STAT 516: Basic Probability and its Applications Lecture 2: Probability Set Function Prof. Michael Levine January 20, 2015 Levine STAT 516: Basic Probability and its Applications Axioms of Probability I For all C ∈ B, P(C ) ≥ 0 I P(C) = 1 I If {Cn } is a sequence of events in B and Cm ∩ Cn = ∅ for all m 6= n, ∞ X C ) = P(Cn ) P (∪∞ n n=1 n=1 Levine STAT 516: Basic Probability and its Applications Countable Additivity Axiom I The last axiom above is called the countable additivity axiom I The following is implied by the countable additivity axiom... I Let A1 ⊃ A2 ⊃ A3 ⊃ · · · be an infinite family of subsets of a sample space Ω s.t. An ↓ A I Then, P(An ) → P(A) as n → ∞ Levine STAT 516: Basic Probability and its Applications Countable Additivity Axiom I To prove, note if Bi = Aci , we have B1 ⊂ B2 ⊂ B3 · · · .. and Bn ↑ B I It is enough to show that P(Bn ) → P(B) as n → ∞ I For any fixed n, Bn = ∪ni=1 (Bi − Bi−1 ) I c In the above, B0 = ∅ and Bi − Bi−1 = Bi ∩ Bi−1 I The needed result immediately follows... I Note that if finite additivity and the above result are assumed as axioms, then countable additivity follows as a consequence Levine STAT 516: Basic Probability and its Applications Simple and Compound Events I An event is simple if it consists of exactly one outcome and compound otherwise. I Example: consider recording the freeway exit (L or R) for each of the 3 vehicles at the end of the exit ramp. The eight possible outcomes form the sample space B = {LLL, RLL, LRL, LLR, LRR, RLR, RRL, RRR}. Examples of simple events would be: I I I E1 = LLL E2 = LRL Levine STAT 516: Basic Probability and its Applications Examples of Compound Events I Examples of compound events would be: I I Exactly one vehicle turns left A = {LRR, RLR, RRL} All three vehicles turn in the same direction B = {LLL, RRR} Levine STAT 516: Basic Probability and its Applications Disjoint Events I I When A and B have no outcomes in common, they are said to be mutually exclusive or disjoint events. I Example: when rolling a die, the event A = {2, 4, 6} (evens ) and B = {1, 3, 5} (odds) are mutually exclusive I Example: when pulling a single card from a standard deck of cards, events A =heart, diamond(red) and B =spade, club(black) are mutually exclusive. Levine STAT 516: Basic Probability and its Applications Disjoint Events II I A collection of events {Cn } is exhaustive if the union of these events is a sample space so that ∞ X P(Cn ) = 1 n=1 I A mutually exclusive and exhaustive collection of events forms a partition of C I Example: events A and B from the card example form a partition of the sample space Levine STAT 516: Basic Probability and its Applications Simplest Properties of Probability I For each C ∈ B 0 P(C ) = 1 − P(C ) I The probability of the null set is zero: P(∅) = 0 I If C1 ⊂ C2 we have P(C1 ) ≤ P(C2 ) I For each event C ∈ B 0 ≤ P(C ) ≤ 1 Levine STAT 516: Basic Probability and its Applications Equilikely Case I I Let C1 , . . . , Ck be mutually exclusive and exhaustive subsets. I Let each of Ci ,i = 1, . . . , k have the same probability P(Ci ) = k1 I For any E = C1 ∪ C2 ∪ . . . ∪ Cr with r ≤ k P(E ) = r X i=1 Levine P(Ci ) = r k STAT 516: Basic Probability and its Applications Example I I During off-peak hours a commuter train has five cars. Suppose a commuter is twice as likely to select the middle car #3 as to select either adjacent car (#2 or #4) , and is twice as likely to select either adjacent car as to select either end car (#1 or #5). I If the probability of selecting car i is pi , we have p3 = 2p2 = 2p4 , and p2 = 2p1 = 2p5 = p4 P Therefore, 1 = pi = 10p1 and p1 = p5 = 0.1, p2 = p4 = 0.2 and p3 = 0.4. I I The probability that one of the three middle cars is selected is, then, p2 + p3 + p4 = 0.8. Levine STAT 516: Basic Probability and its Applications Counting rules I A multiplication rule or the mn-rule I The number of permutations of k elements out of n is Pkn = n(n − 1) · · · (n − (k − 1)) = n! (n − k)! I The number of combinations of k elements out of n is n n! = k k!