Partitions and Exhaustive Case Splits

Why Organize Outcomes Into Cases?

A complicated event can often be described more clearly by dividing the sample space into cases. A useful set of cases has two essential features: no outcome belongs to more than one case, and every outcome belongs to some case. Such cases are called a partition. Partitions let us describe events systematically and, when a finite probability model assigns weights to outcomes, group those weights without omitting or counting any outcome twice.

For example, an inspection outcome might be classified as “no defect,” “scratch,” or “crack.” If these categories cover all possible inspection results and cannot occur together for one item, they form a partition. The event “the item has a defect” can then be described using the scratch and crack cases. The same reasoning applies to card outcomes and survey responses.

Definition and Basic Checks

Definition: Let \(\Omega\) be a nonempty sample space. A collection of nonempty subsets \(C_1,\ldots,C_k\) of \(\Omega\) is a finite partition of \(\Omega\) if (1) \(C_i\cap C_j=\varnothing\) whenever \(i\ne j\), and (2) \(\bigcup_{i=1}^{k} C_i=\Omega\). The sets \(C_i\) are called the cells or cases of the partition.

The first condition is pairwise disjointness: each outcome belongs to at most one cell. The second is exhaustiveness: each outcome belongs to at least one cell. Together they say that every outcome belongs to exactly one cell. A cell must also be nonempty under this definition. If a proposed case is impossible, remove it rather than include an empty cell.

A partition can have just one cell, \(\{\Omega\}\). A binary split into \(B\) and \(B^c\) is a partition when \(B\) is a nonempty proper subset of \(\Omega\). If \(B=\varnothing\) or \(B=\Omega\), one of those two sets is empty; in either case, use the one-cell partition \(\{\Omega\}\) instead.

To check proposed cases, ask two separate questions: “Can one outcome fit into two cases?” and “Can an outcome fit into none of them?” Passing only one check is not enough. Cases that overlap fail disjointness; cases that leave a gap fail exhaustiveness.

Worked Example: Partitioning a One-Card Draw

One card is drawn from a standard 52-card deck, and each individual card is an outcome. Let \(C_1,C_2,C_3,C_4\) be the sets of clubs, diamonds, hearts, and spades, respectively. Verify that these sets partition the sample space, then find the number of outcomes in the event “the card is a king or a queen” by splitting it across the suits.

Disjointness. Each card has exactly one suit. Therefore, a card cannot belong to two of the four sets, so \(C_i\cap C_j=\varnothing\) whenever \(i\ne j\).

Exhaustiveness and nonemptiness. Every card has one of the four suits, so the union of the four sets is the entire sample space. Each suit occurs on 13 cards, so each cell is nonempty. Thus the four suit sets form a partition.

Within each suit, there are two cards in the event: that suit’s king and queen. The event consists of the disjoint pieces formed by its intersection with each suit. Its number of outcomes is therefore \(2+2+2+2=8\), or 8 of the 52 cards in the full sample space.

The partition organizes the count; it does not change the event. It also guards against double-counting: each king or queen belongs to exactly one suit.

Splitting an Event With a Partition

A partition of the sample space can be used to divide any event into disjoint pieces. This is useful even when the original event is not itself one of the cases. For each cell, keep only the outcomes that also belong to the event.

Result: Let \(C_1,\ldots,C_k\) be a finite partition of \(\Omega\), and let \(A\subseteq\Omega\). Then \(\displaystyle A=\bigcup_{i=1}^{k}(A\cap C_i)\), and the sets \(A\cap C_i\) are pairwise disjoint. If \(A\ne\varnothing\), the nonempty sets among these intersections form a partition of \(A\).

Proof. First, suppose \(\omega\in A\). Since the \(C_i\) cover \(\Omega\), the outcome \(\omega\) belongs to some \(C_j\). It also belongs to \(A\), so \(\omega\in A\cap C_j\). This proves that \(A\) is contained in the union of the intersections. Conversely, every point in \(A\cap C_i\) belongs to \(A\), so that union is contained in \(A\). The sets are therefore equal.

Now take distinct indices \(i\ne j\). If an outcome belonged to both \(A\cap C_i\) and \(A\cap C_j\), it would belong to both \(C_i\) and \(C_j\), contrary to the partition’s disjointness. Thus the intersections are pairwise disjoint. Removing any empty intersections leaves nonempty disjoint sets with union \(A\). When \(A\ne\varnothing\), at least one remains, so they partition \(A\). \(\square\)

In a finite model that assigns each outcome \(\omega\) a weight \(p(\omega)\), with an event’s probability calculated as the sum of its outcome weights, this set decomposition gives a corresponding probability calculation:

P(A)=\sum_{\omega\in A}p(\omega) =\sum_{i=1}^{k}\ \sum_{\omega\in A\cap C_i}p(\omega).

The second equality is a regrouping of a finite sum: each outcome in \(A\) occurs in exactly one intersection, and no outcome outside \(A\) occurs in any of them. An empty intersection contributes no terms. This calculation relies on the stated finite-weight model, not on an assumption that the cells are equally likely.

