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Sample Spaces
• Sample Space
– Set of all possible outcomes (or say sample points) of a random
experiment.
第二章
Basic Concepts of Probability
Theory
– Examples: Toss a coin three times and note the sequence of heads and
tails
• The sample spaces S={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
• Discrete sample space
– If S is countable
– Ex: Toss a coin three times and note the number of heads
• S={0, 1, 2, 3}
• Continuous sample space
– If S is not countable
– Ex: pick a number at random between zero and one
• S={x: 0≤ x ≤ 1}=[0,1]
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Events
Random Experiments
• Subset of interested outcomes contained in the
sample space.
• An experiment in which the outcome
varies in an unpredictable fashion
when the experiment is repeated
under the same conditions
– Ex: the three tosses give the same outcome A={HHH,
TTT}
• Two special event
– Certain event S: consists of all outcomes (always
occurs)
– Null event ∅ : contains no outcomes and never occurs
• Simple event: Unique outcome of an experiment
– Ex: Flipping a coin, outcome: Tails
• Compound Event: more than one outcome
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– Ex: the three tosses give the same outcome A={HHH,
4
TTT}
Set Theory
• The intersection of two events that are common to the two events
denoted by the symbol, A ∩ B, or just AB. For example:
E=B∩C={1,3}
• The union of two events denoted as A ∪ B, or A+B.
For example:
S
C={1,2,3}
A
A
D=B∪C={1,2,3,6}
A∪A=S
• Two events are said to be mutually exclusive (or say disjoint) if
they have no sample points in common – that is, if the occurrence
of one event excludes the occurrence of the other. For example:
A={1,2}
B={1,3,6}
A and A are mutually exclusive events (A∩ A = ∅)
• In random experiments we are
interested in the occurrence of
events that are represented by sets
(集合)
• Definition: universal set U: consists
of all possible outcomes, i.e., sample
space
n
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• roll of a die. The Sample space is
S={1,2,3,4,5,6}
∪ Ak = A1 ∪ A2 ∪ ... ∪ An
k =1
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Venn Diagrams
S
• In depicting multiple events, Venn
diagrams are excellent visual tools.
– The six outcomes are the sample points of the experiment.
• An event is a subset of S, and may consist of any number of sample
points. For example:
S
A={2,4}
A
A
• The complement of the event A, denoted by A , consists of all the
sample points in S that are not in A:
S
A
B
A
A={1,3,5,6}
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Venn Diagram Example 1
Complement of an Event A
• is the subset of all elements of sample
space (Ω) that are not in A.
• Denoted as A' or Ac
• A = Cars with Sunroofs
B = Cars with Air conditioning
• What does the shaded area
represent ?
A
A
B
B
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Venn Diagram Example 2
• A = Cars with Sunroofs
B = Cars with Air conditioning
• What does the shaded area represent ?
A
B
Corollary 2
Axioms of Probability
P[A] ≤1
P(A): the probability of event A
Corollary 3
Axiom 1 P ( A) ≥ 0 for any event A
P[∅] = 0
Axiom 2 P ( S ) = 1
Corollary 4
If all Ai’s are mutually exclusive, then
k
Axiom 3
P ( A1 ∪ A2 ∪ ... ∪ Ak ) = ∑ P( Ai )
(finite set)
∞
If A1, A2, …, An are pairwise mutually exclusive
n
n
(兩兩間互斥), then
P( ∪ Ak ) = ∑ P[ Ak ]
k =1
i =1
for n ≥ 2
k =1
P ( A1 ∪ A2 ∪ ...) = ∑ P( Ai )
(infinite set)
i =1
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Corollary 5
Corollary 1:
‘’Additive Rule” of Probability
For any event A, P[Ac]=1-P[A]
• In the case that two events are not
mutually exclusive, we can use the
additive rule below:
If A and B are mutually exclusive, then
P ( A ∩ B ) = 0.
P (A ∪ B ) = P (A) + P (B ) − P (A ∩ B )
If any two events A and B, are collectively
exhaustive, then P(A+B)= 1 (where “+”
means “Union”, i.e. A+B=Sample space)
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A
B
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Corollary 6
n
n
P[∪Ak ] = ∑ P[ A j ] − ∑ P[ Aj ∩ Ak ] +
k =1
j =1
+ (−1) n +1 P[ A1 ∩
j <k
Probability Example 1
∩ An ]
• Prob Student has Visa Card = 0.5
• Prob Student has Master Card = 0.15
• Prob Student has Both Cards = 0.1.
Corollary 7
If
A⊂ B
, P[A] <= P[B]
A
B
• What is the probability that a student does
not have a Master Card ?
• What is the probability that a student has
neither card? (Draw a Venn Diagram)
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Ex. A card is drawn from a well-shuffled deck
of 52 playing cards. What is the probability
that it is a queen or a heart?
