Sample Spaces and Events
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Venn Diagrams:
e.g.
A∪B
A complement
A∩B
mutually disjoint
Axiomatic Probability
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The objective of probability is to assign to each event A a
number P(A), called the probability of the event A, which will
give a precise measure of the chance thtat A will occur.
Axiomatic Probability
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The objective of probability is to assign to each event A a
number P(A), called the probability of the event A, which will
give a precise measure of the chance thtat A will occur.
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Probability Axioms:
Axiomatic Probability
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The objective of probability is to assign to each event A a
number P(A), called the probability of the event A, which will
give a precise measure of the chance thtat A will occur.
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Probability Axioms:
AXIOM 1 For any event A, P(A) ≥ 0.
Axiomatic Probability
I
The objective of probability is to assign to each event A a
number P(A), called the probability of the event A, which will
give a precise measure of the chance thtat A will occur.
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Probability Axioms:
AXIOM 1 For any event A, P(A) ≥ 0.
AXIOM 2 P(S) = 1.
Axiomatic Probability
I
The objective of probability is to assign to each event A a
number P(A), called the probability of the event A, which will
give a precise measure of the chance thtat A will occur.
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Probability Axioms:
AXIOM 1 For any event A, P(A) ≥ 0.
AXIOM 2 P(S) = 1.
AXIOM 3 If A1 , A2 , A3 , . . . is an infinite collection
of disjoint even
P
then P(A1 ∪ A2 ∪ A3 ∪ · · · ) = ∞
P(A
i)
i=1
Axiomatic Probability
Proposition
P(∅) = 0 where ∅ is the null event. This in turn implies that the
property contained in Axiom 3 is valid for finite collection of
events, i.e. if A1 , A2 , . . . , An is P
a finite collection of disjoint events,
then P(A1 ∪ A2 ∪ · · · ∪ A3 ) = ni=1 P(Ai )
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and the event {H} and {T} are
mutually disjoint, by Axiom 3, we have
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and the event {H} and {T} are
mutually disjoint, by Axiom 3, we have
P({H}) + P({T }) = P({H} ∪ {T }) = P(S) = 1
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and the event {H} and {T} are
mutually disjoint, by Axiom 3, we have
P({H}) + P({T }) = P({H} ∪ {T }) = P(S) = 1
If the coin is fair, we should assign 0.5 to P({H}) and 0.5 to
P({T }).
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and the event {H} and {T} are
mutually disjoint, by Axiom 3, we have
P({H}) + P({T }) = P({H} ∪ {T }) = P(S) = 1
If the coin is fair, we should assign 0.5 to P({H}) and 0.5 to
P({T }).
If the coin is more likely to give a Head, then 0.8 for P({H}) and
0.2 for P({T }) may be suitable.
Axiomatic Probability
Examples:
1. Consider the coin tossing experiment and we are only interested
in tossing the coin one time. Then S = {H, T}.
Since P(S) = 1 (Axiom 1), and the event {H} and {T} are
mutually disjoint, by Axiom 3, we have
P({H}) + P({T }) = P({H} ∪ {T }) = P(S) = 1
If the coin is fair, we should assign 0.5 to P({H}) and 0.5 to
P({T }).
If the coin is more likely to give a Head, then 0.8 for P({H}) and
0.2 for P({T }) may be suitable.
In fact, if p is any fixed number between 0 and 1, then P({H}) = p
, and P({T }) = 1 − p is an assignment consistent with the axioms.
Axiomatic Probability
Examples:
2. Consider again the coin tossing example. However, this time we
are interested in getting a Head, i.e. we toss a coin many times
untill we get a Head. Then S = {H, TH, TTH, TTTH, TTTTH,
. . . }.
Axiomatic Probability
Examples:
2. Consider again the coin tossing example. However, this time we
are interested in getting a Head, i.e. we toss a coin many times
untill we get a Head. Then S = {H, TH, TTH, TTTH, TTTTH,
. . . }.
If P({H}) = 0.4 then P({T }) = 0.6, P({TH}) = (0.4)0.6,
P({TTH}) = (0.4)(0.6)2 , P({TTTH}) = (0.4)(0.6)3 , . . . .
Axiomatic Probability
Examples:
2. Consider again the coin tossing example. However, this time we
are interested in getting a Head, i.e. we toss a coin many times
untill we get a Head. Then S = {H, TH, TTH, TTTH, TTTTH,
. . . }.
If P({H}) = 0.4 then P({T }) = 0.6, P({TH}) = (0.4)0.6,
P({TTH}) = (0.4)(0.6)2 , P({TTTH}) = (0.4)(0.6)3 , . . . .
Since {H}, {TH}, {TTH}, {TTTH}, {TTTTH}, . . . are mutually
disjoint and
S = {H} ∪ {TH} ∪ {TTH} ∪ {TTTH} ∪ {TTTTH} ∪ . . . , we have
Axiomatic Probability
Examples:
2. Consider again the coin tossing example. However, this time we
are interested in getting a Head, i.e. we toss a coin many times
untill we get a Head. Then S = {H, TH, TTH, TTTH, TTTTH,
. . . }.
If P({H}) = 0.4 then P({T }) = 0.6, P({TH}) = (0.4)0.6,
P({TTH}) = (0.4)(0.6)2 , P({TTTH}) = (0.4)(0.6)3 , . . . .
Since {H}, {TH}, {TTH}, {TTTH}, {TTTTH}, . . . are mutually
disjoint and
S = {H} ∪ {TH} ∪ {TTH} ∪ {TTTH} ∪ {TTTTH} ∪ . . . , we have
1 = 0.4 + (0.4)(0.6) + (0.4)(0.6)2 + (0.4)(0.6)3 + · · ·
Axiomatic Probability
More Probability Properties
Axiomatic Probability
More Probability Properties
Proposition
For any event A, P(A) + P(A0 ) = 1, from which P(A) = 1 − P(A0 ).
Axiomatic Probability
More Probability Properties
Proposition
For any event A, P(A) + P(A0 ) = 1, from which P(A) = 1 − P(A0 ).
Example 2.13
Consider a system of five identical components connected in series,
as illustrated below.
Denote a component failure by F and success by S. Let A be the
event that the system fails. For A to occur, at least one of the
individual components must fail. If we know P({F }) = 0.1, then
what is P(A)?
Axiomatic Probability
Proposition
For any event A, P(A) ≤ 1 .
Axiomatic Probability
Proposition
For any event A, P(A) ≤ 1 .
Proposition
For any two events A and B,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Axiomatic Probability
Proposition
For any event A, P(A) ≤ 1 .
Proposition
For any two events A and B,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
A Venn Diagram proof:
Axiomatic Probability
Proposition
For any event A, P(A) ≤ 1 .
Proposition
For any two events A and B,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
A Venn Diagram proof:
=
+
Axiomatic Probability
Example 2.14
In a certain residential suburb, 60% of all households subscribe to
the metropolitan newspaper published in a nearby city, 80%
subscribe to the local paper, and 50% of all households subscribe
to both papers. If a househlld is selected at random, what is the
probability that it subscribes to (1)at least one of the two
newspapers and (2) exactly one of the two newspapers?