1.5 Probability Rules
=====================

:bolditalic:`Key Terms`

:bolditalic:`Overlap（重疊）`
   Outcomes that belong to both events.

:bolditalic:`Exhaustive events（窮盡事件）`
   Events that together cover the whole sample space.

:bolditalic:`At least one（至少一個）`
   One or more occurrences.

:bolditalic:`Neither（兩者皆非）`
   Not in either of two events.

:bolditalic:`De Morgan's laws（德摩根定律）`
   Rules for rewriting complements of unions and intersections.


1.5.2 Learning Outcomes
-----------------------

After completing this section, students should be able to:

- use the complement rule;

- use a complement to solve an ``at least one`` problem;

- use the addition rule for two events;

- explain why an overlap must be subtracted;

- recognize mutually exclusive events and use the simplified addition rule;

- split one event into two non-overlapping parts;

- use complements of combined events;

- apply De Morgan's laws;

- add probabilities for several mutually exclusive events;

- explain what it means for events to be exhaustive;

- use the addition rule for three overlapping events;

- translate words such as ``not``, ``or``, ``at least one``, and ``neither`` into event notation.

1.5.3 From Event Language to Probability Rules
----------------------------------------------

In Section 1.2, we used complements, unions, and intersections to describe

events.

For example:

- NOT :math:`A` means :math:`A^c`;

- :math:`A` OR :math:`B` means :math:`A\cup B`;

- :math:`A` AND :math:`B` means :math:`A\cap B`.

In this section, we use the same event notation to calculate probabilities.

The main process is:

    Words -> Event Notation -> Probability Rule

Before calculating, first decide what event the words describe.

The geometry of the event should be clear before a probability rule is
applied. The following diagram illustrates several increasingly complex set
operations.

.. image:: ../_static/figures/probability/figure_1_5_1_diagram_set_operations.png
   :width: 92%
   :align: center
   :alt: Four Venn diagrams showing two events A and B, their intersection, a three-event expression, and a complemented intersection.

Read the four panels in terms of the region selected from the sample space:

- **Panel (a)** shows the basic setting. Events :math:`A` and :math:`B` are
  subsets of the sample space :math:`S` and may overlap.
- **Panel (b)** isolates :math:`A\cap B`. Only outcomes satisfying both
  events are selected. This overlap is the region used by ``A AND B``.
- **Panel (c)** shows :math:`(A\cup B)\cap C`. First form
  :math:`A\cup B`; then keep only the part that also belongs to :math:`C`.
  This illustrates why parentheses matter: a compound event can be built in
  stages.
- **Panel (d)** shows :math:`(A\cap C)^c`. First locate
  :math:`A\cap C`; then take its complement. The event therefore contains
  every point in :math:`S` except the overlap of :math:`A` and :math:`C`.

These shaded regions are the events whose probabilities we calculate. For
example, :math:`P(A\cap B)` assigns probability to the overlap in panel (b),
while :math:`P((A\cap C)^c)` assigns probability to the shaded complement in
panel (d). The probability rules in this section provide efficient ways to
calculate such regions without listing every sample point.

1.5.4 Complement Rule
---------------------

An event :math:`A` and its complement :math:`A^c` cannot occur together.

Also, one of them must occur.

Therefore,

.. math::

   A\cup A^c=S

and

.. math::

   P(A)+P(A^c)=1.

This gives the complement rule.

.. note::

   **Complement Rule**

   .. math::

      P(A^c)=1-P(A).

   Equivalently,

   .. math::

      P(A)+P(A^c)=1.

The complement rule is useful when the opposite event is easier to

calculate.

Common words that may suggest a complement include:

- ``not``;

- ``none``;

- ``at least one``;

- ``neither``.

1.5.5 Example 1: Roll One Fair Die
----------------------------------

**Question:** One fair die is rolled. Let :math:`A` be the event that the

number is greater than 4. Find the probability that :math:`A` does not

occur.

The sample space is

.. math::

   S=\{1,2,3,4,5,6\}.

The event is

.. math::

   A=\{5,6\}.

Therefore,

.. math::

   P(A)=\frac{2}{6}=\frac{1}{3}.

The event that :math:`A` does not occur is the complement :math:`A^c`.

Using the complement rule,

.. math::

   P(A^c)=1-P(A)=1-\frac{1}{3}=\frac{2}{3}.

We can also check this directly.