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Probability Example 2
• Given the following VENN diagram,
– What is the prob that events A and B
will occur?
– What is the prob that events A and B
and C will occur?
Q = Queen and H = Heart
P (Q ) =
4
13
1
, P ( H ) = , P (Q ∩ H ) =
52
52
52
P (Q ∪ H ) = P (Q) + P ( H ) − P(Q ∩ H )
=
4 13 1
+ −
52 52 52
=
16 4
=
52 13
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P
P
P
P
P
P
P
P
(A) = 0.7
(B) = 0.8
(C) = 0.75
(A or B) = 0.85
(A and B) = ?
(A and C) = 0.55
(B and C) = 0.60
(A U B U C) = 0.98
A
B
C
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2.2.1 Discrete Sample Space
(countable sample space)
• For Discrete sample spaces, we can assign
probabilities for outcomes (elementary
events)
• Ex: if sample space S={a1, a2, …, an}
event B={a1’, a2’, …, am’}
Î P(B)=P[a1’]+P[a2’]+…+P[am’]
• If the element in S is equally likely
outcomes Î P[a1]=P[a2]=…=P[an]=1/n
Î P(B)=P[a1’]+P[a2’]+…+P[am’]=m/n
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Ex: 2.9 An urn contains 10 identical balls numbered 0, 1,…,9. A
random experiment involves selecting a ball from the urn and noting
the number of the ball. Find the probability of the following events:
A=“number of ball selected is odd”
B=“number of ball selected is a multiple of 3,”
C=“number of ball selected is less than 5,”
and of A∪B and A ∪B ∪C.
•
Î
• Ex:2.10 suppose that a coin is tossed three
times. If we observe the sequence of heads
and tails, then there are eight possible
outcomes S={HHH, HHT, HTH, HTT, THH, THT,
TTH, TTT}.
• If we assume that the outcomes of S are equiprobable,
then the probability of each of the eight elementary
events is 1/8.
Î P[“2 heads in 3 tosses”]=P[{HHT, HTH, THH}]
=3/8
• If we count the number of heads in three tosses
instead of observing the sequence of heads and tails.
The sample space is now S={0, 1, 2, 3}. If we assume the
outcomes of S to be equiprobable, the each of the
elementary events of S has probability ¼
Î P[“2 heads in 3 tosses”]=P[{2}]=1/4
• The two assignments are not consistent with each other
Î which one of the assignments is acceptable?
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• Ex 2.11 A fair coin is tossed repeatedly until the
first heads shows up; the outcome of the
experiment is the number of tosses required until
the first heads occurs. Find the probability law for
this experiment
Îthe sample space S={1, 2, 3, …}. Suppose the experiment
is repeated n times. Let Nj be the number of trials in which
the jth toss results in the first heads. If n is very large, we
expect N1 to be approximately n/2 since the coin is fair. This
implies that a second toss is about n-N1 ≅ n/2 times.
Î For large n, the relative frequencies fj ≅ Nj/n = (1/2)j, j = 1,
2, …
Î P[j tosses till first heads] = (1/2)j , j = 1, 2, …
Î Verify that these probabilities add up to one with α = 1/2
sample space S = {0, 1,…,9}
A= {1, 3, 5, 7, 9} B={3, 6, 9} C={0, 1, 2, 3, 4}
P[A]=5/10, P[B]=3/10, P[C]= 5/10
P[A ∪B ]=P[A]+P[B]-P[A∩B]=5/10+3/10-P[3,9]=6/10
P[A ∪B ∪C]=P[A]+P[B]+P[C]-P[A ∩B]-P[A ∩C]-P[B ∩C]+P[A ∩B ∩C]
=5/10+3/10+5/10-2/10-2/10-1/10+1/10
= 9/10
or X=A ∪ B Î P[A ∪B ∪C]=P[X∪C]
=P[X]+P[C]-P[X∩C]
∞
=6/10+5/10-2/10
=9/10
∑α
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j =1
j
=
α
1−α
=1
α =1/ 2
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• Ex: 2.12 Consider the random experiment
“pick a number x at random between zero
and one.” The sample space S for this
experiment is the unit interval [0, 1], which
is uncountably infinite. If we suppose that
all the outcomes S are equally likely to be
selected, then we would guess that the
probability that the outcome is in the
interval [0, 1/2] is the same as the
probability that the outcome is in the
interval [1/2, 1]. We would also guess that
the probability of the outcome being exactly
equal to ½ would be zero since there are an
uncountably infinite number of equally likely
outcomes.
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2.2.2 Continuous Sample Space
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Homeworks
• The outcomes are numbers that can
assume a continuum of values
• The sample space S be the entire
real line R
• For continuous sample spaces, we
can assign probabilities for
• 2.1, 2.2, 2.14, 2.15
• 2.21, 2.23, 2.24, 2.33, 2.35
– Intervals of the real line
– Rectangular regions in the plane
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