Why Partitions Matter
A finite experiment can have many possible outcomes, and a question about those outcomes often asks for a useful classification. For example, a quality inspection might classify each possible result as having no defective components, one defective component, or two defective components. A good classification assigns every possible outcome to one category—and only one.
A collection of events with exactly these properties is called a partition of the sample space. Partitions help organize outcomes before probabilities or other summaries are calculated. They are useful only when the categories are both complete and nonoverlapping.
(1) the events are pairwise disjoint: \(A_i\cap A_j=\varnothing\) whenever \(i\ne j\); and
(2) their union is the entire sample space: \(A_1\cup A_2\cup\cdots\cup A_k=S\).
Equivalently, every outcome in \(S\) belongs to exactly one event in the collection.
The two conditions serve different purposes. Pairwise disjointness prevents any outcome from being assigned to two categories. The union condition prevents any possible outcome from being left unclassified. Neither condition alone is enough. A collection may have no overlaps but omit some outcomes; it may cover all outcomes but assign some to more than one event.
The events in a partition are often called its cells. Each cell is a subset of the sample space, and the cells together provide a classification of all possible outcomes. A partition does not require cells to have equal numbers of outcomes or equal probabilities.
A Reliable Partition Check
For a finite sample space, a direct outcome-by-outcome check is usually the safest method. First list the outcomes in each proposed event. Then check for overlaps and confirm that every outcome in the sample space appears somewhere. A useful final question is: Can I point to exactly one cell for every possible outcome?
Make the set of possible outcomes explicit. A classification cannot be checked correctly if the sample space is misunderstood or incomplete.
Compare the proposed events. If any outcome belongs to two different events, the collection is not pairwise disjoint.
Compare the union of the events with the sample space. Every possible outcome must appear in at least one event.
Under the standard definition used here, an empty set is not a cell of a partition.
Worked Examples
Worked Example: Classifying the Result of a Spinner
A spinner has eight labeled sectors, numbered \(1\) through \(8\), and one spin is recorded. Consider the proposed events \(E=\{2,4,6,8\}\) and \(O=\{1,3,5,7\}\). Determine whether \(\{E,O\}\) is a partition of the sample space.
Method: check disjointness and exhaustiveness. The sample space is \(S=\{1,2,3,4,5,6,7,8\}\). The event \(E\) contains the even-numbered results, while \(O\) contains the odd-numbered results.
Check for overlap. No number in \(E\) is in \(O\), so \(E\cap O=\varnothing\).
Check for omissions. The union is \(E\cup O=\{1,2,3,4,5,6,7,8\}=S\). Both events are nonempty. Thus every result belongs to exactly one of the two events, and \(\{E,O\}\) is a partition.
Common mistake avoided: Calling two events a partition just because they do not overlap. In this example, the union check is also necessary; without it, some spinner result could be unclassified.
Worked Example: Grouping Outcomes by the Number of Defects
A hypothetical inspection records the condition of two components. Each component is labeled \(G\) if it passes inspection and \(D\) if it is defective. The recorded outcome lists the first component's condition followed by the second component's condition. Thus \(S=\{GG,GD,DG,DD\}\). Define \(C_0\) as the event of no defective components, \(C_1\) as the event of exactly one defective component, and \(C_2\) as the event of two defective components. Write out these events and verify that they form a partition.
Method: apply the classification rule to each listed outcome. The outcome \(GG\) has no defective components, so \(C_0=\{GG\}\). The outcomes \(GD\) and \(DG\) each have exactly one defective component, so \(C_1=\{GD,DG\}\). Finally, \(DD\) has two defective components, so \(C_2=\{DD\}\).
Check for overlap. The four possible outcome labels are distinct. The event \(C_0\) contains only \(GG\), \(C_1\) contains \(GD\) and \(DG\), and \(C_2\) contains only \(DD\). No outcome appears in more than one event; each pair of different cells has empty intersection.
Check for omissions. The union is \(C_0\cup C_1\cup C_2=\{GG,GD,DG,DD\}=S\). Each cell is nonempty. Therefore, \(\{C_0,C_1,C_2\}\) is a partition.
The order in the recorded labels matters for distinguishing \(GD\) from \(DG\), even though both outcomes belong to the same cell when the classification is based on the number of defects. Grouping outcomes does not erase the sample space's distinctions; it places outcomes into broader categories.
Common mistake avoided: Replacing \(C_1=\{GD,DG\}\) with one outcome, as though the two recorded orders were identical. They are distinct outcomes in \(S\), even though the classification puts them in the same cell.
Worked Example: Checking a Review-Outcome Classification
A hypothetical review process has the finite sample space \(S=\{A,B,C,D,E,F\}\), where the labels are six distinct possible recorded outcomes. A supervisor proposes three categories: \(R_1=\{A,B\}\) for routine closure, \(R_2=\{C,D\}\) for further review, and \(R_3=\{E,F\}\) for closure after a second check. Determine whether these categories form a partition. Then consider a revised category \(R'_2=\{B,C,D\}\) in place of \(R_2\), with \(R_1\) and \(R_3\) unchanged.