The complement is

.. math::

   A^c=\{1,2,3,4\},

so

.. math::

   P(A^c)=\frac{4}{6}=\frac{2}{3}.

Both methods give the same result.

1.5.6 Example 2: At Least One Head
----------------------------------

**Question:** A fair coin is tossed three times. Find the probability of

getting at least one head.

The phrase **at least one head** means one, two, or three heads.

We could list all outcomes that contain a head, but it is easier to use the

opposite event.

The opposite of ``at least one head`` is

    no heads.

For three coin tosses, no heads occurs only when

.. math::

   TTT.

There are

.. math::

   2^3=8

equally likely outcomes in total.

Therefore,

.. math::

   P(\text{no heads})=\frac{1}{8}.

Using the complement rule,

.. math::

   P(\text{at least one head})
   =
   1-P(\text{no heads})
   =
   1-\frac{1}{8}
   =
   \frac{7}{8}.

.. note::

   For an ``at least one`` problem, it is often easier to calculate

   ``none`` first and subtract from 1.

This is a useful strategy, but it is still the same complement rule.

1.5.7 Addition Rule
-------------------

Suppose we want the probability that event :math:`A` **or** event

:math:`B` occurs.

In probability, ``A or B`` normally includes three possibilities:

- :math:`A` occurs;

- :math:`B` occurs;

- both :math:`A` and :math:`B` occur.

This is the union

.. math::

   A\cup B.

If we simply add :math:`P(A)` and :math:`P(B)`, any outcome in both events

is counted twice.

The outcomes in both events form the intersection

.. math::

   A\cap B.

Therefore, we subtract the overlap once.

.. note::

   **Addition Rule**

   .. math::

      P(A\cup B)

      =

      P(A)+P(B)-P(A\cap B).

A simple way to remember the rule is:

    add both probabilities -> subtract the overlap once

The shaded lens in panel (b) of the set-operations diagram is exactly this
overlap. When :math:`P(A)` and :math:`P(B)` are added, that region is counted
once through :math:`A` and once again through :math:`B`. Subtracting
:math:`P(A\cap B)` once leaves each outcome in :math:`A\cup B` counted
exactly once.

1.5.8 Example 3: Roll One Fair Die
----------------------------------

**Question:** One fair die is rolled. Let :math:`A` be the event of rolling

an even number and :math:`B` the event of rolling a number greater than 3.

Find the probability that :math:`A` or :math:`B` occurs.

The sample space is

.. math::

   S=\{1,2,3,4,5,6\}.

The events are

.. math::

   A=\{2,4,6\}

and

.. math::

   B=\{4,5,6\}.

Therefore,

.. math::

   P(A)=\frac{3}{6}

and

.. math::

   P(B)=\frac{3}{6}.

The two events overlap at 4 and 6:

.. math::

   A\cap B=\{4,6\}.

Thus,

.. math::

   P(A\cap B)=\frac{2}{6}.

If we calculate only

.. math::

   \frac{3}{6}+\frac{3}{6},

the outcomes 4 and 6 are counted twice.

Using the addition rule,

.. math::

   P(A\cup B)
   =
   \frac{3}{6}
   +
   \frac{3}{6}
   -
   \frac{2}{6}
   =
   \frac{4}{6}
   =
   \frac{2}{3}.

We can check the answer directly because

.. math::

   A\cup B=\{2,4,5,6\},

which contains 4 of the 6 equally likely outcomes.

1.5.9 Mutually Exclusive Events
--------------------------------

Two events are **mutually exclusive** when they cannot occur together.

This means

.. math::

   A\cap B=\varnothing.

Therefore,

.. math::

   P(A\cap B)=0.

The general addition rule then becomes simpler.

.. note::

   **Mutually Exclusive Events**

   If

   .. math::

      A\cap B=\varnothing,

   then

   .. math::

      P(A\cup B)=P(A)+P(B).

The important point is to check whether the events can occur together

before using the simplified rule.

1.5.10 Example 4: Sum of Two Dice
---------------------------------

**Question:** Two fair dice are rolled. Let :math:`A` be the event that the

sum is 7 and :math:`B` the event that the sum is 11. Find

:math:`P(A\cup B)`.

Each die has 6 possible outcomes, so there are

.. math::

   6\times6=36

equally likely ordered outcomes.

A sum of 7 occurs for

.. math::

   (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).

Therefore,

.. math::

   P(A)=\frac{6}{36}=\frac{1}{6}.

A sum of 11 occurs for

.. math::

   (5,6),(6,5).

