Prop erties of ProbabilityJuana [email protected] Department of StatisticsRecommended reference: Chapter 2 in Ross.J. Sanchez Properties of ProbabilityOutlineProperties of Probabilities of eventsJ. Sanchez Properties of ProbabilityProp erties of ProbabilitiesTheoremLet P be a probability function and let A be any event. Then(a) P(∅) = 0(b) P(Ac) = 1 − P(A)J. Sanchez Properties of ProbabilityPro of of prop erty (a)If we consider a sequence of events E1, E2, ..... whereE1= S , Ei= ∅, i > 1, then as the events are mutually exclusive and asS =S∞i =1Ei, we have from Axiom 3 thatP(S) =∞Xi =1P(Ei) = P(S) +∞Xi =2P(∅)implying thatP(∅) = 0.That is, the null event has probability 0 of occurring.J. Sanchez Properties of ProbabilityIt also follows that for any finite sequence of mutually exclusive eventsE1, E2, ...., Enthe probability of the union of n events is the sum of theprobabilities of each event.P (∪Ei) =nXi =1P(Ei).This follows from Axiom 3 by defining Eito be the null event for allvalues of i greater than n. Axiom 3 is equivalent to the equation abovewhen the sample space is finite. However, the added generality of Axiom3 is necessary when the sample space consists of an infinite number ofpoints.J. Sanchez Properties of ProbabilityTheoremIf P is a probability function and A and B are any sets, thenP(B ∩ Ac) = P(B) − P(A ∩ B). Probability that only B happens.P(A ∪ B) = P(A) + P(B) − P (A ∩ B). Additive ruleIf A ⊂ B → P (A) < P(B)Proof: Can you prove the first statement of this theorem?Make Bunion of two mutually exclusive eventsB = AB ∪ BAcSo by axiom 3,P(B) = P(AB ) + P(BAc)Solve for P(BAc)P(BAc) = P(B) − P(AB)J. Sanchez Properties of ProbabilityTheoremIf P is a probability function and A and B are any sets, thenP(B ∩ Ac) = P(B) − P(A ∩ B). Probability that only B happens.P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Additive ruleIf A ⊂ B → P(A) < P(B)Proof: Can you prove the first statement of this theorem? Make Bunion of two mutually exclusive eventsB = AB ∪ BAcSo by axiom 3,P(B) = P(AB ) + P(BAc)Solve for P(BAc)P(BAc) = P(B) − P(AB)J. Sanchez Properties of ProbabilityTheoremIf P is a probability function and A and B are any sets, thenP(B ∩ Ac) = P(B) − P(A ∩ B). Probability that only B happens.P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Additive ruleIf A ⊂ B → P(A) < P(B)Proof: Can you prove the first statement of this theorem? Make Bunion of two mutually exclusive eventsB = AB ∪ BAcSo by axiom 3,P(B) = P(AB ) + P(BAc)Solve for P(BAc)P(BAc) = P(B) − P(AB)J. Sanchez Properties of ProbabilityTheoremIf P is a probability function and A and B are any sets, thenP(B ∩ Ac) = P(B) − P(A ∩ B). Probability that only B happens.P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Additive ruleIf A ⊂ B → P(A) < P(B)Proof: Can you prove the first statement of this theorem? Make Bunion of two mutually exclusive eventsB = AB ∪ BAcSo by axiom 3,P(B) = P(AB ) + P(BAc)Solve for P(BAc)P(BAc) = P(B) − P(AB)J. Sanchez Properties of ProbabilityCan you prove the second statement?ProofJ. Sanchez Properties of ProbabilityCan you prove the generalization of the second statement to 3events?We must prove now thatP(A∪B ∪C ) = P(A)+P(B)+P(C )−P(AB)−P(AC )−P(CB)+P(ABC )P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) but byearlierproof= P(A ∪ B) + P(C) − P((A ∪ B)C) use distributive law in last term= P(A) + P(B) − P(AB) + P(C) − P(AC ∪ BC)= P(A) + P(B) − P(AB) + P(C) − P(AC ) − P(BC ) + P(ACBC)= P(A) + P(B) + P(C) − P(AC) − P(BC) − P(AB) + P(ABC)J. Sanchez Properties of ProbabilityCan you prove the generalization of the second statement to 3events?We must prove now thatP(A∪B ∪C ) = P(A)+P(B)+P(C )−P(AB)−P(AC )−P(CB)+P(ABC )P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) but byearlierproof= P(A ∪ B) + P(C) − P((A ∪ B)C) use distributive law in last term= P(A) + P(B) − P(AB) + P(C) − P(AC ∪ BC)= P(A) + P(B) − P(AB) + P(C) − P(AC ) − P(BC ) + P(ACBC)= P(A) + P(B) + P(C) − P(AC) − P(BC) − P(AB) + P(ABC)J. Sanchez Properties of ProbabilityCan you prove the generalization of the second statement to 3events?We must prove now thatP(A∪B ∪C ) = P(A)+P(B)+P(C )−P(AB)−P(AC )−P(CB)+P(ABC )P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) but byearlierproof= P(A ∪ B) + P(C) − P((A ∪ B)C) use distributive law in last term= P(A) + P(B) − P(AB) + P(C) − P(AC ∪ BC)= P(A) + P(B) − P(AB) + P(C) − P(AC ) − P(BC ) + P(ACBC)= P(A) + P(B) + P(C) − P(AC) − P(BC) − P(AB) + P(ABC)J. Sanchez Properties of ProbabilityCan you prove the generalization of the second statement to 3events?We must prove now thatP(A∪B ∪C ) = P(A)+P(B)+P(C )−P(AB)−P(AC )−P(CB)+P(ABC )P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) but byearlierproof= P(A ∪ B) + P(C) − P((A ∪ B)C) use distributive law in last term= P(A) + P(B) − P(AB) + P(C) − P(AC ∪ BC)= P(A) + P(B) − P(AB) + P(C) − P(AC ) − P(BC ) + P(ACBC)= P(A) + P(B) + P(C) − P(AC) − P(BC) − P(AB) + P(ABC)J. Sanchez Properties of ProbabilityCan you prove the generalization of the second statement to 3events?We must prove now thatP(A∪B ∪C ) = P(A)+P(B)+P(C )−P(AB)−P(AC )−P(CB)+P(ABC )P(A ∪ B ∪ C) = P((A ∪ B) ∪ C) but byearlierproof= P(A ∪ B) + P(C) − P((A ∪ B)C) use distributive law in last term= P(A) + P(B) − P(AB) + P(C) − P(AC ∪ BC)= P(A) + P(B) − P(AB) + P(C) − P(AC ) − P(BC ) + P(ACBC)= P(A) + P(B) + P(C) − P(AC) − P(BC) − P(AB) + P(ABC)J. Sanchez Properties of ProbabilityMore general resultThe probability of the union of n events equals the sum of theprobabilities of these events taken one at a time, minus the sum of theprobabilities of these events taken two at a time, plus the sum of theprobabilities of these events taken three at a time, and so on.P(E1∪ E2∪ · · · ∪ En) =nXi =1P(Ei) −Xi1<i2P(Ei1Ei2)+ (−1)r +1Xi1<i2<....<irP(Ei1, Ei2, ...., Eir)+ · · · + (−1)n+1P(Ei1Ei2, · · · , Ein)J. Sanchez Properties of ProbabilityThe summationXi1<i2<....<irP(E1, E2, ...., Er)is taken over all of thenrpossible subsets of size r of the set {1, 2, ..., n}.J. Sanchez Properties of ProbabilityExampleA total of 28 percent of American males smoke cigarettes, 7 percentsmoke cigars, and 5 percent smoke both cigars and cigarettes.What percentage of males smoke neither cigars nor cigarettes?Let A be the event that a randomly chosen person is a cigarettesmoker and let B be the event that she
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