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Probability
Chapter Outline
hapter two
ombinatorial Methods
2.1 Introduction
2.2
Counting
Principle2.3 Permutations
2.4 Combinations
2.5 Stirling’s Formula
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Probability
2.1 Introduction
Recall Theorem 1.3: If
– sample space is finite and
– sample points are all equally likely,
then P( A)= N ( A)/ N (S ) Some probability problems can be solved simply by
counting.
We study combinatorial
analysis
in chapter 2.
Combinatorial analysis deals with methods of counting.
– enable us to count systematically
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Probability
2.2 Counting Principle
Theorem 2.1 (Counting Principle)
If set E contains n elements and set F contains m elements
there are nm ways in which we can choose,
• First, an
element
of
E and
• then an element of F .
Theorem 2.2 (Generalized Counting Principle)
Let ,
, . . . ,
be sets with
,
, . . . ,
elements,
respectively.
– There are × × ×∙ ∙ ∙×ways in which we can • first, choose an element of ,
• then an
element
of
, • then an element of , • . . . ,
• and
finally
an
element
of
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Probability
Examples 2.1 and 2.2
Ex. 2.1 How many outcomes are there if we throw 5 dice?
Sol:
– Let , 1 5: set of all possible outcomes of th dice
– = {1, 2, 3, 4, 5, 6}
– Total outcomes of throwing 5 dice: 6 x 6 x 6 x 6 x 6 6
Ex.
2.2
In
tossing
4
fair
dice,
P(at least one 3 among these 4 dice)?
Sol:
– Let
A be
the
event
of
at
least
one
3
among
the
4
dice – Let be the event of no 3 in tossing 4 dice
– ( ) 5 x 5 x 5 x 5
– (
) (
)/ =
5
/6
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Probability
Examples 2.3 and 2.4
Ex. 2.3 Virginia wants to give her son, Brian, 14 different
baseball cards within a 7‐day period.
– If Virginia gives Brian cards no more than once a day, in
how many way can this be done? (7x7x …x7 = 7
) Ex. 2.4 Rose has invited n friends to her birthday party.
– All attend, and each one shakes hands with everyone else
at the party exactly once,
– What is the number of handshakes?
Sol 1:
– 1 Peoples, each shakes hands with other peoples – Each handshake counted twice
– ( 1)/2
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Probability
Number of Subsets of a Set
( A): Power set of A; set of all subsets of A.
Theorem 2.3 A set with n elements has 2 subsets. Proof :
– Let , , , ⋯ , be a set with n elements. Exists a one‐to‐one correspondence between subsets of A
and sequences of 0’s and 1’s of length n:
• To
a
subset
B
of A ,
associate
a
sequence ⋯ , – where = 0 if ∉ B, and = 1 if ∈ B. • Ex. n = 3, association for the subsets of A:
– ∅: 000, {}: 100, {}: 010, …, {, }: 011, …
By Generalized Counting Principle (Thm 2.2),
number of sequences of 0’s and 1’s of length n is
2 2 ⋯ 2 2.
Number of subsets of A is 2
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Probability
Tree Diagrams
Three diagrams systematically identify all possible cases of an
experiment
Ex. 2.8 Bill and John keep playing chess until
– one of them wins 2 games in a row or 3 games altogether. – In what percent of all possible cases does the game end
because Bill wins 3 games without winning 2 in a row?
Sol:
– 10 cases total
– Bill wins 3 games without
winning 2 in a row: 1 case
10% of cases – But probability 0.1
(not equiprobable)
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Probability
Example of Tree Diagram
Ex. 2.9 Mark has $4.
– He decides to bet $1 on the flip of a fair coin 4 times.
– What is the probability that (a) he breaks even; (b) he wins
money?
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Probability
Examples of Permutations (1/2)
Ex. 2.10 3 people, Brown, Smith, and Jones, must be
scheduled for job interviews.
– In how many different orders can this be done?
Sol: 3! Ex. 2.11 2 anthropology, 4 computer science, 3 statistics, 3
biology, and 5 music books are put on a bookshelf with a
random arrangement.
– P( A): Prob. that books of the same subject are together?
Sol:
– Total number of possible arrangements: 17!
– Anthropology books first: 2! X 4! X 3! X 3! X 5! ways
– Subjects can be ordered in 5! ways
!!!!!!
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Probability
Examples (2/2)
Ex. 2.12 If 5 boys and 5 girls sit in a row in a random order,
P(no two children of the same sex sit together)?
Sol:
– 10 persons to sit in a row: 10! ways – No two of the same sex sit together:
• Boys in positions 1, 3, 5, 7, 9, and girls in 2, 4, 6, 8, 10,
or vice versa.
– 5! × 5! possibilities for each case.
Desired probability is
!!!
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Probability
Distinguishable Permutations of Alike Objects
Theorem 2.4 (Distinguishable permutations of alike objects)
The number of distinguishable permutations of n objects of k
different types is
!! ! ⋯ !
– where
are alike ,
are alike , . . . ,
are alike , and
– n = + +∙ ∙ ∙+ ,
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Probability
Examples of Permutations with Alike Objects
Ex. 2.13 How many different 10‐letter codes can be made
using 3 a’s, 4 b’s, and 3 c’s?
Sol: 10!/(3!x4!x3!)
Ex. 2.14 In how many ways can we paint 11 offices so that
– 4 of them will be painted green, 3 yellow, 2 white, and the
remaining 2 pink?
Sol: 11!/(4!x3!x2!x2!)
Ex. 2.15 A fair coin is flipped 10 times.
