Download Review Chapter1-3

Review Chapter 1 - 3
Chapter 1 Combinatorial Analysis
• Basic principle of counting
• Permutation
• Combination
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Basic Principle of Counting
• Suppose that two experiments are to be performed. Then
if experiment 1 can result in any one of m possible
outcomes and if for each outcome of experiment 1 there
are n possible outcomes of experiment 2, then together
there are mn possible outcomes of the two experiments.
• If experiment 1 has two possible outcomes, for the first
outcome there are m possible outcomes for experiment 2,
and for the second outcome, there are n possible
outcomes for experiment 2, then there are m+n total
possible outcomes for the two experiments.
• Useful when computing probabilities.
– P(EF) = P(E)P(F)
– P(E) = P(EFEFc) = P(EF) + P(EFc)
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Permutations
• Possible linear ordering of n items.
– n!
– A class in probability theory consists of 6 men and
4 women with 2 graduate and 8 undergraduate
students. An examination is given and the students
are ranked according to their performance. Assume
that no two students obtain the same score. How
many different rankings are possible?
– 10!
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Variations of Permutation Problems
• Ordering subset of students separately
– If the graduate students are ranked just among themselves and
undergraduate students among themselves, how many different
rankings are possible?
• 2!8!
• Some students are indistinguishable
– If we just list the gender of the students how many different
rankings are possible?
• 10!/(6!4!)
• Some students need to be placed together
– If we know that student X and Y rank next to each other, how
many different rankings are possible?
• 9!2!
• Some students cannot be placed together
– If we know that student X and Y are not ranked next to each
other, how many different rankings are possible?
• 10! - 9!2!
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Combination
• The number of different groups of r objects that could
be formed from a total of n objects.
n
n!
  
 r  (n  r )! r!
• A group of 20 people, from which a committee of 5 is
to be formed. How many different committees are
possible?
– 20!/(15!5!)
• Multiple groups
– If two committee of 5 and 4, respectively, are to be formed
from this group of 20 people, how many different
committees are possible?
• 20!/(5!4!11!)
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Chapter 2 Axioms of Probability
• Sample space and events
• Axioms of probability
• Propositions
• Sample spaces having equally likely outcomes
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Sample Space and Events
• Sample space and events are important concepts.
– List all the outcomes in a sample space or an event.
– HW 2.3, 2.5, 2.7
• Venn diagram and operators of events such as
intersection, union, and complement.
– Chap-2 slides page 13-22, 32-33, 38
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Axioms and Propositions
• Axioms 1-3
– 0 ≤ P(E) ≤ 1
– P(S) = 1
– If E and F are mutually exclusive events, then
P( E  F )  P( E )  P( F )
• Propositions
– 4.1: P(Ec) = 1-P(E)
– 4.2: If E  F , then P( E )  P( F ).
– 4.3: P( E  F )  P( E )  P( F )  P( EF )
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Inclusion-exclusion identity
P( E  F )  P( E )  P( F )  P( EF )
P( E  F  G )
 P( E )  P( F )  P(G )  P( EF )  P( EG )  P( FG )  P( EFG )
• Chap-2: Movie recommendation system, ex. 4a, 5l,
• HW 2.8, HW 2.9, HW 2.12, HW 2.14.
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Sample Space Having Equally Likely
Outcomes
• If in an experiment, all outcomes in the sample space
are equally likely to occur
– P(E) = (number of outcomes in E) / (number of outcomes
in S)
• Most problems use the counting techniques we
learned in Chapter 1.
– A committee of 8 is to be selected from a group of people
with 9 men and 11 women. If the selection is made
randomly, what is the probability that the committee
consists of 3 men and 5 women?
 9 11
  
 3  5 
 20 
 
8
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Sample Space Having Equally
Likely Outcomes
• P(E) = (number of outcomes in E) / (number of outcomes in S)
• Chap-2: ex. 5a, 5b, 5d, 5e, 5f, 5g, 5h, 5i, 5j
• HW 2.31, 2.37, 2.41
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Chapter 3. Conditional Probability and
Independence
• Conditional probabilities
• Bayes’ formula
• Independent events
• Conditional probability is a probability
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Conditional Probabilities
• Conditional probabilities are very important concept.
– P(E|F) = P(EF)/P(F)
– P(EF) = P(E|F)P(F)
– P(E) = P(EFEFc) = P(EF) + P(EFc) = P(E|F)P(F) +
P(E|Fc)P(Fc)
• Bayeis Formula
P( AB)
P( A | B) P( B)
P( B | A) 

P( A)
P( A | B) P( B)  P( A | B c ) P( B c )
P( F j | E ) 

P( EFj )
P( E )
P( E | F j ) P( F j )
n
 P( E | F ) P( F )
i 1
i
i
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Independent Events
• P(EF) = P(E)P(F)
• Proposition 4.1
If E and F are independent, then so are E and Fc.
– Chap-3: Ex 4f, 4g, 4h, 4i,
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• Ex 4f. An infinite sequence of independent
trials is to be performed. Each trial results in a
success with probability p and a failure with
probability 1-p. What is the probability that
(a) all trials result in successes in the first n
trials?
(b) at least 1 success occurs in the first n trials;
(c) exactly k successes occur in the first n
trials;
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Conditional Probability is a Probability
P( E1 | F )  P( E1 | E2 F ) P( E2 | F )  P( E1 | E2c F ) P( E2c | F )
• Ex 5a. Insurance company problem.
• What is the conditional probability that a new policyholder
will have an accident in his or her second year of policy
ownership, given that the policyholder has had an accident in
the first year?
• A: the event that the policyholder is accident prone.
• Ai, i = 1,2, be the event that he or she has had an accident in
the i-th year.
• P(A2|A1) = ?
• Conditioning on whether or not the policyholder is accident
prone:
• P(A2|A1) = P(A2|AA1)P(A|A1) + P(A2|AcA1) P(Ac|A1)
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P(E)?
• P(E) = (number of outcomes in E) / (number of outcomes in S)
– Sample space with equally likely events
• P(E) = P(EFEFc) = P(EF) + P(EFc) = P(E|F)P(F) + P(E|Fc)P(Fc)
– Chap-3: ex 3a part 1,
• P(E) = 1 - P(Ec)
• Conditioning on a particular event
– Chap-3: ex 4h, HW 3.74
• P(EF)?
– P(EF) = P(E|F)P(F) = P(F|E)P(E)
–
Chap-3: ex 2e
– P(EF) = P(EFGEFGc) = P(EFG) + P(EFGc) = P(EF|G)P(G) +
P(EF|Gc)P(Gc)
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P(E|F)?
• P(E|F) = P(EF) / P(F)
• = P(F|E)P(E) / P(F)
– Chap-3 ex. 3a part 2
• = P(F|E)P(E) / (P(F|E)P(E) + P(F|Ec)P(Ec))
– Chap-3 ex. 3d, 3f, …
P( E1 | F )  P( E1 | E2 F ) P( E2 | F )  P( E1 | E2c F ) P( E2c | F )
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