Therefore,

.. math::

   P(B)=\frac{2}{36}=\frac{1}{18}.

One pair of dice cannot have a sum of both 7 and 11 at the same time.

Thus,

.. math::

   A\cap B=\varnothing.

The events are mutually exclusive.

Therefore,

.. math::

   P(A\cup B)
   =
   P(A)+P(B)
   =
   \frac{1}{6}
   +
   \frac{1}{18}
   =
   \frac{2}{9}.

1.5.11 Splitting an Event into Two Parts
----------------------------------------

Sometimes an event can be separated into two smaller parts.

Suppose :math:`A` is an event and :math:`B` is another event.

Every outcome in :math:`A` must either:

- also belong to :math:`B`; or

- not belong to :math:`B`.

Therefore,

.. math::

   A=(A\cap B)\cup(A\cap B^c).

The two parts cannot overlap because an outcome cannot be in both

:math:`B` and :math:`B^c`.

Thus,

.. math::

   (A\cap B)\cap(A\cap B^c)=\varnothing.

Because the two parts are mutually exclusive, their probabilities can be

added.

.. note::

   **Splitting an Event**

   .. math::

      A=(A\cap B)\cup(A\cap B^c).

   Therefore,

   .. math::

      P(A)
      =
      P(A\cap B)+P(A\cap B^c).

This is useful when it is easier to understand or calculate the two parts

separately.

1.5.12 Example 5: Roll One Fair Die
-----------------------------------

**Question:** One fair die is rolled. Let :math:`A` be the event that the

number is greater than 3, and let :math:`B` be the event that the number is

even. Find :math:`P(A)` by separating :math:`A` according to whether

:math:`B` occurs.

The event :math:`A` is

.. math::

   A=\{4,5,6\}.

The outcomes in :math:`A` that are also in :math:`B` are

.. math::

   A\cap B=\{4,6\}.

The outcomes in :math:`A` that are not in :math:`B` are

.. math::

   A\cap B^c=\{5\}.

These two parts do not overlap, and together they form :math:`A`.

Therefore,

.. math::

   P(A)
   =
   P(A\cap B)+P(A\cap B^c).

Because the die is fair,

.. math::

   P(A)
   =
   \frac{2}{6}
   +
   \frac{1}{6}
   =
   \frac{3}{6}
   =
   \frac{1}{2}.

This agrees with the direct calculation because :math:`A` contains three of

the six equally likely outcomes.

1.5.13 Complements of Combined Events
-------------------------------------

Sometimes the complement applies to a union or an intersection.

De Morgan's laws tell us how to rewrite these events.

.. note::

   **De Morgan's Laws**

   NOT (:math:`A` OR :math:`B`):

   .. math::

      (A\cup B)^c=A^c\cap B^c.

   NOT (:math:`A` AND :math:`B`):

   .. math::

      (A\cap B)^c=A^c\cup B^c.

These formulas are easier to understand when translated into words.

``Neither A nor B`` means that :math:`A` does not occur and :math:`B` does

not occur:

.. math::

   A^c\cap B^c=(A\cup B)^c.

``Not both A and B`` means that the intersection does not occur:

.. math::

   (A\cap B)^c=A^c\cup B^c.

The words ``neither`` and ``not both`` are different.

- ``Neither`` means neither event occurs.
- ``Not both`` allows one event to occur, but not both together.

The first De Morgan law can be verified directly from the shaded regions.

.. image:: ../_static/figures/probability/figure_1_5_2_demorgan_law.png
   :width: 70%
   :align: center
   :alt: Venn diagrams showing that the complement of A union B is the same region as A complement intersection B complement.

Start with the left side,

.. math::

   (A\cup B)^c.

The union :math:`A\cup B` contains every point inside either circle. Taking
its complement removes both circles and leaves only the part of the sample
space outside both events.

Now consider the right side,

.. math::

   A^c\cap B^c.

The event :math:`A^c` contains everything outside :math:`A`, and
:math:`B^c` contains everything outside :math:`B`. Their intersection keeps
only outcomes satisfying both conditions: outside :math:`A` **and** outside
:math:`B`. This is exactly the same region obtained from
:math:`(A\cup B)^c`.

Therefore,

.. math::

   (A\cup B)^c=A^c\cap B^c.

The diagram explains why ``OR`` changes to ``AND`` when the complement is
moved inside. To be outside the union, an outcome must be outside both
individual events.

