Why Branch Probabilities Need a Check
A probability tree lays out possible stages of a process. Each branch leaving a node represents one possible outcome at that stage, given the outcomes that led to the node. A tree may give some branch probabilities and leave others blank. Completing it is not just a matter of subtracting numbers: the calculation must use the right condition, and the outgoing branches must describe all the possible outcomes at that node without overlap.
The key idea is that probabilities of complementary outcomes add to one when they are considered under the same condition. For example, if an event has probability \(0.72\) given a particular starting condition, its complement has probability \(1-0.72=0.28\) given that same condition. A probability from a different node cannot be substituted just because it describes a similarly named event.
At the first node, the branches describe outcomes without a preceding condition. At later nodes, the probability on a branch is conditional on the path to that node. For instance, the label \(P(B\mid A)=0.3\) means the probability of \(B\) among outcomes for which \(A\) has occurred; it is not necessarily the overall probability \(P(B)\). This distinction is essential when filling in blanks.
Complements at a Node
Suppose a node represents the condition \(C\), and its two branches are \(A\) and \(A^c\), where \(A^c\) means that \(A\) does not occur. Provided \(P(C)>0\), the conditional outcomes \(A\) and \(A^c\) are exhaustive and mutually exclusive within that node. Their conditional probabilities therefore add to one:
Rearranging gives the missing-branch rule:
The condition \(C\) remains in both terms. If the known branch is \(P(A\mid C)\), subtract it from one to find the probability of “not \(A\) given \(C\),” not the probability of “not \(A\)” under some other condition.
Sometimes a node has more than two branches. If its outcomes \(A_1,\ldots,A_k\) are mutually exclusive and exhaustive under condition \(C\), then their probabilities satisfy
If exactly one branch probability is missing, subtract the known branch probabilities from one. If two or more probabilities are missing, the sum-to-one condition alone does not generally determine each one separately. Additional information is needed.
Worked Examples
Worked Example: Completing a Two-Branch Outcome
In a hypothetical shipment process, a package is recorded as inspected, \(I\), with probability \(0.30\). Otherwise it is not inspected, \(I^c\). If a package is inspected, the probability that it is flagged for review, \(F\), is \(0.08\). Complete the missing branch probabilities at the first two nodes.
First node. The outcomes “inspected” and “not inspected” are complementary. Therefore,
The first branch set is \(0.30\) and \(0.70\), and its probabilities sum to \(1.00\).
Node after inspection. The two outcomes are “flagged” and “not flagged,” conditional on inspection. Thus,
The branch set after \(I\) is \(0.08\) and \(0.92\), which also sums to \(1.00\). The condition matters: \(0.92\) is the probability of not being flagged among inspected packages. It does not describe all packages in the shipment.
Worked Example: Completing Branches After Either First Outcome
In a hypothetical two-stage process, an item first follows route \(R\) with probability \(0.40\), or route \(R^c\). At the next stage, the probability of a delay, \(D\), is \(0.12\) given route \(R\), and \(0.25\) given route \(R^c\). Fill in all the missing branches and check each node.
At the first node, the route probabilities are complementary:
At the node reached after \(R\), delay and no delay are complementary conditional outcomes:
At the node reached after \(R^c\), the condition is different, so use the rate given for that node:
The checks are \(0.40+0.60=1.00\) at the first node, \(0.12+0.88=1.00\) after \(R\), and \(0.25+0.75=1.00\) after \(R^c\). The two probabilities of delay are not interchangeable: one is conditional on \(R\), the other on \(R^c\).
Worked Example: A Node With Three Outcomes
A hypothetical laboratory classifies each sample, after a particular preparation step \(C\), into exactly one of three categories: low, moderate, or high. Given \(C\), the probability of low is \(0.18\), and the probability of moderate is \(0.27\). Find the probability of high, and check the branch set.
The categories are stated to be mutually exclusive and exhaustive, so their conditional probabilities must sum to one. Let \(H\) denote the high category. Then
The branch set checks because \(0.18+0.27+0.55=1.00\). Each probability is between zero and one. This is a residual calculation for one missing outcome; it works because the other two outcomes are known and all three categories cover the possibilities at that node.
Worked Example: Detecting Inconsistent Branch Probabilities
A hypothetical report gives two complementary outcomes at a node: \(P(Q\mid S)=0.64\) and \(P(Q^c\mid S)=0.39\). Determine whether these values can both be correct as stated.
Because \(Q\) and \(Q^c\) are complementary under the same condition \(S\), their probabilities must sum to one. Here,
Equivalently, the complement of \(0.64\) is \(1-0.64=0.36\), not \(0.39\). The reported values cannot both be exact probabilities for this branch set. One should check the source values, their labels, and whether rounding was used; the tree should not be “fixed” by choosing an arbitrary adjustment. If the numbers were rounded, the underlying unrounded values may require examination before deciding whether the discrepancy is only due to rounding.
A Reliable Completion and Checking Process
For each node, pause before calculating and identify the condition represented by the path leading to it. Then identify what the outgoing branches mean. A branch set is valid only if those outcomes are mutually exclusive and exhaustive under that condition.
Write down the event or path already reached. Carry that condition into every conditional probability at this node.
Confirm that the branches do not overlap and that one of them must occur. A complement pair is a common two-branch case.
For a complementary pair, subtract the known probability from one. For several exhaustive outcomes with one blank, subtract the sum of the known probabilities from one.
Verify that each probability lies between zero and one and that the probabilities leaving the node sum to one. Keep enough digits during calculations and round only when reporting.
A probability below zero or above one signals that the inputs or the interpretation need attention. So does a branch sum that differs substantially from one. A small difference can arise if reported probabilities were rounded, but rounding does not justify ignoring an obvious inconsistency.
Common Mistakes
- Dropping the condition. If a branch is \(P(A\mid C)\), its complement at that node is \(P(A^c\mid C)\), not an unconditional probability or a probability conditioned on another event.
- Using a rate from a different node. Conditional rates can differ across paths. Calculate each missing complement from the known rate at its own node.
- Assuming a branch probability is the overall chance of an outcome. A label such as \(P(D\mid R)\) applies only after \(R\). It is not automatically \(P(D)\).
- Subtracting from one when outcomes are not complementary. With three or more outcomes, a missing probability is one minus the sum of all the other known probabilities only if the outcomes are mutually exclusive and exhaustive and only one probability is missing.
- Forcing an impossible answer. If known probabilities already sum to more than one, the missing probability would be negative. That is evidence of inconsistent information, not a valid branch value.
Completing branch probabilities carefully makes a tree easier to read and less likely to carry an error into later calculations. The essential habit is local checking: at every node, use the condition attached to that node, and verify that its outgoing branches form a complete probability distribution.
Check Your Understanding
For each question, state the missing probability and the check you would use.
- At a node conditioned on \(A\), the branches are \(B\) and \(B^c\), and \(P(B\mid A)=0.37\). Find the other branch probability.
- At a node conditioned on \(C\), three exhaustive categories have probabilities \(0.22\), \(0.31\), and an unknown value. Find the unknown value.
- A tree gives \(P(D\mid R)=0.16\). What does the complement of this branch describe, and what is its probability?
- Two complementary branches at the same node are reported as \(0.55\) and \(0.48\). Explain why the values need to be checked.
- A tree has two missing probabilities among three exhaustive branches, and the one known probability is \(0.40\). Does the sum-to-one condition determine both missing values uniquely? Explain.