Chapter 2: Axioms of Probability
1"
Spring, 2020
Xiugang Wu
University of Delaware
2"
Example Re-Revisited
A communication system consists of four antennas. Assume that this system
will be functional if no two consecutive antennas are defective.
Thinking process:
1) List all the possiblities: 0110, 0101, 1010, 0011, 1001, 1100
2) If all the cases are equally likely, then the desired probability is
3
6
=
1
2
.
Question: If there are exactly two antennas defective, what is the probabi l i ty
that the resulting system will be functional?
3"
Outline
• Sample Space and Events
• Axioms of Probability
• Some Simple Propositions
• Sample Space with Equally Likely Outcomes
4"
Outline
• Sample Space and Events
• Axioms of Probability
• Some Simple Propositions
• Sample Space with Equally Likely Outcomes
5"
Sample Space
S — Sample space: the set of all possible outcomes of an experiment
1) Flip a coi n. S = {tails, heads}
2) Flip two coi n s. S = {hh, tt, ht, th}
3) Toss two dices. S = {(i, j):i, j =1, 2, 3, 4, 5, 6}
4) The order of finish in a race among 7 horses. S = {all 7! permutations of (1,2,3,4,5,6,7)}
5) Measuring (in hours) the life time of a transistor. S = {x :0 x<1}
6"
Event
1) Flip a coin. S = {tails, heads}.
E = {h ead s},F = {tails}.
2) Flip two coins. S = {hh, tt, ht, th}.
E = {hh, ht}, F = {th, ht}.
3) Toss two dices. S = {(i, j):i, j =1, 2, 3, 4, 5, 6}.
E = {(1, 6), (2, 5) , (3, 4), (4, 3), (5, 2) , (6, 1)}.
4) The order of finish in a race among 7 horses . S = {all 7! permutations of (1,2,3,4,5,6,7) }.
E = {al l permutations start i n g with 3}.
5) Measuring (in hours) t h e life time of a tran si s t or. S = {x :0 x<1}.
E = {x :0 x 5}
Event: any subset E of the sample space is known as an event; an event is a set
of some possible outcome. If th e out com e of the experiment is contained in E,
then we say that E has occurred.
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Set Operations
1) Union: E [ F ; [
n
i=1
E
i
; [
1
i=1
E
i
2) Intersection: E \ F or EF; \
n
i=1
E
i
; \
1
i=1
E
i
E and F are said to be mutually exclusive or disjoint if EF = ;
3) Complement: E
c
= {all outcomes not in E}; S
c
= ;
4) Subset: E ✓ F i ↵ all elements in E are also in F .
If E ✓ F and F ✓ E,thenE = F .
Can you demonst r at e these operations by using Venn D iagr ams?
8"
Laws of Set Theory
1) Commutative l aws: E [ F = F [ E; EF = FE
2) Associat ive laws: (E [ F ) [ G = E [ (F [ G);(EF ) G = E(FG)
3) Distrib ut i ve laws: (E [ F )G =(EG) [ (FG); (EF) [ G =(E [ G)(F [ G)
4) DeMorgan’s law: ([
n
i=1
E
i
)
c
= \
n
i=1
E
c
i
;(\
n
i=1
E
i
)
c
= [
n
i=1
E
c
i
Can you justify these laws by using Venn Diagrams?
9"
Outline
• Sample Space and Events
• Axioms of Probability
• Some Simple Propositions
• Sample Space with Equally Likely Outcomes
10"
Axioms of Probability
Example: Toss a coin. If heads is as likely as tails, then P ({h})=P ({t})=0.5.
Or P ({h})=1 P ({t})=0.25 for a biased coin.
Example: Roll a dice. If all sides ar e equally likely, t h en P ({i})=1/6 for any
i =1, 2, 3, 4, 5, 6.
1) Non-negativi ty: 0 P (E) 1, for any E
2) Normalization : P (S)=1
3) Additivity: For any sequence of mutually exclusive events E
1
,E
2
,...,i.e.
E
i
E
j
= ; for i 6= j,
P ([
1
i=1
E
i
)=
1
X
i=1
P (E
i
)
11"
Outline
• Sample Space and Events
• Axioms of Probability
• Some Simple Propositions
• Sample Space with Equally Likely Outcomes
12"
Some Simple Propositions
1) P (E
c
)=1 P (E)
2) If E ✓ F ,thenP (E) P (F )
3) P (E [ F )=P (E)+P (F ) P (EF)
4) P (E[F [G)=P (E)+P (F )+P (G)P (EF )P (EG)P (FG)+P (EFG)
Using Venn Diagrams can help understand these propositi on s.
13"
Inclusion-Exclusion Identity
P (E
1
[ E
2
[ ···[ E
n
)=
X
i
P (E
i
)
X
i
1
<i
2
P (E
i
1
E
i
2
)
+ ···+(1)
r+1
X
i
1
<i
2
<···<i
r
P (E
i
1
E
i
2
···E
i
r
)
+ ···+(1)
n+1
P (E
1
E
2
···E
n
)
14"
Example
Example: Mike is going to take two cour s e A and B next semester. With proba-
bility 0.8 he will like course A; with probability 0.7 he will like B; with probabil-
ity 0.6 he will like both. What is t he probability that he will like n ei t he r course?
Solution: Let A = {Mike will like course A} and B = {Mike will like course B}.
Then we have
A [ B = {Mi ke will like course A or B}
which happens with probability
P (A [ B)=P (A)+P (B) P (AB)=0.8+0.7 0.6=0.9.
Therefore,
P ({Mike will like neither course A nor B})=1 P (A [ B)=1 0.9=0.1
15"
Outline
• Sample Space and Events
• Axioms of Probability
• Some Simple Propositions
• Sample Space with Equally Likely Outcomes
16"
Sample Space with Equally Likely Outcomes
Suppose S = {1, 2,...,N} and P ({i})=1/N for any i =1, 2,...,N. Then for
any E,
P (E)=
number of outcomes in E
total number of outcomes in S
Example: If two dice are rolled. What is the probability that the sum of the
dice will equal 7?
Solution: The sample space S = {(i, j):i, j =1, 2, 3, 4, 5, 6}. The event of
interest is
E = {the s um of the dice will equal 7} = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}.
Therefore,
P (E)=
6
36
=
1
6
.
17"
Example
Example: Su ppose we randomly draw 3 bal ls from an urn containing 6 white
and 5 black b al ls . What is the probability that of the 3 balls, one is white and
two are black?
Solution:
6
1
5
2
11
3
=
4
11
What is the probability that of the 3 balls, one is black and two are white?
Solution:
6
2
5
1
11
3
=
15
33
18"
Example
Example: There are n people in a room. What is the probability that no two
of them cele br at e their birthday on the same day of the year?
Solution:
p
n
=
365 ⇥ 364 ⇥ 363 ⇥ ···⇥ (365 n + 1)
365
n
How big should n be such that this probability is less than 1/2?
Solution: Obviously p
n
is decreasing with n.Whenn 23, p
n
is less than 1/2.
Also when n = 50, p
n
⇡ 3%; when n = 100, p
n
⇡
1
3,000,000
.