The second De Morgan law follows the same logic:

.. math::

   (A\cap B)^c=A^c\cup B^c.

Here, ``not both`` means an outcome fails at least one condition. It may be
outside :math:`A`, outside :math:`B`, or outside both, so the complements
are joined by a union.

Thus the ordinary-language translations are:

- ``neither A nor B`` :math:`\rightarrow (A\cup B)^c=A^c\cap B^c`;
- ``not both A and B`` :math:`\rightarrow (A\cap B)^c=A^c\cup B^c`.

1.5.14 Example 6: Neither a Heart nor a King
--------------------------------------------

**Question:** One card is drawn from a standard 52-card deck. Find the

probability that the card is neither a heart nor a king.

Let :math:`A` be the event of drawing a heart.

Let :math:`B` be the event of drawing a king.

There are 13 hearts and 4 kings.

However, the king of hearts belongs to both events.

Therefore,

.. math::

   P(A\cup B)
   =
   \frac{13}{52}
   +
   \frac{4}{52}
   -
   \frac{1}{52}
   =
   \frac{16}{52}
   =
   \frac{4}{13}.

The event ``neither a heart nor a king`` is the complement of

:math:`A\cup B`.

Using the complement rule,

.. math::

   P((A\cup B)^c) = 1-P(A\cup B) = 1-\frac{4}{13} = \frac{9}{13}.

By De Morgan's law,

.. math::

   (A\cup B)^c=A^c\cap B^c.

Therefore, the event can also be described as

    not a heart AND not a king.

1.5.15 Several Mutually Exclusive Events
----------------------------------------

The addition idea also works for more than two mutually exclusive events.

Suppose

.. math::

   A_1,A_2,\ldots,A_k

are mutually exclusive.

No two of these events can occur together.

Therefore,

.. math::

   P(A_1\cup A_2\cup\cdots\cup A_k)
   =
   P(A_1)+P(A_2)+\cdots+P(A_k).

There are no overlap terms to subtract because all intersections are empty.

1.5.16 Exhaustive Events
------------------------

A group of events is **exhaustive** when the events together cover all

possible outcomes.

If

.. math::

   A_1\cup A_2\cup\cdots\cup A_k=S,

then the events are exhaustive.

If the events are both mutually exclusive and exhaustive, exactly one of

them must occur.

Because their union is the whole sample space,

.. math::

   P(A_1)+P(A_2)+\cdots+P(A_k)=P(S)=1.

For example, on one die roll, the six events

.. math::

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

are mutually exclusive and exhaustive.

Only one can occur, and together they contain every possible die result.

1.5.17 Addition Rule for Three Events
-------------------------------------

For three events :math:`A`, :math:`B`, and :math:`C`, the events may

overlap in several places.

If we first add

.. math::

   P(A)+P(B)+P(C),

the pairwise overlaps are counted more than once.

So we subtract

.. math::

   P(A\cap B),\qquad
   P(A\cap C),\qquad
   P(B\cap C).

However, an outcome in all three events was originally added three times and

then subtracted three times.

It has now been counted zero times.

Therefore, we must add the triple intersection back once.

.. note::

   **Three-Event Addition Rule**

   .. math::

      \begin{aligned}

      P(A\cup B\cup C)
      ={}&
      P(A)+P(B)+P(C)\\
      &-P(A\cap B)-P(A\cap C)-P(B\cap C)\\
      &+P(A\cap B\cap C).

      \end{aligned}

A useful way to remember the pattern is:

    add the three events

    subtract the pairwise overlaps

    add the triple overlap back once

Panel (c) of the set-operations diagram helps explain why three-event
problems require more bookkeeping. Three events can create several pairwise
overlap regions and, potentially, a central region belonging to all three.
The three-event addition rule corrects the counting of these regions
systematically.

If the three events are mutually exclusive, all intersection probabilities

are 0, so the formula reduces to adding the three probabilities.

1.5.18 Choosing a Probability Rule
----------------------------------

Probability problems often become easier after translating the words into

event notation.

Use the following guide.

