1
Lecture 2
ECE 514
¾
Using the
coin tossing experiment,
and from
experience we know that if we keep tossing a coin,
eventually, a head must show up, i.e.,
But
and using axiom 4
(generalization of axiom 3) of
probability
.
1
)
(
=
A
P
∪
∞
=
=
1
,
n
n
A
A
).
(
)
(
1
1
∑
∞
=
∞
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
n
n
n
n
A
P
A
P
A
P
∪
¾
In a fair coin since only one in 2
n
outcomes is in favor of
A
n
, we have
Agreeing
with
axiom 2 of probability
¾
In summary, we have a nonempty set
Ω
of outcomes (
elementary events), a
σ
field
F
of subsets of
Ω
(combining
to form events), and a probability measure
P
on the sets in
F
subject to the four axioms forming a triplet
(
Ω
,
F, P
)
i.e.
a
probability model
.
¾
The probability of more complicated events must follow
from this framework by deduction.
,
1
2
1
)
(
and
2
1
)
(
1
1
=
=
=
∑
∑
∞
=
∞
=
n
n
n
n
n
n
A
P
A
P
Conditional Probability and Independence
¾
In
N
independent trials, suppose
N
A
,
N
B
,
N
AB
denote the
number of times events
A
,
B
and
AB
occur respectively.
¾
Using
the frequency interpretation of probability, for
large
N
¾
In
N
A
occurrences of
A,
only
N
AB
of them are also found
in the
N
B
occurrences of
B,
giving
the ratio
.
)
(
,
)
(
,
)
(
N
N
AB
P
N
N
B
P
N
N
A
P
AB
B
A
≈
≈
≈
)
(
)
(
/
/
B
P
AB
P
N
N
N
N
N
N
B
AB
B
AB
=
=
as a measure of “the event
A
given that
B
has already
occurred”. We denote this conditional probability by
P
(
A

B
) =
Probability of “the event
A
given
that
B
has occurred”.
¾
We define
provided
¾
For a conditional probability to be a probability, it has to
satisfy all probability axioms discussed earlier.
,
)
(
)
(
)

(
B
P
AB
P
B
A
P
=
.
0
)
(
≠
B
P
We have
(i)
(ii)
since
Ω
B
=
B
.
(iii)
Suppose
Then
But
hence
satisfying all probability axioms , making a
cond.
Prob.
a legitimate probability measure.
,
0
0
)
(
0
)
(
)

(
≥
>
≥
=
B
P
AB
P
B
A
P
.
)
(
)
(
)
(
)
)
((
)

(
B
P
CB
AB
P
B
P
B
C
A
P
B
C
A
P
∪
=
∩
∪
=
∪
,
1
)
(
)
(
)
(
)
(
)

(
=
=
Ω
=
Ω
B
P
B
P
B
P
B
P
B
P
),

(
)

(
)
(
)
(
)
(
)
(
)

(
B
C
P
B
A
P
B
P
CB
P
B
P
AB
P
B
C
A
P
+
=
+
=
∪
.
0
=
∩
C
A
AB
BC
,
∩=
φ
).
(
)
(
)
(
CB
P
AB
P
CB
AB
P
+
=
∪