Product (Multiplication) Rule for Independent Events, Addition Rule for Mutually Exclusive Events

**PROBABILITY RULES**

If *A* and *B* are independent events, their joint
probability of occurrence is given by the Formula 5.1:

*P*(*A **∩** B*) =* P*(*A*) ×*
P*(*B*) (5.1)

For a clear application of this rule, consider the
experiment in which we roll two dice. What is the probability of a 1 on the
first roll and a 2 or 4 on the second roll?

First of all, the outcome on the first roll is
independent of the outcome on the second roll; therefore, define *A* = {get a 1 on one die and any outcome
on the sec-ond die}, and let *B* = {any
outcome on one die and a 2 or a 4 on the second die}. We can describe *A* as the following set of elementary
events: *A* = [{1, 1}, {1, 2}, {1, 3},
{1, 4}, {1, 5}, {1, 6}] and *B* = [{1,
2}, {1, 4}, {2, 2}, {2, 4}, {3, 2}, {3, 4}, {4, 2}, {4, 4}, {5, 2}, {5, 4}, {6,
2}, {6, 4}].

The event *C*
= *A **∩** B* = [{1, 2}, {1, 4}]. By the law of multiplication for inde-pendent
events, *P*(*C*) = *P*(*A*) × *P*(*B*) = (1/6) × (1/3) = 1/18. You can check
this by considering the elementary events associated with C. Since there are
two events, each with probability 1/36, *P*(*C*) = 2/36 = 1/18.

If *A* and *B* are mutually exclusive events, then
the probability of their union (i.e., the probability that at least one of the
events, *A* or *B*, occurs) is given by Formula 5.2. Mutually exclusive events are
also called disjoint events. In terms of symbols, event *A *and event* B *are
disjoint if* A **∩** B *=* *Ø.

*P*(*AU B*) =* P*(*A*)
+* P*(*B*) (5.2)

Again, consider the example of rolling the dice; we
roll two dice once. Let *A* be the
event that both dice show the same number, which is even, and let *B* be the event that both dice show the
same number, which is odd. Let *C* = *A U B*. Then *C* is the event in which the roll of the dice produces the same
number, either even or odd.

For the two dice together, *C* occurs in six elementary ways: {1, 1}, {2, 2}, {3, 3}, {4, 4},
{5, 5}, and {6, 6}. A occurs in three elementary ways, namely, {2, 2}, {4, 4},
and {6, 6}. *B* also occurs in three
elementary ways, namely, {1, 1}, {3, 3}, and {5, 5}.

*P*(*C*) =
6/36 = 1/6, whereas* P*(*A*) = 3/36 = 1/12 and* P*(*B*) = 3/36 = 1/12. By
the* *addition law for mutually
exclusive events, *P*(*C*) = *P*(*A*) + *P*(*B*) = (1/12) + (1/12) = 2/12 = 1/6. Thus,
we see that the addition rule applies.

An application of the rule of addition is the rule
for complements, shown in For-mula 5.3. Since *A* and *A ^{c}* are
mutually exclusive and complementary, we have Ω =

*P*(*A ^{c}*)
= 1 –

In general, the addition rule can be modified for
events *A* and *B* that are not disjoint. Let *A*
and *B* be the sets identified in the
Venn diagram in Figure 5.3. Call the overlap area *C* = *A **∩ **B*. Then, we can divide the set *A U B* into three mutually exclusive sets as labeled in the diagram,
namely, *A **∩ **B ^{c}*,

When we compute *P*(*A*) + *P*(*B*), we obtain *P*(*A*) = *P*(*A
**∩ **B*) + *P*(*A
**∩ **B ^{c}*) and

The problem with the summation formula is that *P*(*C*)
is counted twice. We remedy this error by subtracting *P*(*C*) once. This
subtraction yields the generalized ad-dition formula for union of arbitrary
events, shown as Formula 5.4:

*P*(*A U B*) =* P*(*A*)
+* P*(*B*) –* P*(*A **∩ **B*) (5.3)

Note that Formula 5.4 applies to mutually exclusive
events *A* and *B* as well, since for mutually exclusive events, *P*(*A
**∩** B*) = 0.
Next we will generalize the mul-tiplication rule, but first we need to define
conditional probabilities.

Suppose we have two events, *A* and *B*, and we want to
define the probability of *A* occurring
given that *B* will occur. We call this
outcome the conditional probability of *A*
given *B* and denote it by *P*(*A*|*B*). Definition 5.4.1 presents the formal
mathe-matical definition of *P*(*A*|*B*).

*Definition 5.4.1: Conditonal
Probability of ***A Given B. **Let

*Figure 5.3. **Decomposition of AU B into disjoint sets.*

Consider rolling one die. Let *A* = {a 2 occurs} and let *B*
= {an even number occurs}. Then *A* =
{2} and *B* = [{2}, {4}, {6}]. *P*(*A
B*) = 1/6 because *A **∩ **B* = *A* =
{2} and there is 1 chance out of 6 for 2 to come up. *P*(*B*) = 1/2 since there
are 3 ways out of 6 for an even number to occur. So by definition, *P*(*A*|*B*) = *P*(*A **∩ **B*)/*P*(*B*) = (1/6)/(1/2) = 2/6 = 1/3.

Another way to understand
conditional probabilities is to consider the restricted outcomes given that *B* occurs. If we know that *B* occurs, then the outcomes {2}, {4},
and {6} are the only possible ones and they are equally likely to occur. So
each outcome has the probability 1/3; hence, the probability of a 2 is 1/3.
That is just what we mean by *P*(*A*|*B*).
Directly from Definition 5.4.1, we derive Formula 5.5 for the general law of
conditional probabilities:

*P*(*A*|*B*) =*
P*(*A **∩ **B*)/*P*(*B*) (5.5)

Multiplying both sides of the
equation by *P*(*B*), we have *P*(*A*|*B*)
*P*(*B*)
= *P*(*A **∩** B*). This equation, shown as Formula 5.6, is the generalized
multiplication law for the in-tersection of arbitrary events:

*P*(*A **∩ **B*) =* P*(*A*|*B*)*P*(*B*)
(5.6)

The generalized multiplication formula holds for
arbitrary events *A* and *B*. Conse-quently, it also holds for
independent events.

Suppose now that *A* and *B* are independent;
then, from Formula 5.1, *P*(*A **∩ **B*) = *P*(*A*)* P*(*B*). On the other hand, from Formula 5.6,* P*(*A **∩ **B*) =* P*(*A*|*B*)* P*(*B*).
So* P*(*A*|*B*)* P*(*B*) =* P*(*A*)* P*(*B*).

Dividing both sides of *P*(*A*|*B*) *P*(*B*) = *P*(*A*) *P*(*B*) by *P*(*B*) (since *P*(*B*)
> 0), we have *P*(*A*|*B*)
= *P*(*A*). That is, if *A* and *B* are independent, then the probability
of *A* given *B* is the same as the unconditional probability of *A*.

This result agrees with our intuitive notion of
independence, namely, condition-ing on *B*
does affect the chances of *A*’s
occurrence. Similarly, one can show that if *A
*and* B *are independent, then* P*(*B*|*A*) =*
P*(*B*).

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