Why Organize Outcomes into Cases?
An event can have different probabilities under different circumstances. For example, a hypothetical insurer might have different claim frequencies across customer groups. To find the probability of a claim for a randomly selected policy, it is not enough to average the group-specific claim probabilities: the sizes of the groups matter. A partition organizes the possible cases so that each outcome belongs to exactly one of them.
The key calculation is a weighted sum. Each case contributes its conditional probability of the event multiplied by the probability of that case. This idea is useful whenever a model separates outcomes into distinct categories, such as policy types, routes, locations, or stages of a process.
A partition must meet both conditions. Mutually exclusive events do not overlap, but they might leave some outcomes out. Exhaustive events cover the sample space, but they might overlap. Both properties are needed for a clean division into cases.
The Overall Probability Across a Partition
Let \(A\) be an event of interest and let \(B_1,\ldots,B_k\) be a partition. For any case \(B_i\) with \(P(B_i)>0\), the conditional probability is defined by \(P(A\mid B_i)=P(A\cap B_i)/P(B_i)\). Since the cases cover the sample space, the event \(A\) can be divided into its parts in those cases.
The events \(A\cap B_i\) are mutually exclusive because the \(B_i\) are mutually exclusive. Finite additivity therefore gives \(P(A)=\sum_{i=1}^{k}P(A\cap B_i)\). Substituting \(P(A\cap B_i)=P(A\mid B_i)P(B_i)\) gives the weighted-sum formula.
Each term \(P(A\mid B_i)P(B_i)\) is the probability that both the case \(B_i\) occurs and the event \(A\) occurs. The overall probability is the sum of these disjoint contributions. If a partition event has probability zero, its conditional probability is not defined by the usual ratio formula; its intersection with any event also has probability zero, so it contributes zero and can be omitted from the sum.
Worked Example: Checking Cases Before Weighting Them
A hypothetical policy review classifies each policy as standard, enhanced, or specialty. The proportions are \(0.50\), \(0.30\), and \(0.20\), respectively. Conditional on the classification, the probability of a late report during the year is \(0.02\), \(0.08\), and \(0.25\). Find the probability that a randomly selected policy has a late report. Assume each policy belongs to exactly one classification and that the stated proportions describe the selected policy’s classification.
Method: verify the partition, then apply the weighted-sum formula. The three classifications are mutually exclusive by the stated model. They are exhaustive because every policy belongs to one of them. Their probabilities are nonnegative and sum to \(0.50+0.30+0.20=1.00\), which checks the stated proportions.
Let \(A\) be the event of a late report, and let \(B_1,B_2,B_3\) denote the standard, enhanced, and specialty cases. The three weighted contributions are:
Standard: \(P(A\mid B_1)P(B_1)=0.02(0.50)=0.010\).
Enhanced: \(P(A\mid B_2)P(B_2)=0.08(0.30)=0.024\).
Specialty: \(P(A\mid B_3)P(B_3)=0.25(0.20)=0.050\).
Add the disjoint contributions: \(P(A)=0.010+0.024+0.050=0.084\). The probability of a late report is \(0.084\), or \(8.4\%\). The calculation is already exact to four decimal places, so no further rounding is needed.
Check: The result lies between the smallest and largest conditional probabilities, \(0.02\) and \(0.25\), as a weighted average with nonnegative weights summing to one must. The specialty case contributes the most, \(0.050\), even though it is the smallest group, because its conditional probability is comparatively high.
Common mistake avoided: Taking the unweighted average \((0.02+0.08+0.25)/3\) would give \(0.1167\), which treats the three groups as equally likely. They are not equally likely; the correct weights are \(0.50,0.30,0.20\).
Using the Formula Carefully
The calculation has three separate ingredients: a valid partition, the probabilities of its cases, and the conditional probability of the event within each case. Before substituting numbers, check that the case probabilities sum to one and that the conditional probabilities match the event and case named in the problem.
Define the event of interest \(A\) and each partition event \(B_i\) in the problem’s context.
Confirm that exactly one case occurs, or equivalently that the cases are pairwise disjoint and exhaustive. Check that their probabilities sum to one.
For each case, multiply its probability by the conditional probability of \(A\) in that case.
The intersections of \(A\) with the partition cases are disjoint, so their probabilities add to \(P(A)\).
If the event of interest is being considered only within a union of partition cases, first calculate the probability of the event and the probability of that union. Their ratio gives the conditional probability within the union. This is still a direct use of conditional probability; the cases in the union must be weighted according to their probabilities.
Worked Example: Combining Two Cases Within a Restricted Group
A hypothetical portfolio has three mutually exclusive policy segments: segment \(A_1\) contains \(40\%\) of policies, segment \(A_2\) contains \(35\%\), and segment \(A_3\) contains \(25\%\). The probabilities of a claim during the year, conditional on the segment, are \(0.03\), \(0.06\), and \(0.10\), respectively. Find (a) the overall probability of a claim and (b) the probability of a claim given that the policy is in segment \(A_1\) or \(A_2\). Assume the segments cover all policies, and the conditional claim probabilities refer to the same one-year period.
Method: apply the partition formula for (a), then use the definition of conditional probability for (b). The segment proportions sum to \(0.40+0.35+0.25=1.00\), and the segments are stated to be mutually exclusive and exhaustive.
Let \(C\) be the event of a claim. The contributions to the overall claim probability are:
\(P(C\cap A_1)=0.40(0.03)=0.012\), \(P(C\cap A_2)=0.35(0.06)=0.021\), and \(P(C\cap A_3)=0.25(0.10)=0.025\).
