Sets. Review of basic probability. Tuples. E = {all even integers} S = {x E : x is a multiple of 3} CSE 101. I = [0, 1] = {x : 0 x 1}

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1 Sets A = {a, b, c,..., z} A = 26 Review of basic probability = {0, 1} = 2 E = {all even integers} E = CSE 1 S = {x E : x is a multiple of 3} I = [0, 1] = {x : 0 x 1} In a set, the order of elements doesn t matter: {0, 1, 2} = {2, 0, 1} and there are no duplicates. Tuples Let C = {H, T }. All pairs of elements from C: {(H, H), (H, T ), (T, H), (T, T )} = C C = C 2 All triples of elements of C: {(H, H, H), (H, H, T ), (H, T, H),...} = C C C = C 3 All sequences of elements from C: denoted C = C C C. How many sequences of length are there? C = C = 2. In a sequence, the order of elements matters: Let A = {a, b, c,..., z}. How many sequences of length 2? 26 2 How many sequences of length? 26 How many sequences of length n? 26 n An alien language has an alphabet of size. Every sequence of 5 of these characters is a valid word. How many words are there in this language? = = (H, T ) (T, H).

2 Union and intersection A A = {any element in A or in or in both} A = {any element in A and in } M = {2, 3, 5, 7, 11} and N = {1, 3, 5, 7, 9} M N = {1, 2, 3, 5, 7, 9, 11} M N = {3, 5, 7} S = {all even integers} and T = {all odd integers} S T = {all integers} S T = Permutations How many ways to order the three letters A,, C? AC, AC, AC, CA, CA, CA 3 choices for the first, 2 choices for the second, 1 choice for the third = 6. Call this 3! How many ways to order A,, C, D, E? = 5! = 120 How many ways to place 6 men in a line-up? = 6! = 720 How many possible outcomes of shuffling a dec of cards? 52! General rule: The number of ways to order n distinct items is: n! = n(n 1)(n 2) 1. Combinations An ice-cream parlor has flavors {chocolate, vanilla, strawberry, pecan}. You are allowed to pic two of them. How many options do you have? CV, CS, CP, VS, VP, SP In general, the number of ways to pic items out of n is: ( ) n n! n(n 1) (n + 1) = = (n )!!! For instance, ( ) 4 2 = 4 3 2! = 6. How many ways to pic three ice-cream flavors? ( ) 4 = 4 3 Pic any 4 of your favorite 0 songs. How many ways to do this? ( ) = Probability spaces How to interpret a statement lie: The chance of getting a flush in a five-card poer hand is about 0.20%. (Flush = five of the same suit.) The underlying probability space has two components: 1. The sample space (the space of outcomes). In the example, Ω = {all possible five-card hands}. 2. The probabilities of outcomes. In the example, all hands are equally liely: probability 1/ Ω. Note: ω Ω Pr(ω) = 1. Event of interest: the set of outcomes A = {ω : ω is a flush} Ω. Pr(A) = ω A Pr(ω) = A Ω Ω A

3 Examples Roll three dice. What is the chance that their sum is 3? Sample space Ω = {(1, 1, 1), (1, 1, 2), (1, 1, 3),..., (6, 6, 6)} Roll a die. What is the chance of getting a number > 3? Sample space Ω = {1, 2, 3, 4, 5, 6}. Probabilities of outcomes: Pr(ω) = 1 6. Event of interest: A = {4, 5, 6} Pr(A) = Pr(4) + Pr(5) + Pr(6) = 1 2. = Ω o Ω o Ω o where Ω o = {1, 2, 3, 4, 5, 6}. Probabilities of outcomes: Pr(ω) = 1 Ω = Event of interest: A = {(1, 1, 1)}. Pr(A) = Roll n dice. Then Ω = Ω o Ω o = Ω n o, where Ω o = {1, 2, 3, 4, 5, 6}. What is Ω? 6 n. Probability of an outcome: Pr(ω) = 1 6 n. Socs in a drawer. A drawer has three blue socs and three red socs. You put your hand in and pull out two socs at random. What is the probability that they match? Thin of grabbing one soc first, then another. Ω = {(, ), (, R), (R, ), (R, R)} = {, R} 2. Probabilities: Pr((, )) = = 1 5 Pr((, R)) = = 3 Pr((R, )) = = 3 Pr((R, R)) = = 1 5 Event of interest: A = {(, ), (R, R)}. Pr(A) = 2 5. Socs in a drawer, cont d. This time the drawer has three blue socs and two red socs. You put your hand in and pull out two socs at random. What is the probability that they match? Sample sample space, Ω = {(, ), (, R), (R, ), (R, R)} = {, R} 2. Different probabilities: Pr((, )) = = 3 Pr((, R)) = = 3 Pr((R, )) = = 3 Pr((R, R)) = = 1 Event of interest: A = {(, ), (R, R)}. Pr(A) = 2 5.

