Sample Spaces and Events

Example A constructed routing record notes a parcel’s zone, North, East, South or West, followed by its size, small, medium or large. Each outcome is an ordered pair such as $(N, s)$, zone first and size second. Every combination is possible, so the sample space $\Omega$ consists of the 12 pairs in the grid below, and $\lvert \Omega \rvert = 4 \times 3 = 12$.

Let $A$ be the event “North or East” and $B$ the event “large”. Then $A$ consists of the 6 outcomes in the North and East rows, and $B$ of the 4 outcomes in the large column. The outcome $(E, m)$ is an element of $A$, while $\{(E, m)\}$ is the event that contains only that outcome.

The level of detail is a choice. A roll of a red die and a blue die can be recorded as an ordered pair, which gives 36 outcomes, or as the sum, which gives the 11 outcomes 2 to 12. The sum record discards information. The pairs $(1, 2)$ and $(2, 1)$ both give the sum 3, so a recorded sum of 3 does not say which face the red die showed. A list of the 11 sums also says nothing about whether they are equally likely. In the usual model for two fair dice, where all 36 ordered pairs are equally likely, the sum 2 arises only from the pair $(1, 1)$ and the sum 7 from six pairs, so those two sums are not equally likely.

Example In the routing record both conditions hold only for North-large and East-large, so $A \cap B = \{(N, l), (E, l)\}$. The union starts with the 6 outcomes of $A$ and adds the two large outcomes that are not already there, $(S, l)$ and $(W, l)$. This gives 8 outcomes, in agreement with $6 + 4 - 2 = 8$. The count does not make the probability of $A \cup B$ equal to $8/12$. That value would follow from the counts in a model that makes the 12 outcomes equally likely, and nothing in the list of outcomes says they are.

Example In the routing record the complement $A^{c}$ is “South or West”, the 6 outcomes with zone $S$ or $W$. If the record allowed a fifth zone, the same condition “neither North nor East” would pick out a larger set. The two differences are

Their union is the event “exactly one of $A$ and $B$”, called the symmetric difference $A \mathbin{\triangle} B$, and it has $4 + 2 = 6$ outcomes. Ordinary “or” includes the overlap, while “exactly one” leaves it out.

Example In the routing record, “neither $A$ nor $B$” means that the zone is neither North nor East and the size is not large. That leaves $(S, s)$, $(S, m)$, $(W, s)$ and $(W, m)$. Removing the 8 outcomes of $A \cup B$ from all 12 gives the same 4 outcomes, as the first law says. The event “not both” removes only the 2 outcomes of $A \cap B$, so it is the set $A^{c} \cup B^{c}$ with 10 outcomes. Four outcomes against ten shows how much the exact wording matters.

Example In the routing record the events “small” and “large” are disjoint, because every parcel has exactly one size. The events $A$ and $B$ are not disjoint, since $(N, l)$ belongs to both. The events “small or medium” and “medium or large” are exhaustive, since every size is covered, but they are not disjoint, since they share all four medium outcomes.

Example The size events “small”, “medium” and “large” partition the routing sample space into three sets of 4 outcomes, and $4 + 4 + 4 = 12$. Any two events also split $\Omega$ into four pairwise disjoint pieces, $A$ only, $B$ only, both and neither. For the events $A$ and $B$ above these pieces have 4, 2, 2 and 4 outcomes, which add to 12, and since none of them is empty they form a partition. By contrast, “North or East” and “North or South” have $6 + 6 = 12$ outcomes between them, but they are not a partition, because both contain the three North outcomes and the West outcomes are in neither.