Partitions#

If two events sets \(A\) and \(B\) are disjoint, e.g. if \(A\cap B = \emptyset\), then

\[P(A\cup B)=P(A)+P(B)\]

If the intersection of two or more event sets is non-empty, life gets complicated quickly. Disjoint event sets are extremely helpful, so we will often use partitions as tools to break down complicated events into digestible pieces.

Definition

A partition is a collection of subsets of the probability space \(S\) such that the subsets \(A_1, A_2, \dots , A_n\) have the following properties:

  1. They are non-empty

\[A_i\neq \emptyset \hspace{5mm}\forall i\]

  1. They are pairwise disjoint

\[A_i \cap A_j = \emptyset \hspace{5mm}\forall i\neq j\]

  1. Their union is \(S\)

\[A_1 \cup A_2 \cup \cdots \cup A_n = S\]

Property (2) guarantees that every possible intersection of these subsets is empty:

\[A_i\cap A_j=\emptyset \hspace{5mm}\forall\ i\neq j\in\{1,2,\dots,n\}\]

Property (3) ensures that every outcome in the probability space can be found in one of the subsets in the partition. Both are extremely helpful when simplifying probability problem-solving situations.

We have two primary partitions in the standard deck of playing cards: values and suits. Consider the four suits. Each suit is a non-empty set because it has 13 elements. A card can be only one suit, so spades and hearts are disjoint as is every pair of possible suits. Finally,.

\[S=\{\spadesuit,\heartsuit, \clubsuit,\diamondsuit\}\]

Thus, the suits are subsets that form a partition. Let us define a few additional subsets in the standard deck of playing cards:

  • Even(\(E\)): \(2,4,6,8,T,Q\) in all suits

  • Odd(\(O\)): \(A,3,5,7,9,J,K\) in all suits

  • Faces(\(F\)): \(J, Q,K\) (cards with person pictured) in all suits

  • Aces (\(A\)): An \(A\) in all suits

  • Numbered(\(N\)): \(2,3,4,5,6,7,8,9,T\)

Note two more partitions are readily available: \(E\cup O=S\) and \(F\cup A \cup N=S\).

Example 5

A card is drawn at random from a deck of playing cards. Find the probability that the card is:

  1. A heart \emph{or} even.

  2. A heart \emph{and} even.

  3. Even but not a heart.

Solution. Let \(S\) represent the probability space, and let event sets \(E\) and \(H\) represent drawing an even card or a heart respectively. The situation in part (a) is shown in Figure \ref{fig:EunionH}:

The intersection of \(E\) and \(H\) is non-empty, namely, the six even hearts, so we cannot just add. When we try to calculate:

\[P(H\text{ or }E)\overset{??}{=}P(H)+P(E)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\]

We arrive at an incorrect answer. The 6 even hearts would be counted twice, once in the first fraction and again in the second. We must add a third term to deal with the double-counting:

\[\begin{split} P(H\text{ or }E)\hspace{5mm}=\hspace{15mm}P(H\cup E)\\ =P(H)+P(E)-P(H\cap E)\\ =\frac{13}{52}+\frac{26}{52}-\frac{6}{52}\\ =\frac{33}{52} \end{split}\]

For part (b), we have:

\[\begin{split}P(H\text{ and } E)=P(H\cap E)\\ =\frac{6}{52}=\frac{3}{26}\end{split}\]

For part (c) we will need set subtraction notation: \(A-B=\) is the set of all elements in \(A\) that are not in \(B\). In terms of set operations we have:

\[A-B=A\cap \bar B\]

which is shown below.

The blue shaded area is \(\bar B\), so when we take \(A\cap \bar B\), we gather all elements of \(A\) that are not in \(B\), as required. The situation for our example is demonstrated below. Using set operations, we have:

\[P(E\text{ and not }H)=P(E - H)\]
\[=P(E\cap \bar H)\]
\[=P(E)-P(H\cap E)\]
\[=\frac{26}{52}-\frac{6}{52}\]
\[=\frac{5}{13} \]

Example 6

A card is drawn at random from a deck of playing cards. Find the probability that the card is a heart (\(\heartsuit\)) or is even or is a face card.

Solution. Clearly, counting all the non-heart, non-even, non-face would be easy,of which cards there are 15.

\[P(H\cup E \cup F)=1-P(\overline{H\cup E \cup F})=\frac{52-15}{52}=\frac{37}{52}\]

The situation in Example 3 is known as the principle of inclusion-exclusion, and the explicit formula will be developed in the next section.