4 Toss a fair coin times. What is the chance none are heads? Shuffle a pac of cards. Sample space Ω = {all possible orderings of 52 cards}. What is Ω? 52! = Sample space Ω = {H, T }. It includes, for instance, (H, T, H, T, H, T, H, T, H, T ). What is Ω? 2 = 24. For any sequence of coin tosses ω Ω, we have Pr(ω) = Event of interest: A = {(T, T, T, T, T, T, T, T, T, T )}. Pr(A) = What is the probability of exactly one head? Event of interest: A = {ω Ω : ω has exactly one H}. What is A?. Each sequence in A can be specified by the location of the one H, and there are choices for this. What is Pr(A)? 24. Toss a fair coin times. What is the chance of exactly two heads? Again, sample space Ω = {H, T }, with Ω = 2 = 24. For any sequence of coin tosses ω Ω, we have Pr(ω) = Event of interest: A = {ω Ω : ω has exactly two H s}. What is A? ( ) 2 = 9 2 = 45. Each sequence in A can be specified by the locations of the two H s and there are ( ) 2 choices for these locations. What is Pr(A)? What is the probability of exactly heads? Event of interest: A = {ω Ω : ω has exactly H s}. What is A? ( ). What is Pr(A)? ( ) /24. irthday paradox. A room contains people. What is the chance that they all have different birthdays? Number the people 1, 2,...,. Number the days of the year 1, 2,..., 5. Let ω = (ω 1,..., ω ), where ω i {1, 2,..., 5} is the birthday of person i. Thus Ω = {1, 2,..., 5}. What is Ω? 5. Event of interest: A = {(ω 1,..., ω ) : all ω i different}. What is A? (5 + 1). Therefore, Pr(A) = 5 4 (5 + 1) 5 = For = 23, this is less than 1/2. In other words, in a group of 23 random people, chances are some pair of them have a common birthday!

5 People s probability judgements Experiment by Kahneman-Tversy. Subjects were told: Linda is 31, single, outspoen, and very bright. She majored in philosophy in college. As a student, she was deeply concerned with racial discrimination and other social issues, and participated in anti-nuclear demonstrations. They were then ased to ran three possibilities: (a) Linda is active in the feminist movement. (b) Linda is a ban teller. (c) Linda is a ban teller and is active in the feminist movement. Over 85% respondents chose (a) > (c) > (b). ut: Pr(ban teller, feminist) Pr(ban teller). F In a city, 60% of people have a car, 20% of people have a bie, and % of people have a motorcycle. Anyone without at least one of these wals to wor. What is the minimum fraction of people who wal to wor? Let Ω = {people in the town}. Let C = {has car}, = {has bie}, M = {has motorcyle}, W = {wals}. General picture: C M Pr(W ) 1 Pr(C M) Pr(C M) Pr(C) + Pr() + Pr(M) and thus Pr(W ) 0.1. = = 0.9 M C Complements and unions The complement of an event. Let Ω be a sample space and E Ω an event. Write E c for the event that E does not occur, that is, E c = Ω \ E. The union bound. For any events E 1,..., E : Pr(E c ) = 1 Pr(E). Pr(E 1 E ) Pr(E 1 ) + + Pr(E ). This inequality is exact when the events are disjoint. Coupon-collector problem Each cereal box has one of action figures. How many boxes do you need to buy so that you are liely to get all figures? Say we buy n boxes. Let A i be the event that the ith action figure is not obtained. Pr(A i ) = Pr(not in 1st box) Pr(not in 2nd box) Pr(not in nth box) ( = 1 1 ) n e n/ y union bound, the probability of missing some figure is Pr(A 1 A ) Pr(A 1 ) + + Pr(A ) e n/. Setting n ln 2 maes this 1/2. Therefore: enough to buy O( log ) cereal boxes.

