Slide set 2
Stat 330 (Spring 2015)
Last update: January 13, 2015
Stat 330 (Spring 2015): slide set 2
Probability
Example:
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Consider the Event C (a successful transmission) defined earlier.
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From our experience with the network provider, we can decide that the
chance that the next message gets through is 90 %.
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We write: P (C) = 0.9
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To be able to work with probabilities, in particular, to be able to compute
probabilities of events, a mathematical foundation is necessary.
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Stat 330 (Spring 2015): slide set 2
Sets and Venn Diagrams
1. Review symbols ∈, ∈,
/ ⊂, ⊆, ⊃, ⊇,.
For e.g.
•
•
If a is a member of B, this is denoted a ∈ B
If every member of set A is also a member of set B, then A is said to
be a subset of B, written A ⊆ B (A is said to be contained in B).
2. Union (∪): A union of events is an event consisting of all the outcomes
in these events.
A ∪ B = {ω | ω ∈ A or ω ∈ B}
3. Intersection (∩): An intersection of events is an event consisting of the
common outcomes in these events.
A ∩ B = {ω | ω ∈ A and ω ∈ B}
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Stat 330 (Spring 2015): slide set 2
4. Complement (Ā): A complement of an event A is an event that occurs
when event A does not happen.
Ā = {ω | ω ∈
/ A}
5. Demorgan’s Law:
(A ∪ B) = Ā ∩ B̄
6. Empty Set ∅ is a set having no elements, usually denoted by {}
The empty set is a subset of every set:
∅⊆A
7. Disjoint sets: Sets A, B are disjoint if their intersection is empty:
A∩B =∅
8. Mutually exclusive sets: Sets A1, A2, . . . are mutually exclusive if any
two of these events are disjoint:
Ai ∩ Aj = ∅ for any i 6= j.
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Stat 330 (Spring 2015): slide set 2
Sets and Venn Diagrams (contd.)
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Stat 330 (Spring 2015): slide set 2
Kolmogorov’s Axioms
To be able to work with probabilities properly - to compute with them - one
must lay down a set of postulates:
A system of probabilities (a probability model) is an assignment of numbers
P (A) to events A ⊂ Ω (requires further mathematical complications for the
events that we can define probability on) in such a manner that
(i) 0 ≤ P (A) ≤ 1 for all A
(ii) P (Ω) = 1.
(iii) if A1, A2, . . . are (possibly, infinite many) mutually exclusive events (i.e.
Ai ∩ Aj = ∅ for all i 6= j) then
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P (A1 ∪ A2 ∪ . . .) = P (A1) + P (A2) + . . . =
P (Ai).
i
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Stat 330 (Spring 2015): slide set 2
Kolmogorov’s Axioms (continued)
These are the basic rules of operation of a probability model
• every valid model must obey these,
• any system that does, is a valid model
Whether or not a particular model is realistic or appropriate for a specific
application is another question.
Example:
Draw a single card from a standard deck of playing cards: Ω = {red, black}
Two different, equally valid probability models are:
Model 1
P (Ω) = 1
P (red) = 0.5
P (black) = 0.5
Model 2
P (Ω) = 1
P (red) = 0.3
P (black) = 0.7
Mathematically, both schemes are equally valid. But, of course, our real
world experience would favor to pick model 1 over model 2 as the ‘correct’
model.
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Stat 330 (Spring 2015): slide set 2
Useful Consequences of Kolmogorov’s Axioms:
Let A, B ⊂ Ω.
1. Probability of the Complementary Event: P (A) = 1 − P (A)
Corollary: P (∅) = 0
2. Addition Rule of Probability
P (A∪B) = P (A)+P (B)−P (A∩B)
3. If A ⊂ B, then P (A) ≤ P (B).
Corollary: For any A, P (A) ≤ 1.
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Stat 330 (Spring 2015): slide set 2
Example: Using Kolmogorov’s Axioms
We attempt to sign on to AOL using dial-up. We connect successfully if
and only if the phone number works and the AOL network works. Assume
P ( phone up ) = .9
P ( network up ) = .6, and
P ( phone up and network up ) = .55.
1. What is the probability that the phone is up or the network is up?
2. What is the probability that both the phone and the network are down?
3. What is the probability that we fail to connect?
4. What is the probability that only the phone is up?
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Stat 330 (Spring 2015): slide set 2
Solution
Let A ≡ phone up; B ≡ network up
1) What is the probability that the phone is up or the network is up?
P ( phone up or network up) = P (A ∪ B) = 0.9 + 0.6 − 0.55 = 0.95
2) What is the probability that both the phone and the network are down?
P ( phone down and network down) = P (Ā ∩ B̄) = P (A ∪ B)
= 1 − .95 = .05
3) What is the probability that we fail to connect?
P ( phone down or network down) = P (Ā∪ B̄) = P (Ā)+P (B̄)−P (Ā∩ B̄)
P (Ā ∪ B̄) = (1 − .9) + (1 − .6) − .05 = .1 + .4 − .05 = .45
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