(n − k)! I The binomial expansion n X n k n−k (a + b) = a b k n k=0 Levine STAT 516: Basic Probability and its Applications Probability set function for a finite set and loaded dice I Let C consist of k distinct points (outcomes) I Assign the probability of I For any event C ∈ C define P(C ) = 1 k to each of these outcomes number of points in C X = f (x) k x∈C with f (x) = 1 k for x ∈ C I With loaded dice, with C = {1, 2, 3, 4, 5, 6} we may have x f (x) = 21 I Then, for C = {1, 2, 3} we have P(C ) = 1 2 3 2 + + = 21 21 21 7 Levine STAT 516: Basic Probability and its Applications Counting rules I The number of ways of linearly arranging n distinct objects with order of arrangement matters is n! I The number of ways of choosing r objects from n distinct objects if the same object could be chosen repeatedly is nr I The number of ways of distributing n distinct objects into k distinct categories when order in which the distributions are made is not important, ni being the number of objects allocated to the ith category is n n−n1 n−n1 −n2 −···−nk−1 n! ··· n1 !n2 !···nk ! = n1 n2 nk Levine STAT 516: Basic Probability and its Applications Example I A carton of eggs has 12 eggs of which three are bad. We don’t know that fact in advance I We want to make a three-egg omelet; we can select three eggs in 12 3 = 220 ways I Assume that three eggs are selected completely at random; the number of ways to select them out of 9 good eggs is 93 = 84 I The probability of the favorable outcome is Levine 84 220 = 0.38 STAT 516: Basic Probability and its Applications Example I Six different cookies are distributed completely at random to six children. It is possible for a child to get more than one cookie I This implies that there are 66 = 46, 656 possible outcomes (sample points). I Let A = {Each child receives exactly one cookie}. There are 720 = 0.015 6! = 720 ways of doing so and thus P(A) = 46656 I Then, P(A ) = 1 − 0.015 = 0.985 0 Levine STAT 516: Basic Probability and its Applications Example I I I Out of 5 pairs of shoes in the closet, we select 4 shoes at random. What is the probability that we will have at least one complete pair among them? The total number of sample points is 10 4 = 210 We can have either two complete pairs, selected in 52 = 10 ways or... I Or one pair and two nonconforming shoes selected complete in 51 42 × 2 × 2 = 120 ways I The probability of the event of interest is, then, 10 + 120 = .62 210 Levine STAT 516: Basic Probability and its Applications Example-a five card poker I A player is given 5 cards from a full deck of 52 cards at random I Events of interest are A = { two pairs } and B = a flush I Two pairs is a hand with two cards each of two different denominations and the fifth card of some other denomination I Thus, h 4i2 P(A) = 13 2 Levine 2 52 5 44 1 = 0.04754 STAT 516: Basic Probability and its Applications Example - a 5 card poker I A flush is a hand with 5 cards of the same suit that are not in a sequence I There are 10 ways to select five cards from a