Method: test each proposed collection separately. For the original categories, no label is repeated: \(R_1\), \(R_2\), and \(R_3\) are pairwise disjoint. Their union is \(R_1\cup R_2\cup R_3=\{A,B,C,D,E,F\}=S\). All three are nonempty, so the original categories form a partition.
For the revised collection, \(R_1\cap R'_2=\{B\}\), so the first two categories overlap. Also, \(R_1\cup R'_2\cup R_3=\{A,B,C,D,E,F\}=S\), so the revised categories still cover the sample space. Coverage does not repair the overlap: the revised collection is not a partition.
Common mistake avoided: Checking only that every outcome is included. In the revised collection every outcome is included, but \(B\) belongs to two categories, so the classification fails the pairwise-disjoint condition.
Why the Definition Guarantees a Unique Classification
The equivalence between the two-condition definition and “exactly one cell” follows directly from set membership. Suppose the cells are pairwise disjoint and their union is \(S\). If \(\omega\in S\), the union condition means \(\omega\) belongs to at least one cell. If \(\omega\) belonged to two different cells, say \(A_i\) and \(A_j\), then \(\omega\in A_i\cap A_j\), contradicting pairwise disjointness. Therefore, \(\omega\) belongs to exactly one cell.
Conversely, suppose every outcome in \(S\) belongs to exactly one cell. No outcome can belong to two different cells, so each pair of cells has empty intersection. And because every outcome belongs to a cell, the union of all the cells is \(S\). This proves why both conditions are necessary: together they make the classification complete and unambiguous.
Recognizing Different Classifications
The same sample space can be partitioned in different ways, depending on the question being asked. In the two-component example, classifying outcomes by the number of defects gave three cells. A different rule could classify them by whether the first component is defective. The resulting cells would be \(\{GG,GD\}\) and \(\{DG,DD\}\). These are different categories, but they still form a partition because every outcome is in exactly one of them.
A more detailed classification can split a cell from a broader one while retaining the same sample space. For instance, “zero, one, or two defects” is more detailed than “zero or at least one defect”: the latter uses cells \(\{GG\}\) and \(\{GD,DG,DD\}\). The finer classification separates outcomes that the broader classification groups together. In either case, test the actual cells against the same partition conditions.
Common Mistakes and Exam Traps
- Mutually exclusive but not exhaustive. Disjoint events may leave some possible outcomes out. Check that their union equals the whole sample space, not merely a subset of it.
- Exhaustive but not mutually exclusive. A collection may cover every outcome and still assign an outcome to two cells. Check intersections as well as the union.
- Confusing event labels with outcomes. An event is a set of outcomes. A label such as “one defect” describes a set; it is not itself a sample-space outcome unless the experiment defines it that way.
- Ignoring how outcomes are recorded. If order or other features distinguish outcomes in the stated sample space, retain those distinctions when listing cells—even when the classification groups several outcomes together.
- Assuming equally sized cells are required. A partition has no equal-size requirement. Its cells may contain different numbers of outcomes.
Check Your Understanding
For each problem, decide whether the proposed events form a partition, or construct the requested partition. Check both coverage and overlap.
- A finite experiment has \(S=\{a,b,c,d,e\}\). Do \(A=\{a,c,e\}\) and \(B=\{b,d\}\) form a partition of \(S\)? (Answer: Yes; they are disjoint, nonempty, and their union is \(S\).)
- For \(S=\{1,2,3,4,5,6\}\), a student proposes \(A=\{1,2,3\}\), \(B=\{3,4\}\), and \(C=\{5,6\}\). Do these events form a partition? (Answer: No; outcome \(3\) belongs to both \(A\) and \(B\).)
- A hypothetical device test has outcomes \(S=\{LL,LH,HL,HH\}\), where each letter records the test result for one of two stages and \(H\) means a high reading. Give a partition by the number of high readings. (Answer: \(\{\{LL\},\{LH,HL\},\{HH\}\}\).)
- Let \(S=\{p,q,r,s,t,u\}\). The categories are \(A=\{p,q\}\), \(B=\{r,s\}\), and \(C=\{t\}\). Identify the defect in this proposed partition. (Answer: It is not exhaustive; \(u\) is omitted from the union.)
- A finite survey records one of \(S=\{N_1,N_2,Y_1,Y_2\}\), where \(N\) or \(Y\) is the response and the subscript is the survey version. Construct a partition by response, then explain why its cells are disjoint. (Answer: \(\{\{N_1,N_2\},\{Y_1,Y_2\}\}\); no outcome has both an \(N\) and a \(Y\) response.)