.. list-table::
   :header-rows: 1
   :widths: 35 30 35

   * - Words
     - Event notation
     - Main idea
   * - NOT :math:`A`
     - :math:`A^c`
     - Use :math:`1-P(A)`
   * - :math:`A` OR :math:`B`
     - :math:`A\cup B`
     - Add and subtract the overlap
   * - :math:`A` or :math:`B`, but they cannot occur together
     - :math:`A\cup B`
     - Mutually exclusive; add directly
   * - At least one
     - Complement of none
     - Often calculate none first
   * - Neither :math:`A` nor :math:`B`
     - :math:`(A\cup B)^c`
     - Find the union, then use the complement
   * - Not both :math:`A` and :math:`B`
     - :math:`(A\cap B)^c`
     - Complement of the intersection

The general decision process is:

    Words -> Event Notation -> Probability Rule

For example:

- ``not A`` -> :math:`A^c` -> complement rule;

- ``A or B`` -> :math:`A\cup B` -> addition rule;

- ``A or B`` with no overlap -> mutually exclusive addition rule;

- ``at least one`` -> often use the complement of none;

- ``neither A nor B`` -> :math:`(A\cup B)^c`;

- ``not both A and B`` -> :math:`(A\cap B)^c`.

1.5.19 Common Mistakes
----------------------

A common mistake is to add :math:`P(A)` and :math:`P(B)` without checking

for overlap.

For overlapping events,

.. math::

   P(A\cup B)
   =
   P(A)+P(B)-P(A\cap B).

Another mistake is to subtract the overlap when the events are mutually

exclusive.

If

.. math::

   A\cap B=\varnothing,

then

.. math::

   P(A\cap B)=0,

so the probabilities can be added directly.

A third mistake is to read ``at least one`` as ``exactly one``.

For three coin tosses, at least one head includes outcomes with one, two,

or three heads.

A fourth mistake is to confuse ``neither`` with ``not both``.

``Neither A nor B`` means

.. math::

   (A\cup B)^c.

``Not both A and B`` means

.. math::

   (A\cap B)^c.

These are not the same event.

Another mistake is to split an event into parts that overlap.

When using

.. math::

   A=(A\cap B)\cup(A\cap B^c),

the two parts are useful because they are non-overlapping.

For three events, a common mistake is to subtract the three pairwise

intersections but forget to add the triple intersection back once.

1.5.20 Summary
--------------

The complement rule is

.. math::

   P(A^c)=1-P(A).

The general addition rule for two events is

.. math::

   P(A\cup B)
   =
   P(A)+P(B)-P(A\cap B).

If :math:`A` and :math:`B` are mutually exclusive,

.. math::

   A\cap B=\varnothing

and

.. math::

   P(A\cup B)=P(A)+P(B).

An event can be split into two non-overlapping parts:

.. math::

   A=(A\cap B)\cup(A\cap B^c),

so

.. math::

   P(A)=P(A\cap B)+P(A\cap B^c).

De Morgan's laws are

.. math::

   (A\cup B)^c=A^c\cap B^c

and

.. math::

   (A\cap B)^c=A^c\cup B^c.

For several mutually exclusive events,

.. math::

   P(A_1\cup\cdots\cup A_k)

   =

   P(A_1)+\cdots+P(A_k).

If those events are also exhaustive,

.. math::

   P(A_1)+\cdots+P(A_k)=1.

For three overlapping events,

.. math::

   \begin{aligned}

   P(A\cup B\cup C)

   ={}&

   P(A)+P(B)+P(C)\\

   &-P(A\cap B)-P(A\cap C)-P(B\cap C)\\

   &+P(A\cap B\cap C).

   \end{aligned}

The main strategy is:

    Words -> Event Notation -> Probability Rule

Before calculating, identify whether the problem involves a complement,

a union, an overlap, mutually exclusive events, or a combined-event

complement.

.. 1.5.21 Short Review Questions

.. -----------------------------

.. 1. What is the complement rule?

.. 2. Why is the complement useful for an ``at least one`` problem?

.. 3. State the addition rule for two events.

.. 4. Why must :math:`P(A\cap B)` be subtracted in the addition rule?

.. 5. What does it mean for two events to be mutually exclusive?

.. 6. When can :math:`P(A\cup B)=P(A)+P(B)` be used?

.. 7. How can :math:`A` be split using :math:`B` and :math:`B^c`?

.. 8. Why can the probabilities of the two parts be added?

.. 9. State De Morgan's two laws.

.. 10. What does ``neither A nor B`` mean in event notation?

.. 11. What does ``not both A and B`` mean in event notation?

.. 12. What does it mean for several events to be exhaustive?

.. 13. If mutually exclusive events are also exhaustive, what must their probabilities add to?

.. 14. State the addition rule for three events.

.. 15. Why is the triple intersection added back in the three-event addition rule?