Notes for Math 324, Part 12

Transcription

1 72 Notes for Math 324, Part 12

2 Chapter 12 Definition and main properties of probability 12.1 Sample space, events We run an experiment which can have several outcomes. The set consisting by all possible outcomes of the experiment is called the sample space S. An (subset) event is a collection of outcomes of the sample space S. Let S denote the collection of all possible events (subsets) of S. We denote events by A, B,... The event with no elements on it is called the empty set and it is denoted by. Example We throw a coin 3 times, the sample space is S = {(H, H, H), (H, H, T ), (H, T, H), (H, T, T ), (T, H, H), (T, H, T ), (T, T, H), (T, T, T )}, where for example (H, T, H) denotes the outcome that the first throw landed heads, the second throw landed tails and the third throw landed heads. An event is A = {(H, H, T ), (H, T, H), (T, H, H)}. We say that the event A is included in the event B if every outcome in A is also in B. We denote this by A B. There are several operations which can be done with events. Given two events A and B, the union of A and B is the event consisting by the outcomes which are either in A, or in B or in both A and B. The union of the events A and B is denoted by A B. Given events A i, i I, where I is an index set, the union of A i, i I, is the event consisting by all the outcomes which are in A i for some i I. The union of the events A i, i I, is denoted by i I A i. Given two events A and B, the intersection of A and B is the event consisting by all the outcomes which are in both A and B. The intersection of the events A and B is denoted by A B. Given sets A i, i I, where I is an index set, the intersection of A i, i I, is defined as the event consisting by all the outcomes which are in A i for each i I. The intersection of the events A i, i I, is denoted by i I A i. 73

3 74 CHAPTER 12. DEFINITION AND MAIN PROPERTIES OF PROBABILITY The complementary of an event A is the event consists by all outcomes which are not in A. The complementary of an event A is denoted by A c or by A. The main properties of unions and intersections are A B = B A. A (B C) = (A B) C. A B = B A. A (B C) = (A B) C. commutative property of the union associative property of the union commutative property of the intersection associative property of the intersection A (B C) = (A B) (A C). distributive property of the union... A ( i I B i ) = i I (A B i ).... with respect to the intersection A (B C) = (A B) (A C). distributive property of the intersection... A ( i I B i ) = i I (A B i ). (A B) c = A c B c. ( i I A i ) c = i I A c i. (A B) c = A c B c. ( i I A i ) c = i I A c i.... with respect to the union De Morgan s law De Morgan s law De Morgan s law De Morgan s law Two events A and B are said to be disjoint (or mutually exclusive) if A B =. Events A 1,..., A n are said to be disjoint (or mutually exclusive) if for each i j, A i A j = Definition of Probability Definition A probability function P is a function P : S [0, 1] such that: (i) P [S] = 1 (ii) If {A i } i=1 is a sequence of disjoint events, then P [ i=1a i ] = P [A i ]. Some probabilities are defined as followed: for each event A, P[A] = i=1 number of elements in A number of elements in S. In order to being able to define a probability in this way. S has be the finite. This type of probability functions are called probabilities withs equally likely outcomes. The main properties of a probability function are: 1. P [ ] = 0. 1 This definition is not rigorous. The real definition of probability function restricts the domain of P to a σ-field of S.

4 12.2. DEFINITION OF PROBABILITY If A B, then P [A] P [B]. 3. P [A c ] = 1 P [A]. 4. If A B =, then P [A B] = P [A] + P [B]. 5. P [A B] = P [A B c ] + P [A B] + P [A c B]. 6. P [A B] = P [A] + P [B] P [A B]. 7. P [A B C] = P [A B C] +P [A B C c ] + P [A B c C] + P [A c B C] +P [A c B c C] + P [A c B C c ] + P [A B c C c ] +P [A c B c C c ]. 8. P[A B C] = P[A] + P[B) + P [C] P (A B) P [A C] P [B C] + P [A B C]. Working with events, it is easier to set up their probabilities in either a table of a Venn diagram: Figure 12.1: Sets A and B B B c total A P [A B] P [A B c ] P [A] A c P [A c B] P [A c B c ] P [A c ] total P [B] P [B c ] 1 In the following problems it is easier to visualize what is going on using the diagram above. Example Prove that P [A B] = P [A] + P [B] P [A B].

