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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– 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.– 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