Therefore \(P(C)=0.012+0.021+0.025=0.058\), or \(5.8\%\).
For part (b), the probability of being in segment \(A_1\) or \(A_2\) is \(0.40+0.35=0.75\). A claim and membership in this union have probability \(0.012+0.021=0.033\). Thus
The conditional claim probability within segments \(A_1\) and \(A_2\) combined is \(0.044\), or \(4.4\%\). As a check, weighting their conditional claim probabilities by their relative proportions within the union gives \((0.40/0.75)(0.03)+(0.35/0.75)(0.06)=0.016+0.028=0.044\).
Common mistake avoided: The average \((0.03+0.06)/2=0.045\) incorrectly gives the two segments equal weight. Within the restricted group, their relative weights are \(0.40/0.75\) and \(0.35/0.75\), not one-half each.
Interpreting the Weights and Avoiding Traps
The weights in the formula are the probabilities of the cases, not the conditional probabilities of the event. They are nonnegative and sum to one. Consequently, when all cases have positive probability, \(P(A)\) must lie between the smallest and largest of \(P(A\mid B_i)\). This is a useful arithmetic check, though it does not replace verifying the model.
A case description must also be precise. “Urban” and “rural” might form a partition if every location is classified as exactly one of those types. “Urban” and “near a highway” generally do not: an outcome could belong to both, and an outcome might belong to neither. Adding conditional probabilities across non-partition cases can double-count overlaps or omit possibilities.
Worked Example: An Outcome Across Three Operating Conditions
A hypothetical service network operates under three mutually exclusive and exhaustive conditions: normal load with probability \(0.80\), elevated load with probability \(0.15\), and emergency load with probability \(0.05\). The probability of a service interruption during a specified day is \(0.01\), \(0.12\), and \(0.50\), respectively, conditional on those conditions. Find the probability of an interruption that day. Assume the probabilities apply to the same network and day, and that each day has exactly one of the three conditions.
Method: use the conditional probabilities within the three cases and add their joint contributions. The case probabilities sum to \(0.80+0.15+0.05=1.00\), and the conditions are stated to form a partition.
Let \(I\) denote an interruption. Under normal load, \(P(I\cap B_1)=0.80(0.01)=0.008\). Under elevated load, \(P(I\cap B_2)=0.15(0.12)=0.018\). Under emergency load, \(P(I\cap B_3)=0.05(0.50)=0.025\).
Adding the three contributions gives \(P(I)=0.008+0.018+0.025=0.051\). The interruption probability is \(0.051\), or \(5.1\%\). The result is exact to three decimal places.
Check: The value \(0.051\) lies between the smallest conditional probability, \(0.01\), and the largest, \(0.50\). Although emergency load has the highest conditional interruption probability, its contribution to the overall probability is \(0.025\), because that condition occurs only \(5\%\) of the time.
Common mistake avoided: Reporting \(0.054\) would not follow from the weighted contributions. Their sum is \(0.008+0.018+0.025=0.051\); adding the products carefully is essential.
Common Mistakes
- Checking only mutual exclusivity. Disjoint cases may still leave outcomes uncovered. Confirm exhaustiveness as well.
- Using an unweighted average. The conditional probabilities must be multiplied by the probabilities of their cases.
- Adding conditional probabilities directly. The sum \(\sum_i P(A\mid B_i)\) is not generally \(P(A)\); the weighted terms are \(P(A\mid B_i)P(B_i)\).
- Mixing up a case probability and a conditional probability. \(P(B_i)\) describes how likely the case is; \(P(A\mid B_i)\) describes the event’s probability within that case.
- Applying the full partition after restricting the population. If conditioning on a union of cases, use the probability of the event within that union divided by the probability of the union.
- Ignoring a missing or overlapping category. If cases do not form a partition, the weighted-sum argument does not apply as stated.
Check Your Understanding
For each problem, identify the cases and keep the case probabilities separate from the conditional event probabilities. All scenarios are hypothetical.
- A delivery network uses three mutually exclusive routes with probabilities \(0.45\), \(0.35\), and \(0.20\). The probability of a late delivery conditional on each route is \(0.01\), \(0.04\), and \(0.09\), respectively. Find the overall probability of a late delivery. (Answer: \(0.0365\))
- A portfolio is divided into three exhaustive risk bands with proportions \(0.50\), \(0.30\), and \(0.20\). Their conditional probabilities of a claim are \(0.02\), \(0.08\), and \(0.25\). Find the overall claim probability, and state the contribution from the third band. (Answer: overall \(0.084\); third-band contribution \(0.050\))
- A hypothetical power system is in stable, strained, or critical condition with probabilities \(0.70\), \(0.20\), and \(0.10\). Conditional outage probabilities are \(0.005\), \(0.03\), and \(0.12\). Find the overall outage probability. (Answer: \(0.0215\))
- A set of policies is divided into three mutually exclusive segments of sizes \(0.20\), \(0.50\), and \(0.30\). Their conditional annual claim probabilities are \(0.04\), \(0.02\), and \(0.10\). Find the probability of a claim conditional on membership in the first or third segment. (Answer: \(0.076\))
- A hypothetical equipment pool has three exhaustive maintenance classes with proportions \(0.60\), \(0.25\), and \(0.15\). The conditional probability of failure during a month is \(0.01\), \(0.05\), and \(0.20\), respectively. Find the overall failure probability, then find the probability of no failure. (Answer: failure \(0.0485\); no failure \(0.9515\))