0 0 Correction: Problem 18 Alice and Bob have 2n+1 FAIR coins, each with probability of a head equal to 1/2

1 1 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

3 3 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?

4 4 The Counting Principle An r stage process –(a) n 1 possible results at the first stage –(b) For every possible result of the first stage, there are n 2 possible results at the second stage –(c) n i ….. –Total number of possible results : n 1 n 2 … n r (proof by induction)

5 5 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=2 2 subsets –S={1,2,3} Subsets : Ф, {1}, {2}, {1,2}, {3}, {1,3}, {2,3}, {1,2,3}, 8=2 3 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 =2 n

6 6 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

7 7 Permutations (continued) Selection of k objects out of n objects (k<=n), order matters –k-stages, n 1 =n, n i+1 = n i -1, … n k = 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)

8 8 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.

9 9 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}

10 Combinations (continued) Properties of n_choose_k

11 Expansion of (a+b) n

12 Combinations (continued) Binomial formula –Let p =1/2

13 Partitions Partitions of n objects into r groups, with the ith group having n i objects, sum of n i 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 n 1 = 2, n 2 =2, n 3 =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)

14 Partitions (continued) Example : partitions of 4 objects into 3 groups, –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}

15 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 n i objects –Order does not matter within a group, the groups are labeled

16 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

17 Pascal’s Triangle