hypothetical insurance application, for example, might be complete, satisfy an underwriting standard, and be issued within a specified period. The probability of all three events occurring can be calculated by starting with the first event and then accounting for each additional event given that the earlier ones occurred.
This method is the chain rule for intersections. It extends the multiplication rule for two events by adding one conditional probability for each successive event. The order used to write the factors matters: each factor must condition on all the events that precede it. The rule does not require the events to be independent.
Why the Chain Rule Holds
The rule follows by repeatedly using the definition of conditional probability. The proof also shows why the positive-probability condition is necessary: conditional probabilities are defined by a ratio whose denominator must be positive.
Let \(C_j=A_1\cap\cdots\cap A_j\). The full intersection is \(C_n\), while \(C_{j-1}\) is the condition that the first \(j-1\) events have all occurred.
For each \(j\geq 2\), the assumption \(P(C_{j-1})>0\) gives \(P(A_j\mid C_{j-1})=P(C_j)/P(C_{j-1})\). Thus \(P(C_j)=P(C_{j-1})P(A_j\mid C_{j-1})\).
Starting with \(P(C_n)=P(C_{n-1})P(A_n\mid C_{n-1})\), substitute the corresponding expression for \(P(C_{n-1})\), then for \(P(C_{n-2})\), continuing until \(P(C_1)=P(A_1)\).
The substitutions produce the chain-rule product. Each intermediate prefix probability cancels from the denominator of one ratio and the numerator of the preceding ratio. For example, with four events, the conditional factors are \(P(A_2\mid A_1)\), \(P(A_3\mid A_1\cap A_2)\), and \(P(A_4\mid A_1\cap A_2\cap A_3)\), in addition to \(P(A_1)\).
If a prefix has probability zero, a conditional probability given that prefix is undefined under the usual definition. In that case, the full intersection also has probability zero, because it is a subset of the zero-probability prefix. Do not try to use a chain-rule product containing an undefined conditional term; identify the zero-probability prefix directly.
Applying the Rule in the Given Order
Read a sequence of conditional probabilities as a sequence of increasingly specific conditions. “The second event given the first” is not the same information as “the second event given that the first and third occurred.” A reliable calculation writes each conditioning event explicitly before multiplying.
Worked Example: Three Conditions for an Application
Hypothetical scenario. An insurer reviews a group of applications. Let \(A\) be the event that an application is complete, \(B\) the event that it meets an underwriting standard, and \(C\) the event that it is issued within the review period. Suppose \(P(A)=0.70\), \(P(B\mid A)=0.80\), and \(P(C\mid A\cap B)=0.90\). Find the probability that all three conditions hold.
Method. Apply the chain rule in the order \(A,B,C\). The given conditional probabilities match the successive event prefixes, so no additional independence assumption is needed.
First, the probability of a complete application is \(P(A)=0.70\). The probability that it is complete and meets the underwriting standard is \(0.70\times0.80=0.56\). Including timely issue gives \(0.56\times0.90=0.504\). Equivalently,
The probability that all three conditions hold is \(0.5040\), or \(50.40\%\). The result is no greater than any one of the three given probabilities, as expected for an intersection contained within each event.
Common mistake avoided. Multiplying is appropriate here because the second and third factors are conditional probabilities with the required prefixes. The justification is the chain rule—not an assumption that the application events are independent.
Worked Example: Four Stages in a Hypothetical Review
Hypothetical scenario. A review process has four required stages. Let \(A\) mean that a file is received by the deadline, \(B\) that its documents pass an initial check, \(C\) that its recorded values fall within an allowed range, and \(D\) that a decision is issued on time. Suppose \(P(A)=0.65\), \(P(B\mid A)=0.80\), \(P(C\mid A\cap B)=0.75\), and \(P(D\mid A\cap B\cap C)=0.90\). Find the probability that a file meets all four requirements.
Method. Use the four-event chain rule, checking that every supplied condition includes all the earlier stages.
After the first stage, the probability is \(0.65\). After both the first and second stages, it is \(0.65\times0.80=0.52\). After the third stage, it is \(0.52\times0.75=0.39\). After the fourth stage, it is \(0.39\times0.90=0.351\). Therefore,
Thus the probability of meeting all four requirements is \(0.3510\), or \(35.10\%\). The intermediate values are the probabilities of the successive prefixes; each one is the previous prefix probability multiplied by the next conditional factor.
