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Correction: Problem 18. Alice and Bob have 2 n+1 FAIR coins, each with probability of a head equal to 1/2. 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)
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Correction: Problem 18 Alice and Bob have 2n+1FAIR coins, each with probability of a head equal to 1/2
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
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?
The Counting Principle • An r stage process • (a) n1 possible results at the first stage • (b) For everypossible 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)
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
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
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)
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.
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 • 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}
Combinations (continued) • Properties of n_choose_k
Combinations (continued) • Binomial formula • Let p =1/2
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)
Partitions (continued) • Example : partitions of 4 objects into 3 groups, 2-1-1. • 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}
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
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