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Chapter 2. Foundations of Probability Section 2.3. Definition of Probability Jiaping Wang Department of Mathematical Science 01/16/2013, Wednesday The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Outline Introduction Definition of Probability Inclusive-Exclusive Principle Homework #1 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 1. Examples The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Introduction Intuitive idea of probability is related to relative frequency of occurrence. For example, in flipping a coin, we will think the probability to obtain head is ½ and same for obtaining tail. Another example like rolling a die, the probability to obtain a number among {1,2,3,4,5,6} would be 1/6 for each number, so that the probability to obtain the even number is ½ = 1/6+1/6+1/6 and same for obtaining the odd number. Kolmogorov proposed the three axioms to define the probability in Foundations of the Calculus of Probabilities. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 2. Definition of Probability The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Definition 2.3 Suppose that a random experiment has associated with a sample space S. A probability is a numerically valued function that assigned a number P(A) to every event A so that the following axioms hold: (1) P(A) ≥ 0 (2) P(S) = 1 (3) If A1, A2, … is a sequence of mutually exclusive events (that is Ai∩Aj=ø for any i≠j), then The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Some Basic Properties 1. P( ø ) = 0, P(S) = 1. 2. 0≤ P(A) ≤1for any event A. 3. P(AUB) = P(A) + P(B) if A and B are mutually exclusively. 4. P(AUB) = P(A) + P(B) – P(A∩B) for general events A and B. 5. If A is a subset of B, then P(A) ≤ P(B). 6. P(A) = 1 – P(A). 7. P(A∩B) = P(A) – P(A∩B). The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 3. Inclusive-Exclusive Principle The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Union of More Events We know P(AUB) = P(A) + P(B) – P(A∩B), how about P(AUBUC)? Consider a new event D=BUC, then P(AUBUC) = P(A)+P(D)-P(A∩D) =P(A)+P(BUC)-P((A∩B)U(A∩C)) =P(A)+P(B)+P(C) – P(B∩C) – [P(A∩B) +P(A∩C) - P(A∩B∩A∩C)] =P(A)+P(B)+P(C) – P(B∩C) – P(A∩B)- P(A∩C) + P(A∩B∩C) How about P(AUBUCUD) and P(A1UA2UA3UA4U….. UAn) for any integer n>0? The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Inclusive-Exclusive Principle Theorem 2.1. For events A1, A2, …, An from the sample space S, We can use induction to prove this. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Determine the Probability Values The definition of probability only tells us the axioms that the probability function must obey; it doesn’t tell us what values to assign to specific event. The value of the probability is usually based on empirical evidence or on careful thought about the experiment. For example, if a die is balanced, then we may think P(Ai)=1/6 for Ai={ i }, i = 1, 2, 3, 4, 5, 6 However, if a die is not balanced, to determine the probability, we need run lots of experiments to find the frequencies for each outcome. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Example 2.3 A farmer has decided to plant a new variety of corn on some of his land, and he has narrowed his choice to one of three varieties, which are numbered 1, 2, and 3. All three varieties have produced good yields in variety trials. Which corn variety produces the greatest yield depends on the weather. The optimal conditions for each are equally to choose, the farmer writes the name of each variety on a peace of paper, mixes the pieces, and blindly selects one. The variety that is selected is purchased and planted. Let Ei denote the event that variety i is selected (i = 1, 2, 3), let A denote the event that variety 2 or 3 is selected, and let B denote the event that variety 3 is not selected. Find the probabilities of Ei, A and B. Solution: Sample space S={1,2,3}, so E1={1}, E2={2} and E3={3}. Intuitively, it is reasonable to assign a probability 1/3 to each Ei, that is P(E1)=P(E2)=P(E3)=1/3 And A=E2UE3 , thus, P(A)=P(E2)+P(E3)=2/3, and P(B)=1-P(E3)=1-1/3=2/3. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Example 2.4 The National Crime Victimization Survey (NCVS) is an ongoing survey of a national representative sample of residential households. Twice each year, data are obtained from about 49,000 households, consisting of approximately 100,000 persons, on the frequency, characteristics, and consequences of victimization in the United States. Results for 1995 and 2002 are given in the following table. 19 95 20 02 Rates per 1000 persons Ages 18-24 *Based on 10 or fewer sample cases College NonStudents Stude nts College NonStudent Stude s nts Violent Crime 87.7 101.6 40.6 56.1 Rape/Sexual Assault 4.3 4.4 3.3 4.1 Robbery 8.4 12.1 2.9* 6.8 Aggravated Assault 14.5 22.2 9.1 13.2 Simple Assault 60.5 62.8 25.3 32.0 Serious Violent Crime 27.3 38.8 15.3 24.1 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Q&A Q1: For a randomly selected college student in 1995, what is the probability that the person was a victim of simple assault that year? A1: 60.5/1000=0.0605. Q2: For a randomly selected college student in 2002, what is the probability that the person was a victim of simple assault that year? A2: 25.3/1000=0.0253. Q3: For a randomly selected non-student in 2002 who was in the 18 to 24 age group, what is the probability that the person was not a victim of a violent crime that year? A3: 1-56.1/1000=1-0.0561 = 0.9439. Q4: Note that in table, the category “Violent Crime” has five subcategories, why do the rates in the five subcategories not total to the Violent Crime rate? A4: They are not mutually exclusively. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL More Example In a class of 60 students, 13 could not roll their tongue, 17 had attached earlobes and 10 could roll their tongues and had earlobes. A student is randomly selected from this class. Find the probability that the selected student: 1. Can roll his or her tongue; 2. Could not roll his or her tongue and had attached earlobes; 3. Could either roll his or her tongue or had attached earlobes but not both. Sample space S={60 students}, A={students could roll tongue}, B={students had attached earlobes}. P(A)=13/60 , P(B)=17/60, P(A∩B)=10/60 Answer 1: P(A)= 1-13/60 =47/60= 0.783 Answer 2: P(A ∩B)= P(B)-P(A ∩B)=17/60-10/60=7/60=0.117 Answer3: P[(A ∩B)U(A ∩B)]=P(A ∩B)+P(A ∩B)=7/60+P(A)-P(A ∩B)=7/60+47/60-10/60 =44/60=11/15=0.733 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Homework #1 Page 20: 2.8, 2.10 Page 21: 2.14, 2.15 Page 29: 2.18, 2.20 Page 31: 2.27, 2.28 Due next Wed., 01/23/2013 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL