Partition of a sample space

A collection of events $E_1, E_2, \dots, E_n$ is said to form a partition of the sample space $S$ when all three of the following hold simultaneously.

Mutually exclusive (pairwise disjoint). No two of the events share an outcome:

where: $E_i, E_j$ are any two distinct events from the collection, and $\varnothing$ is the empty set — there is no outcome common to both.

Exhaustive. Together the events use up the entire sample space, leaving nothing out:

where: the union on the left collects every outcome that lies in at least one $E_i$, and this union must equal the full sample space $S$ — every possible outcome belongs to some $E_i$.

Nonzero probability. Every member of the partition is a genuinely possible event, not an impossible one:

where: $P(E_i)$ is the probability of $E_i$; ruling out $P(E_i) = 0$ matters because the total probability theorem and Bayes' theorem later divide by such quantities, so a zero-probability member would make those formulas break down. In plain words: the pieces $E_1, \dots, E_n$ never overlap, together they cover every outcome in $S$, and each piece is something that can actually occur. Think of $S$ as a plot of land and the $E_i$ as fields on it — no field trespasses on another, every square metre belongs to exactly one field, and no field has zero area. The simplest and most common partition of all comes from a single event and its complement. For any event $E$ with $0 < P(E) < 1$, the pair $\{E, E'\}$ always forms a partition of $S$: $E \cap E' = \varnothing$ (an outcome cannot both happen and not happen) and $E \cup E' = S$ (every outcome either is in $E$ or is not). Every richer partition used later in the total probability theorem is just a generalisation of this two-piece idea to $n$ pieces.

Why does the diagram matter beyond geometry? Because it is the exact picture the total probability theorem relies on. If $E_1, \dots, E_n$ partition $S$, then any event $A$ can be written as

where: each $A \cap E_i$ is the part of $A$ that happens "via route $E_i$" — and because the $E_i$ never overlap, the pieces $A \cap E_i$ never overlap either, even though $A$ itself might sprawl across every one of the fields.

A single die is rolled, so the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Consider the events $E_1 = \{1, 2\}$, $E_2 = \{3, 4\}$, $E_3 = \{5, 6\}$. Verify that $\{E_1, E_2, E_3\}$ forms a partition of $S$, and use it to decompose $A = \{2, 3, 5, 6\}$, the event "the die shows an even-position pair member other than 4," which we take here simply as the given set for illustration.

Step 1 — check pairwise disjointness. $E_1 \cap E_2 = \varnothing$, $E_2 \cap E_3 = \varnothing$, $E_1 \cap E_3 = \varnothing$. All three pairwise intersections are empty, so the disjointness condition holds.

Step 2 — check exhaustiveness. $E_1 \cup E_2 \cup E_3 = \{1,2\} \cup \{3,4\} \cup \{5,6\} = \{1,2,3,4,5,6\} = S$. Nothing is left out, so the exhaustive condition holds.

Step 3 — check nonzero probability. $P(E_1) = P(E_2) = P(E_3) = \dfrac{2}{6} = \dfrac{1}{3} > 0$ for each event, so the third condition holds. All three conditions are satisfied, so $\{E_1, E_2, E_3\}$ is indeed a partition of $S$.

Step 4 — decompose $A$ using the partition. With $A = \{2,3,5,6\}$: $A \cap E_1 = \{2\}$, $A \cap E_2 = \{3\}$, $A \cap E_3 = \{5,6\}$. Each piece has probability $\dfrac{1}{6}$, $\dfrac{1}{6}$, and $\dfrac{2}{6}$ respectively, and summing gives $P(A) = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{4}{6} = \dfrac{2}{3}$, which matches direct counting: $A$ has 4 outcomes out of 6. The partition decomposition reproduces $P(A) = \dfrac{2}{3}$ exactly.

Competitive stretch

A box contains bulbs from three factories, A, B and C. Let $E_1$ = "bulb is from Factory A," $E_2$ = "bulb is from Factory B," and $E_3$ = "bulb is defective, regardless of factory." Does $\{E_1, E_2, E_3\}$ form a partition of "which bulb is drawn," given every bulb comes from exactly one of the three factories?

Step 1. $E_1$ and $E_2$ alone cover only Factory A and Factory B bulbs, leaving Factory C bulbs uncovered — so even before considering $E_3$, exhaustiveness already fails for $\{E_1, E_2\}$.

Step 2. Does $E_3$ rescue this? No: $E_3$ ("defective") is not the same as "from Factory C." A defective bulb could come from any factory, so $E_3$ overlaps $E_1$ and $E_2$ (a defective Factory-A bulb lies in both $E_1$ and $E_3$), which breaks disjointness rather than fixing exhaustiveness.

Step 3. Conclusion: $\{E_1, E_2, E_3\}$ is not a valid partition — it fails disjointness (mixed criteria) and, absent a genuine "Factory C" event, exhaustiveness too. The fix is to replace $E_3$ with $E_3' =$ "bulb is from Factory C," giving the valid partition $\{E_1, E_2, E_3'\}$, all defined along the single criterion of origin.

• Assuming any three "categories" automatically partition $S$ — always check disjointness and exhaustiveness explicitly, never just count the listed events.
• Forgetting the nonzero-probability condition. An event with $P(E_i) = 0$ is not a valid partition member, since the total probability theorem and Bayes' theorem later divide by such quantities.
• Mixing criteria, such as partitioning partly by origin and partly by an unrelated attribute like defectiveness — this is the exact failure in the competitive-stretch example above, and it almost always breaks disjointness.
• Thinking $\{E, E'\}$ is the only kind of partition. Any number of pairwise disjoint, exhaustive, positive-probability events can form a partition; the complement pair is simply the smallest, two-piece case.

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