Example 2.10:
Suppose that we accept that when drawing a single card from a deck of cards that the
probability that we get an Ace is
(
)
4 / 52
P Ace
. Then the probability of not getting an Ace is
(
)
48 / 52
P Not an Ace
.
Also, based on the definition of a probability function, we know that if sets A and B are disjoint, then
(
)
( )
( )
P A
B
P A
P B
.
Example 2.11:
Suppose that we accept that when drawing a single card from a deck of cards that the
probability that we get an Ace is
(
)
4 / 52
P Ace
and the same is true for a King
(
)
4 / 52
P King
. Then
the probability of getting an Ace or a King is
(
)
4 / 52
4 / 52
8 / 52
P Ace
King
.
Is it always true that
(
)
( )
( )
P A
B
P A
P B
? That is not part of our definition of a probability function,
but maybe it is always true and we should say so in a theorem. Let’s consider some examples.
Example 2.12:
We will be drawing a card from a deck of cards. Let A be the set of Red cards and let B be
the set of Diamonds. We will assign
( )
.5
P A
and
( )
.25
P B
. Since B is a subset of A, it should be clear
that
(
)
( )
P A
B
P A
. We already know that
( )
.5
P A
. So if
(
)
( )
( )
P A
B
P A
P B
were always true, we
would have
(
)
.5
.25
.75
.5
P A
B
.
Learning to do this sort of investigative thinking will serve you well in any mathematics type course.
Example 2.13:
Let the sets A and B both be
(the sample space).
(
)
(
)
(
)
1
1
2
P
P
P
.
This contradicts our theorem that states all probabilities are less than or equal to 1.
Clearly, we need to work on
(
)
P A
B
when A and B are not disjoint.
We now define the set A-B to be those outcomes in A that are not in B and
similarly for the set B-A.
We can now write the union of two sets as three
disjoint sets as follows:
(
)
(
)
(
)
A
B
A
B
A
B
B
A
. This gives

23
(
)
(
)
(
)
(
)
P A
B
P A
B
P A
B
P B
A
(
)
(
)
(
)
(
)
(
)
P A
B
P A
B
P B
A
P A
B
P A
B
(Adding 0 is a beautiful thing)
( )
( )
(
)
P A
P B
P A
B
(Since
(
)
(
)
A
A
B
A
B
and the last 2 sets are disjoint.)
Now we have our general rule for the probability of the union of two events.
Theorem 2.3:
(
)
( )
( )
(
)
P A
B
P A
P B
P B
A
(We just did the proof above)
Example 2.14:
If
( )
.4
P A
,
( )
.5
P B
and
(
)
.1
P A
B
, determine
(
)
P A
B
.
(
)
.4
.5
.1
P A
B
Exercise 2.5:
If
( )
.25
P A
,
( )
.37
P B
and
(
)
.12
P A
B
, determine
(
)
P A
B
.
At this point in time, some of you may be considering a new rule for intersections based on things from
the past that are partially remembered:
(
)
( )
( )
P A
B
P A
P B
Is this statement true? Do some investigative thinking!
Theorem 2.4:
If the events A and B are disjoint, then
(
)
0
P A
B
Proof:
(
)
( )
( )
(
)
P A
B
P A
P B
P A
B
by the theorem above.
Also by Rule 3, we have
(
)
( )
( )
P A
B
P A
P B
. We see that
(
)
P A
B
is equal to two different quantities
and we will set those two quantities equal to each other.
( )
( )
(
)
( )
( )
(
)
0
P A
P B
P A
B
P A
P B
P A
B
Theorem 2.5:
If
A
B
, then
(
)
( )
P A
B
P A
Proof:
Since
A
B
, the set
A
B
B
. Using our rule for union:
(
)
( )
( )
(
)
P A
B
P A
P B
P A
B
.