Worked Example: Survey Responses by Age Group

A survey model assigns probability \(0.18\) to a “yes” response from a respondent under 30, \(0.27\) to a “no” response from a respondent under 30, \(0.22\) to a “yes” response from a respondent aged 30 or older, and \(0.33\) to a “no” response from a respondent aged 30 or older. Treat these four response-and-age outcomes as the finite outcomes of the model. Find the probability of a “yes” response by using age groups as cases.

The cases “under 30” and “30 or older” are disjoint: no age is both less than 30 and at least 30. They are exhaustive for the age values in the model: every respondent is in one of the two groups. Together they give a partition of the outcomes by age group.

The event of a “yes” response has two outcomes in this model, one in each age group. Their assigned weights are \(0.18\) and \(0.22\), so \(\displaystyle P(\text{yes})=0.18+0.22=0.40.\)

For a check, the four weights sum to \(0.18+0.27+0.22+0.33=1.00\), and the total weight in the two age cases is \(0.18+0.27=0.45\) and \(0.22+0.33=0.55\). The two groups account for the full model, and the “yes” event uses only the two yes outcomes.

Regrouping Cases and Checking Boundaries

Sometimes a useful description requires combining cells of an existing partition. Combining whole cells preserves the case structure: if the original cells are disjoint, unions formed from disjoint groups of cells remain disjoint. If the groups collectively include every original cell, their unions remain exhaustive.

Result: Suppose \(C_1,\ldots,C_k\) partition \(\Omega\), and divide the index set \(\{1,\ldots,k\}\) into nonempty, disjoint groups \(I_1,\ldots,I_m\) whose union is the entire index set. Then the sets \(D_j=\bigcup_{i\in I_j}C_i\), for \(j=1,\ldots,m\), form a partition of \(\Omega\).

Proof. Each \(D_j\) is nonempty because \(I_j\) contains an index \(i\), and \(C_i\) is nonempty and contained in \(D_j\). If \(j\ne r\), the index groups \(I_j\) and \(I_r\) are disjoint. Thus any cell contributing to \(D_j\) differs from every cell contributing to \(D_r\); those original cells are disjoint, so \(D_j\cap D_r=\varnothing\). Finally, the index groups together contain every index \(1,\ldots,k\). Hence the union of the \(D_j\) is the union of all the \(C_i\), which is \(\Omega\). The \(D_j\) form a partition. \(\square\)

Worked Example: Grouping Inspection Outcomes

A finite inspection model has three possible outcomes for one item: no defect with probability \(0.91\), a scratch with probability \(0.06\), and a crack with probability \(0.03\). Find the probability of a defect, and describe two different partitions that can organize this calculation.

The three outcome sets—no defect, scratch, and crack—are nonempty, mutually exclusive, and cover all the model’s outcomes. They are a partition. The defect event consists of the scratch and crack cells. Their weights give \(\displaystyle P(\text{defect})=0.06+0.03=0.09.\)

One can also regroup the three original cells into two cases: “no defect” and “some defect,” where “some defect” is the union of the scratch and crack cells. These new cases are disjoint and exhaustive. Their weights are \(0.91\) and \(0.06+0.03=0.09\), which sum to \(1.00\).

A different question might distinguish “crack” from “no crack.” The cases “crack” and “no crack” also form a partition; “no crack” combines the no-defect and scratch outcomes, with weight \(0.91+0.06=0.97\). The right partition depends on the question, but every proposed split must still be disjoint and exhaustive.

Common Mistakes

Checking only exhaustiveness. “Age under 40” and “age over 30” overlap for ages 31 through 39, so these cases are not disjoint, even though they cover many ages.
Missing a boundary value. “Age under 30” and “age over 30” omit age 30. A complete split is “age under 30” and “age 30 or older,” or “age 30 or younger” and “age over 30.”
Using a two-set complement split without checking nonemptiness. \(B\) and \(B^c\) partition \(\Omega\) only when \(B\) is nonempty and is not all of \(\Omega\). In the two trivial cases, use \(\{\Omega\}\).
Counting an outcome in multiple cases. If cases overlap, adding their weights can count shared outcomes more than once. Redefine the cases so each outcome has exactly one home.
Assuming a partition implies equal probabilities. A partition describes sets of outcomes, not their weights. The survey and inspection examples used unequal probabilities for their cases.

Partitions matter because they turn an informal list of cases into a precise description that can be checked and used in calculations. Before adding case probabilities in a finite model, verify that the cases are disjoint and exhaustive, and make sure the probabilities correspond to the outcomes in the event being calculated.

Check Your Understanding

For each question, check both disjointness and exhaustiveness where relevant.

For whole-number ages, do “under 18” and “18 or older” form a partition of all possible ages? Explain.
A sample space is partitioned into \(C_1,C_2,C_3\). Describe how to split an event \(A\) into disjoint pieces using those cells.
Why do \(B\) and \(B^c\) fail to be a two-cell partition if \(B=\Omega\), and what partition can be used instead?
An inspection model assigns weights \(0.80\), \(0.12\), and \(0.08\) to no defect, scratch, and crack. Calculate the probability of a defect by grouping the relevant cases.
For whole-number ages, explain the overlap in “age under 40” and “age over 30,” and give a disjoint, exhaustive split with a boundary at 30.