suit such that the cards are in a sequence I Thus, The probability of flushes is 13 4 − 10 1 5 = .00197 P(B) = 52 5 Levine STAT 516: Basic Probability and its Applications Inclusion-Exclusion Formula I P(C1 ∪ C2 ) = P(C1 ) + P(C2 ) − P(C1 ∩ C2 ) I The general inclusion-exclusion formula P(C1 ∪ C2 ∪ . . . ∪ Ck ) = n X i=1 + X P(Ci ) − X P(Ci ∩ Cj ) 1≤i<j≤n P(Ci ∩ Cj ∩ Ck ) − · · · + (−1)k+1 P(C1 ∩ C2 ∩ · · · ∩ Ck ) 1≤i<j<k≤n Levine STAT 516: Basic Probability and its Applications Example II I In a residential suburb, 60% of all households get Internet service from a local cable company, 80% get television service from that company, and 50% get both services from that company I The probability that a randomly selected household subscribes to at least one of these two services from the local company is P(C1 ∪ C2 ) = P(C1 ) + P(C2 ) − P(C1 ∩ C2 ) = 0.9 Levine STAT 516: Basic Probability and its Applications Example II I The probability that a household subscribes only to TV service is 0 P(C1 ∩ C2 ) = P(C1 ∪ C2 ) − P(C1 ) = 0.9 − 0.6 = 0.3, I The probability that a household subscribes only to Internet is 0 P(C1 ∩ C2 ) = P(C1 ∪ C2 ) − P(C2 ) = 0.1, I The probability that a household subscribes to exactly one of these services from the local company is 0 0 P(C1 ∩ C2 ) + P(C1 ∩ C2 ) = 0.1 + 0.3 = 0.4 Levine STAT 516: Basic Probability and its Applications Missing faces in dice rolls I A fair die is rolled n times. What is the probability that at least one of six sides never shows up in these n rolls? I Let Ai be the side that never shows up, 1 ≤ i ≤ 6 I If all of 6n outcomes are equally likely, P(Ai ) = n n P(Ai ∩ Aj ) = 46n , P(Ai ∩ Aj ∩ Ak ) = 36n etc I Thus, the needed probability is n n 6 5 6 3 pn = − + ··· 1 6n 2 6n I Manual calculation gives p10 = 0.73, p12 = 0.56, p13 = 0.49, p15 = 0.36 I Note that it takes 13 rolls of a fair die to have better than 50% chance (1 − pn ) that each of the six sides shows up Levine 5n 6n , STAT 516: Basic Probability and its Applications Bonferroni bounds and inequality I Pn P Denote PS1 = i=1 P(Ci ), S2 = 1≤i<j≤n P(Ci ∩ Cj ), S3 = 1≤i<j<k≤n P(Ci ∩ Cj ∩ Ck ) etc I Then, the inclusion-exclusion formula becomes P(∪ni=1 Ci ) = S1 − S2 + S3 − · · · I Let pn = P(∪ni=1 Ci ). The following series of bounds is true: pn ≤ S1 ; pn ≥ S1 − S2 ; pn ≤ S1 − S2 + S3 ; · · · I Also, (Bonferroni inequality) P(∩ni=1 Ci ) ≥1− n X 0 P(Ci ) i=1 Levine STAT 516: Basic Probability and its Applications Example I Let P(Ci ) ≥ 0.95 for i = 1, . . . , 10. This means that 0 P(Ai ) ≤ 0.05 for each i I By Bonferroni inequality, P(∩ni=1 Ai ) ≥ 1 − 10 ∗ 0.05 = 0.5 I Therefore, despite individual probabilities of at least 95%, we can only be 50% sure that all 10 of events will occur I This is a rather crude lower bound - better options are available Levine STAT 516: Basic Probability and its Applications Continuity theorem for probabilities I A sequence of events Cn is non-decreasing if Cn ⊂ Cn+1 for all n I Call limn→∞ Cn = ∪∞ n=1 Cn I Theorem (this is called the continuity from below): lim P(Cn ) = P( lim Cn ) = P (∪∞ n=1 Cn ) n→∞ n→∞ I For a sequence An s.t. An ⊃ An+1 (a non-increasing set sequence) the limit is limn→∞ An = ∩∞ i=1 Ai I The continuity from above: lim P(An ) = P( lim An ) = P (∩∞ n=1 An ) n→∞ n→∞ Levine STAT 516: Basic Probability and its Applications Bool’s inequality I Let Cn be an arbitrary sequence of events. I P (∪∞ n=1 Cn ) ≤ ∞ X P(Cn ) n=1 Levine STAT 516: Basic Probability and its Applications
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