Set Theory

When we perform an experiment, there are many results that we might see. We want to be able to quantify the likelihood of seeing certain results. For this, we need to develop some mathematical terminology.

The sample space, often denoted \Omega, is the set of all possible results of an experiment. A single result is called an outcome, while a collection of results is called an event.

An experiment consists of rolling a standard 6-sided die. The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. One possible event is A = \{2, 4, 6\}, i.e., the event that the result of the roll is even.

What would the sample space look like if we roll the die two times and recorded the results?

\Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} Note that, for example, (1, 2) is a different outcome from (2, 1).

Let A be a set. The symbol \in means "is an element of", as in a \in A. Given another set B, we say A is a subset of B, written A\subset B, to mean that every element of the set A is also an element of the set B.

The subset symbol includes the possibility that A and B are equal sets, i.e., that they contain precisely the same elements.

Consider sets A and B, each contained inside \Omega. We can combine sets in a variety of ways:

  • Union

    The union of A and B is the set A \cup B = \{x \mid x \in A \text{ or } x \in B\}.

  • Intersection

    The intersection of A and B is the set A \cap B = \{x \mid x \in A \text{ and } x \in B\}.

  • Difference

    The set difference A-B is the set A - B = \{x \mid x \in A \text{ and } x \notin B\}.

  • Complement

    The complement of A is the set A^c = \{x \in \Omega \mid x \notin A\}.

  • Empty Set

    The empty set, usually written \emptyset or \{\}, is the set which contains no elements.

    • It's useful sometimes to draw pictures called Venn diagrams representing sets:

    Example Venn Diagram

    Venn diagram showing sets A, B, C with the region representing (A\cup B\cup C) - (A \cap C) shaded.

    \begin{tikzpicture} \def\firstcircle{(90:1.75cm) circle (2.5cm)} \def\secondcircle{(210:1.75cm) circle (2.5cm)} \def\thirdcircle{(330:1.75cm) circle (2.5cm)} \fill[gray!30] \firstcircle; \fill[gray!30] \secondcircle; \fill[gray!30] \thirdcircle; \begin{scope} \clip \firstcircle; \clip \thirdcircle; \fill[white] \firstcircle; \end{scope} \draw \firstcircle node[text=black,above] {$A$}; \draw \secondcircle node [text=black,below left] {$B$}; \draw \thirdcircle node [text=black,below right] {$C$}; \node at (0, -4.5) {$(A\cup B\cup C) - (A \cap C)$}; \end{tikzpicture}

    Two sets A and B are disjoint if A \cap B = \emptyset.

    Given a finite set A, the cardinality of A, written |A|, is the number of elements in A.

    Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, B = \{2, 4, 9, 10, 12, 14, 19\}, and C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of \Omega = \{1, 2, 3, \dotsc, 20\}.

    Find A - (B \cap C).

    \{1, 2, 3, 4, 5, 6, 7, 8\}

    Find |A|, |B|, |C|, |A\cup B|, |A \cap B|, |B\cap C|, |A\cap C|, and |A\cup B\cup C|. Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets?

    |A| = 10, |B| = 7, |C| = 7, |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, |A\cap C| = 2, |A\cup B\cup C| = 17. In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.

    Find A^c and (A\cup B)^c.

    A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}.

    Suppose we have a 6-sided die that's weighted to roll a 6 half of the time. We roll the die two times. List the set of all possible results. [Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result (4, 2)---rolling a 4 and then a 2.]

    \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} Note that \Omega simply lists outcomes with no reference to the probabilities. So the answer here is the same as in .

    Suppose we flip a coin two times. List the set of all possible results. What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be?

    For two flips: \Omega = \{ HH, HT, TH, TT \}.

    For three flips: \Omega = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}.

    For four flips: \Omega = \{ \amp HHHH, HHHT, HHTH, HTHH, \amp THHH, HHTT, HTHT, HTTH, \amp THHT, THTH, TTHH, HTTT, \amp THTT, TTHT, TTTH, TTTT \}.

    Each additional flip doubles the number of outcomes. So, with ten flips, we'll have |\Omega| = 2^{10} = 1024.

    If we roll a 6-sided die ten times, how many possible results will there be?

    Each additional roll will multiply the number of outcomes by 6. So, with 10 rolls, we'll have |\Omega| = 6^{10}.