– P(exactly
3
heads)? Sol: 10! 3! 7!⁄ 2⁄
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─ Set of all sequences of Hs and Ts of length 10: 2 elements ─ Exactly 3 Hs:
/3!=10!/(3! x 7!)
Sec.
2.4
Unordered
arrangement
(Combination)
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Probability
2.4 Combinations
Order in which elements are arranged is immaterial
※ Definition An unordered arrangement of r objects from a set
A containing n objects (r ≤ n) is called
– an r‐element
combination
of
A ,
or
– a combination of the elements of A taken r at a time.
Number of r ‐element combinations of n objects is given by
! ! ! ! Observation: + + + ⋯ + + + = 2
– Total no. of subsets of a set with n elements: 2
– No. of subsets of size r of a set of size n: Total no. of subsets: + + ⋯ + +
+
+
+ ⋯ +
+
+
= 2
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Probability
More Observations
,
,
For any 0 r n, and
(2.4)
̶ (2.4)
can
be
proved
algebraically
or
verified
combinatorially Proof: by combinatorial argument
Consider a set of n+1 objects {, , ⋯ , , }
• Number of r ‐
element
combinations
of
this
set:
– Separate combinations into two disjoint classes:
With : ( r‐1)‐element combinations of
, ⋯ , Without :
r‐element combinations of
, ⋯ ,
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Probability
Examples of Combinations (1/3)
Ex. 2.16 In how many ways can 2 math and 3 biology books be
selected from 8 math and 6 biology books?
Sol:
Ex. 2.17 45 instructors were selected randomly to ask
whether they are happy with their teaching loads.
– Responses of 32 were negative. – If Drs. Smith, Brown, and Jones were
among those questioned.
– P(all three gave negative responses)?
Sol:
– No. of possible groups with 32 negative responses:
– 3 (negative) selected and other 29 from remaining 42:
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Probability
Examples of Combinations (2/3) Ex. 2.18 In a small town, 11 of 25 school teachers are against
abortion, 8 are for abortion, and the rest are indifferent.
– A random sample of 5 schoolteachers is selected.
(a) P(all 5 are for abortion)?
(b) P(all 5 have the same opinion)?
Ex. 2.19 In Maryland’s lottery, player pick 6 integers between
1 and 49, order of selection being irrelevant.
– P(grand prize)?, P(2nd prize)?, P(3rd prize)?
Sol:
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Probability
Examples of Combinations (3/3) Ex. 2.20 7 cards are drawn from 52 without replacement.
– P(at least one of the cards is a king)?
Sol:
– No kings
– Desired probability: 1 No kings 1
Ex. 2.21 5 cards are drawn from 52. – P(full house)
Sol:
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Binomial Expansion (1/2)
Theorem 2.5 (Binomial Expansion)
– For any integer n ≥ 0 ,
Proof:
– Observations:
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iinn
i
n y x
i
n y x
0
)(
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Probability
Binomial Expansion (2/2)
– Expansion of
– Results in only terms of the form , 0 ≤ i ≤ n
– appears
times
Because
emerges only whenever – x’s of n−i of the n factors of ( x + y) multiplied by
– y’s of the remaining i factors of ( x + y).
Hence
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Probability
Examples (Binomial Expansion) (1/4)
Ex. 2.25 What is the coefficient of x2y3 in the expansion of
2 3?
Sol:
– Let 2 and 3
2 3=
– In the expansion of
• Coefficient of : • = 2 ∙ 3
In the expansion of 2 3, coefficient of : 2
∙ 3
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Probability
Example (Binomial Expansion) (2/4)
Ex. 2.26 Evaluate the sum
Sol: Two approaches
By set theory, total no. of subsets of a set of n elements: 2 – A set of n elements has
, 0 ≤ i ≤ n, subsets with i
elements.
Total no. of subsets of a set of n elements:
By binomial expansion, = ∑ – Let x = y = 1,
11=2= ∑ 11 =∑
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Examples (Binomial Expansion) (3/4)
Ex. 2.27 Evaluate the sum
Sol:
So
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Examples (Binomial Expansion) (4/4)
Ex. 2.28 Prove that
Sol:
– Let , , ⋯ , and , , ⋯ , be two disjoint sets, each with n elements .
Number of subsets with n elements of A∪B: . – Any subset with n elements of A∪B is the union of
• a subset with i elements of A and
• a subset with n − i elements of B, for some 0 ≤ i ≤ n.
For each i:
subsets,
Number of subsets with n elements of A∪B:∑
∑
∑
we have the identity
∑
.
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Probability
Multinomial Expansion
Ex. 2.30 Distribute n distinguishable balls into k distinguishable cells so that
• n1 balls are distributed into the first cell,
• n2 balls
into
the
second cell,
…,
• nk balls into the k th cell, where n1 + n2 +…+ nk =n.
– How many possible ways?
Sol:
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!!...!
!
21 k nnn
n
Sol:
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Probability
Theorem 2.6 Multinomial Expansion
Theorem 2.6 (Multinomial Expansion) In the expansion of
– The coefficient of the term
» where
• is
Therefore,
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k
k
nk
nn
nnnn k
x x xnnn
n
21
21
21... 21 !...!!
!
nk x x x )( 21
nk x x x )( 21
k n
k
nn x x x 21
21
nnnn k ...21
!...!!
!
21 k nnn
n
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2.5 Stirling’s Formula
Stirling’s formula can be used to estimate ! Theorem 2.7 (Stirling’s Formula)
Stirling’s formula usually gives excellent approximations in
numerical computations Note:
– !/ 2 becomes 1 at ∞,
– But, it is still close to 1, even for very small values of n.
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