Common mistake avoided. The factor for \(D\) is \(P(D\mid A\cap B\cap C)=0.90\), not an unconditional probability for \(D\). Replacing it with an unconditional value would change the model and would not follow from the supplied data.
Finding a Missing Conditional Probability
The chain rule can also be rearranged to solve for an unknown conditional probability. If the probability of a full intersection and the earlier factors are known, divide the intersection probability by the probability of the preceding prefix. The prefix must have positive probability.
Worked Example: Recovering the Final-Stage Probability
Hypothetical scenario. In a file review, let \(A\) mean that a file passes an initial screen, \(B\) that it passes a document check, and \(C\) that it satisfies a final eligibility criterion. Suppose \(P(A)=0.80\), \(P(B\mid A)=0.60\), and \(P(A\cap B\cap C)=0.216\). Find \(P(C\mid A\cap B)\).
Method. Use the three-event chain rule and isolate the final conditional factor.
The probability of the prefix \(A\cap B\) is \(P(A)P(B\mid A)=(0.80)(0.60)=0.48\). This is positive, so conditioning on \(A\cap B\) is defined. The chain rule gives
Substitute the given intersection probability and the calculated prefix probability: \(0.216=0.48P(C\mid A\cap B)\). Hence
The final-stage conditional probability is \(0.4500\). Checking by substitution gives \((0.80)(0.60)(0.45)=0.48(0.45)=0.216\), exactly the stated probability of all three events.
Common mistake avoided. Divide by \(P(A\cap B)=0.48\), not by \(P(A)=0.80\). The unknown condition is given both \(A\) and \(B\), so its denominator in the conditional-probability definition is the probability of \(A\cap B\).
Order, Interpretation, and Common Traps
An intersection does not depend on the order in which its events are written: \(A\cap B\cap C\) is the same event as \(C\cap A\cap B\). A chain-rule factorization may use either order, but the conditional factors must change to match that order. For example, starting with \(A\) requires the terms \(P(B\mid A)\) and \(P(C\mid A\cap B)\); starting with \(C\) requires \(P(A\mid C)\) and \(P(B\mid C\cap A)\). These conditional probabilities are not interchangeable.
The main exam trap is to multiply probabilities merely because several events are named. The chain rule uses conditional probabilities along a stated sequence. In contrast, multiplying unconditional event probabilities requires additional conditions and cannot be assumed from the fact that the events are listed together. A useful check is to write down the conditioning set for every factor before doing any arithmetic.
| Factor | Correct condition in the order \(A,B,C,D\) | What it represents |
|---|---|---|
| First | \(P(A)\) | Probability that the first event occurs |
| Second | \(P(B\mid A)\) | Probability of \(B\) after the first event is known to occur |
| Third | \(P(C\mid A\cap B)\) | Probability of \(C\) after both earlier events are known to occur |
| Fourth | \(P(D\mid A\cap B\cap C)\) | Probability of \(D\) after all three earlier events are known to occur |
Check Your Understanding
For each problem, write the successive conditional factors in the stated order before calculating.
- A hypothetical application must satisfy events \(A\), \(B\), and \(C\), in that order. Given \(P(A)=0.90\), \(P(B\mid A)=0.80\), and \(P(C\mid A\cap B)=0.50\), find \(P(A\cap B\cap C)\). (Answer: \(0.3600\))
- For four stages \(A,B,C,D\), suppose \(P(A)=0.72\), \(P(B\mid A)=0.75\), \(P(C\mid A\cap B)=0.80\), and \(P(D\mid A\cap B\cap C)=0.90\). Find the probability all four occur. (Answer: \(0.3888\))
- Suppose \(P(A)=0.50\), \(P(B\mid A)=0.60\), and \(P(C\mid A\cap B)=0.40\). Calculate the probability that all three events occur. (Answer: \(0.1200\))
- Suppose \(P(A)=0.70\), \(P(B\mid A)=0.40\), and \(P(A\cap B\cap C)=0.084\). Find \(P(C\mid A\cap B)\). (Answer: \(0.3000\))
- In the order \(A,B,C,D\), the conditional factors are \(P(A)=0.80\), \(P(B\mid A)=0.90\), \(P(C\mid A\cap B)=0.50\), and \(P(D\mid A\cap B\cap C)=0.75\). Find the four-event intersection and state the probability after the first three events. (Answer: Full intersection \(0.2700\); first three events \(0.3600\))