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## Intro to Probability

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**Intro to Probability**Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong**e.g.1 (Page 10)**• Suppose that there is only one question in the final exam. • What is the sample space of all possible patterns of correct answers? Sample space = { T , F }where T corresponds to the answer being true and F corresponds to the answer being false**FTT**TTF TTT TFT Sample Space FFF FTF TFF FFT e.g.2 (Page 10) • Suppose that there are three questions in the final exam. • What is the sample space of all possible patterns of correct answers? Sample space = {TTT, TTF, TFT, FTT, TFF, FTF, FFT, FFF}**Sample space = { , , ,**, , } e.g.3 (Page 11) • Suppose that I want to throw one 6-sided dice. • What is the sample space of all possible outcomes of throwing this dice?**e.g.4 (Page 12)**• Suppose that I want to throw one two-sided coin (One side is a head H and the other side is a tail T) • What is the sample space of all possible outcomes of throwing the coin? Sample space = { H , T }**FTT**TTF TTT TFT Sample Space Event TTF TTT FFF FTF TFF FFT e.g.5 (Page 13) • Suppose that there are three questions in the final exam. • What is the event E that the first two answers are true?**1/8**1/8 1/8 1/8 FTT TTF TTT TFT Sample Space Event TTF TTT FFF FTF TFF FFT e.g.6 (Page 15) • Suppose that there are three questions in the final exam. 1/8 1/8 1/8 1/8 • What is the event E that the first two answers are true? 1/8 1/8 P(E) = 1/8 + 1/8 = 2/8 = 1/4**3/16**1/16 1/16 1/16 FTT TTF TTT TFT Sample Space Event TTF TTT FFF FTF TFF FFT e.g.7 (Page 15) • Suppose that there are three questions in the final exam. 3/16 3/16 3/16 1/16 • What is the event E that the first two answers are true? 3/16 3/16 P(E) = 3/16 + 3/16 = 6/16 = 3/8**P(E U V)**= 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = ½ 1/8 1/8 1/8 1/8 FTT TTF TTT TFT Sample Space Event Event TTF FFF TTT FFT FFF FTF TFF FFT e.g.8 (Page 16) Note that P(E) + P(V) = ¼ + ¼ = ½ Thus, P(E U V) = P(E) + P(V) • Suppose that there are three questions in the final exam. 1/8 1/8 1/8 1/8 These two sets are disjoint. • What is the event E that the first two answers are true? 1/8 1/8 P(E) = 1/8 + 1/8 = 2/8 = 1/4 • What is the event V that the first two answers are false? 1/8 1/8 P(V) = 1/8 + 1/8 = 2/8 = 1/4**FTT**TTF TTT TFT Sample Space FTT TFT Complement of Event E Event TTF TTT FFF FTF TFF FFT FFF FTF TFF FFT e.g.9 (Page 24) • Suppose that there are three questions in the final exam. • What is the event E that the first two answers are true? • What is the complement of event E?**Slot 0**Slot 1 Slot 2 Slot 3 Slot 4 Slot 5 Slot 6 Slot 7 e.g.10 (Page 27) • Consider that we hash a list of 6 keys into a hash table with 8 locations (or slots) (namely, slot 0, slot 1, slot 2, slot 3, …., slot 7)? e.g. we have 6 keys: 3, 12, 15, 8, 11, 5 8 3 12 15 8 11 5 3 11 Slot 3 Slot 4 Slot 7 Slot 0 Slot 3 Slot 5 12 A 6-tuple 3, 12, 15, 8, 11, 5 5 Each li is a number between 0 and 7(or between 1 and 8) Outcome: (3, 4, 7, 0, 3, 5) (l1, l2, l3, l4, l5, l6) 15 86 How many possible 6-tuples are there?**e.g.11 (Page 29)**• There are 20 locations/slots in the hash table. • Suppose that we are given 3 keys. • We can represent the slot no. of these 3 keys by a 3-tuple, (l1, l2, l3) • We want that these 3 keys have no collisions. • That is, these 3 keys are hashed to different locations/slots. • How many possible 3-tuples without collisions are there? We know that l1, l2 and l3 are all not equal.**1**1 1 2 2 2 … … … 20 20 20 l2 l1 l3 20 choices 19 choices 18 choices = 203 Total no. of 3-tuples without collisions = 20 x 19 x 18**e.g.12 (Page 30)**• There are 20 locations/slots in the hash table. • Suppose that we are given 3 keys. • We can represent the slot no. of these 3 keys by a 3-tuple, (l1, l2, l3) • Let A be the event that all 3 keys have no collisions. • What is the probability that A occurs? Total no. of 3-tuples without collisions = 20 x 19 x 18 Total no. of 3-tuples = 20 x 20 x 20 Probability that A occurs = (20 x 19 x 18)/(20 x 20 x 20) = 0.855**Slot 0**Slot 0 Slot 1 Slot 1 Slot 2 Slot 2 Slot 3 Slot 3 Slot 4 Slot 4 Slot 5 Slot 5 … … Slot 19 Slot 19 e.g.13 (Page 30) Case (i) Case (ii) 5 3 15 5 3 15**20n+1**20n+1 20 x 19 x 18 x … x (20 – (n-1)) x (20 – n) = 20n x 20 20n 20 x 19 x 18 x … x (20 – (n-1)) (20 – n) = x 20n 20n 20 20n (20 – n) This is because n x = 1 – (20 – n) 20n n 20 20 < 1 = pn x (1 – ) = pn x 20 20 e.g.14 (Page 33) • Given pn = • show that pn+1 < pn pn+1 = < pn**FTT**TTF TTT TFT Sample Space Event TTF TTT FFF FTF TFF FFT e.g.15 (Page 35) • Suppose that there are three questions in the final exam. • Let E be the event that the first two answers are true. P(E) = 2/8 = 1/4**e.g.16 (Page 36)**• Probability • Rule 1: P(A) 0 for all A S • Rule 2: P(S) = 1 • Rule 3: P(A U B) = P(A) + P(B)for any two disjoint events A and B