Correction: Problem 18
Alice and Bob have 2n+1 FAIR coins, each with
probability of a head equal to 1/2
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Review
 Conditional probability of A given B ,
P(B)>0
– P(A|B)=P(A∩B)/P(B)
– Define a new probability law
– Conditional independence
 P(A∩C|B)=P(A|B) P(C|B)
 Total probability theorem
 Bayes’ rule
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1.6 Counting
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Counting
 To calculate the number of outcomes
 Examples
– To toss a fair coin 10 times, what’s the probability that the first toss was
a head?
– Fair coin
1/2
– To toss a fair coin 10 times, what’s the probability that there was only 1
head?
– 1/10?
– What about the probability that there are 5 heads?
– 1/2?
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The Counting Principle
 An r stage process
– (a) n1 possible results at the first stage
– (b) For every possible result of the first stage, there are n2 possible
results at the second stage
– (c) ni …..
– Total number of possible results : n1 n2 … nr (proof by induction)
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The Counting Principle
 Example1 : fair coin toss 3 times
– 2X2X2 = 8 possible outcomes
 TTT TTH THT THH HTT HTH HHT HHH
 Example2 : number of subsets of an n-element set S
– S={1,2}
 Subsets : Ф, {1}, {2}, {1,2}, 4=22 subsets
– S={1,2,3}
 Subsets : Ф, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, 8=23 subsets
– The choice of a subset as a sequential process of choosing one
element at a time.
 n stages, binary choice at each stage
– 2X2X2…X2 =2n
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Permutations
 Selection of k objects out of n objects (k<=n), order matters
– Sequences: 123≠ 321
– K-permutation: the number of possible sequence
– Example: number of 3-letter words using a,b,c or d at most once
– 4X3X2 =24 3-permutation out of 4 objects
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Permutations (continued)
 Selection of k objects out of n objects (k<=n), order matters
– k-stages, n1=n, ni+1= ni-1,… nk= n-k+1
 Counting principle: n(n-1)(n-2)…(n-k+1)
– number of permutations of n objects out of n objects (k=n)
 n!
(0! = 1)
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Combinations
 Selection of k objects out of n objects (k<=n), NO ordering
– Sets: {1,2,3} = {3,2,1}
– k-combinations: the number of possible different K-element subsets
– Example: number of 3-element subsets of {a,b,c,d}.
 {a,b,c} {a,b,d} {a,c,d} {b,c,d} : 4 3-element subsets
 3-permutations of {a,b,c,d} = abc, acb, bac, bca, cab, cba,
abd, adb, bad,bda, dab, dba,
acd, adc, cad, cda, dac, dca
bcd, bdc, cbd, cdb, dbc, dcb
 k-combinations = k-permutations – Order
– Each set (k-combination) is counted k! times in the k-permutation.
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Combinations (continued)
 Example1: 2-combinations of a 4-object set {a,b,c,d}.
 4 choose 2 = 4!/(2!2!) =6
 {a,b} {a,c} {a,d} {b,c} {b,d}, {c,d} : 6 2-element subsets
 Example2: k-head sequences of n coin tosses
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–
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n=5, k=2
HHTTT HTHTT HTTHT HTTTH THHTT
THTHT THTTH TTHHT TTHTH TTTHH
HHTTT  {1,2}
HTHTT  {1,3}…. TTTHH {4,5}
Number of k-head sequences = number of k-combinations from
{1,2,…n}
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Combinations (continued)
 Properties of n_choose_k
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Expansion of (a+b)n

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Combinations (continued)
 Binomial formula
– Let p =1/2
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Partitions
 Partitions of n objects into r groups, with the ith group having
ni objects, sum of ni is equal to n.
– Order does not matter within a group, the groups are labeled
– Example : partition S={1,2,3,4,5} into 3 groups
n1 = 2, n2 =2, n3 =1
 {1,2}{3,4}{5} = {2,1}{4,3}{5}
 {1,2}{3,4}{5} ≠ {3,4}{1,2}{5}
 Total number of choices (group by group)
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Partitions (continued)
 Example : partitions of 4 objects into 3 groups, 2-1-1.
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–
–
–
–
–
–
4!/(2!1!1!)=12
{ab}{c}{d} {ab}{d}{c}
{ac}{b}{d} {ac}{d}{b}
{ad}{b}{c} {ad}{c}{b}
{bc}{a}{d} {bc}{d}{a}
{bd}{a}{c} {bd}{c}{a}
{cd}{a}{b} {cd}{b}{a}
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Summary
 k-permutation of n objects n!/(n-k)!
– Order matters : 123 ≠ 321
 k-combinations of n objects
– Order does not matter : {1,2,3} = {3,2,1}
 Partitions of n objects into r groups, with the ith group having
ni objects
– Order does not matter within a group, the groups are labeled
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Binomial formula
 Toss an unfair coin (p-head, (1-p)-tail) n times
– The outcome is a n-sequence : THHTHT…H
– Ω={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} for n=3
– Group the sequences according to the number of H
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Pascal’s Triangle
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