6 Independence Two events A, are independent if the probability of occurring is the same whether or not A occurs. Example: toss two coins. A = {first coin is heads} = {second coin is heads} A Not independent Formally, we say A, are independent if Pr(A ) = Pr(A)Pr(). Independent or not? 1. You have two children. A = {first child is a boy}, = {second child is a girl}. Independent. A Possibly independent 2. You throw two dice. A = {first is a six}, = {sum > }. Not independent. 3. You get dealt two cards at random from a dec of 52. A = {first is a heart}, = {second is a club}. Not independent. A Not independent Random variables Roll a die. Define X = { 1 if die is 3 0 otherwise Here the sample space is Ω = {1, 2, 3, 4, 5, 6}. Roll n dice. ω = 1, 2 X = 0 ω = 3, 4, 5, 6 X = 1 X = # of 6 s Y = # of 1 s before the first 6 oth X and Y are defined on the same sample space, Ω = {1, 2, 3, 4, 5, 6} n. For instance, ω = (1, 1, 1,..., 1, 6) X = 1, Y = n 1. The distribution of a random variable Roll a die. Define X = 1 if die is 3, otherwise X = 0. X taes values in {0, 1} and has distribution: Pr(X = 0) = 1 3 and Pr(X = 1) = 2 3. Roll n dice. Define X = number of 6 s. X taes values in {0, 1, 2,..., n}. The distribution of X is: Pr(X = ) = #(sequences with 6 s) Pr(one such sequence) ( ) ( ) ( ) n n 1 5 = 6 6 In general, a random variable (r.v.) is a defined on a probability space. It is a mapping from Ω to R. We ll use capital letters for r.v. s.

7 Expected value, or mean The expected value of a random variable X is E(X ) = x Roll a die. Let X be the number observed. x Pr(X = x). E(X ) = = = 3.5 (average) 6 iased coin. A coin has heads probability p. Let X be 1 if heads, 0 if tails. E(X ) = 1 p + 0 (1 p) = p. Toss a coin with bias p repeatedly, until it comes up heads. Let X be the number of tosses. Pascal s wager Pascal: I thin there is some chance (p > 0) that God exists. Therefore I should act as if he exists. Let X = my level of suffering. Suppose I behave as if God exists (that is, I behave myself). Then X is some significant but finite amount, lie 0 or 00. Suppose I behave as if God doesn t exists (I do whatever I want to). If indeed God doesn t exist: X = 0. ut if God exists: X = (hell). Therefore, E(X ) = 0 (1 p) + p =. The first option is much better! E(X ) = 1 p. Linearity of expectation Linearity: examples Roll 2 dice and let Z denote the sum. What is E(Z)? If you double a set of numbers, how is the average affected? It is also doubled. If you increase a set of numbers by 1, how much does the average change? It also increases by 1. Rule: E(aX + b) = ae(x ) + b for any random variable X and any constants a, b. ut here s a more surprising (and very powerful) property: E(X + Y ) = E(X ) + E(Y ) for any two random variables X, Y. Liewise: E(X + Y + Z) = E(X ) + E(Y ) + E(Z), etc. Method 1 Distribution of Z: z Pr(Z = z) Now use formula for expected value: E(Z) = = 7. Method 2 Let X 1 be the first die and X 2 the second die. Each of them is a single die and thus (as we saw earlier) has expected value 3.5. Since Z = X 1 + X 2, E(Z) = E(X 1 ) + E(X 2 ) = =

8 Toss n coins of bias p, and let X be the number of heads. What is E(X )? Let the individual coins be X 1,..., X n. Each has value 0 or 1 and has expected value p. Since X = X 1 + X X n, E(X ) = E(X 1 ) + + E(X n ) = np. Roll a die n times, and let X be the number of 6 s. What is E(X )? Let X 1 be 1 if the first roll is a 6, and 0 otherwise. Liewise, define X 2, X 3,..., X n. Since X = X X n, we have E(X 1 ) = 1 6. E(X ) = E(X 1 ) + + E(X n ) = n 6. Coupon collector, again Each cereal box has one of action figures. What is the expected number of boxes you need to buy in order to collect all the figures? Suppose you ve already collected i 1 of the figures. Let X i be the time to collect the next one. Each box you buy will contain a new figure with probability ( (i 1))/. Therefore, E(X i ) = i + 1. Total number of boxes bought is X = X 1 + X X, so E(X ) = E(X 1 ) + E(X 2 ) + + E(X ) = ( = ) ln. Independent random variables Random variables X, Y are independent if Pr(X = x, Y = y) = Pr(X = x)pr(y = y). Independent or not? Pic a card out of a standard dec. X = suit and Y = number. Independent. Flip a fair coin n times. X = # heads and Y = last toss. Not independent. X, Y tae values { 1, 0, 1}, with the following probabilities: Y X Independent. X Y