5 76 CHAPTER 12. DEFINITION AND MAIN PROPERTIES OF PROBABILITY Solution: P [A B] counts the probability inside the circles A and B. P [A]+P [B] counts the probability inside the circles A B counting the probability in the intersection twice. So, P [A B] = P [A] + P [B] P [A B]. Example Doing Venn diagrams, prove that (A B) C = (A C) (B C). Figure 12.2: (A B) C Figure 12.3: (A C) (B C) Problem (# 1, May 2000) The probability that a visit to a primary care physician s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP s office, 30% are referred to specialists and 40% require lab work. Determine the probability that a visit to a PCP s office results in both lab work and referral to a specialist. Answer: 0.05 Solution: Let A = {visit results in referral to a specialist} and let B = {visit results in lab work}. We need to find P [A B]. We know that P [A c B c ] =

6 12.2. DEFINITION OF PROBABILITY 77 Figure 12.4: 1, May , P [A] = 0.30 and P [B] = We can put all this information on a graph 0.35 is the probability of the outside of the circles is the probability of the first circle is the probability of the second circle. Since 0.35 is the probability of the outside of the circles, the probability inside the circles is = Since the probability the circles are 0.30 and 0.40 respectively, the probability of the intersection of the two circles is = That is the probability we are looking for. Problem (# 1, Sample Test) A marketing survey indicates that 60% of the population owns an automobile, 30% owns a house, and 20% owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both. Answer: 0.5 Figure 12.5: 1, Sample Test Solution: Let A = {a person owns an automobile} and let B = {a person owns a house}. We need to find P [A B c ] + P [A c B]. We have that P [A] = 0.6, P [B] = 0.3 and P [A B] = 0.2. We are finding the probability inside the two circles removing the intersection. Since P [A B] = 0.2, the probability in the part of the

7 78 CHAPTER 12. DEFINITION AND MAIN PROPERTIES OF PROBABILITY first circle which does not intersect the second circle is = 0.4. The probability in the part of the second circle which does not intersect the first circle is = 0.1. So, the probability we are looking for is = 0.5. We also can do P [A B c ] + P [A c B] = P [A] + P [B] 2P [A B] = (2)(0.2) = Problem (# 9, November 2001) Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist. Determine the probability that a randomly chosen member of this group visits a physical therapist. Answer 0.48 Solution: Let A = {patient visits a physical therapist} and let B = {patient visits a chiropractor}. We need to find P[A]. We know that P [A B] = 0.22 and P [A c B c ] = Let x = P[A B c ], then P[A c B] = x We have that 1 = P [A B] + P(A B c ] + P[A c B] + P [A c B c ] = x + x = x. So, x = = 0.26 and P [A] = P [A B] + P [A B c )] = = Problem (# 12. May 2001) You are given P [A B] = 0.7 and P [A B ] = 0.9. Determine P [A]. Answer: 0.6 Solution: By taking the complementary of the events, P [A c B c ] = 1 P [A B] = = 0.3 and P [A c B] = 1 P [A B ] = = 0.1. So, P [A c ] = P [A c B] + P [A c B c ] = = 0.4 and P [A] = 1 P[A c ] = = 0.6. For three sets, we can proceed similarly. It helps a lot to draw a graph: Figure 12.6: Sets A, B and C Problem (# 3, November 2000) An auto insurance company has 10,000 policyholders. Each policyholder is classified as

8 12.2. DEFINITION OF PROBABILITY 79 (i) young or old; (ii) male or female; and (iii) married or single. Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The policyholders can also be classified as 1320 young males, 3010 married males, and 1400 young married persons. Finally, 600 of the policyholders are young married males. How many of the company s policyholders are young, female, and single? Answer: 880 Figure 12.7: 3, November 2000 Let A = {the policyholder is young}, let B = {the policyholder is male} and let C = {the policyholder is married}. We are looking for the number of elements in A B c C c. There are 600 elements A B C, which is the intersection of the 3 circles. Since there are 600 elements A B C, and 1320 elements in A B, the number of elements in A B C c is = 720. Since there are there are 600 elements A B C, and 3010 elements in B C, the number of elements in A c B C is = Since there are 600 elements A B C, and there are 1400 elements in A C, the number of elements in A B c C is = 800. Since there are 3000 elements in A, 600 elements in A B C, 800 in A B c C, and 720 in A B C c, we have that the number of elements in A B c C is = 880. Problem (# 31, May 2001) An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose exactly two of the supplementary coverages A, B, and C, or they may choose no supplementary coverage. The proportions of the company s employees that choose coverages A, B, and C are 1, 1, and respectively. Determine the probability that a randomly chosen employee will choose no 12 supplementary coverage. Answer: 1, 2 Solution: Doing a graph, we see that the only places with probability are A c B c C c,

9 80 CHAPTER 12. DEFINITION AND MAIN PROPERTIES OF PROBABILITY Figure 12.8: 31, May 2001 A B C c, A B c C and A c B C. So, P [A B C] = P[A] + P[B] + P[C] 2 and P [A c B c C c ] = 1 P [A B C] = 1 2. = = Problems 1. Doing Venn diagrams, prove that A (B C) = (A B) (A C). 2. Doing Venn diagrams, prove that (A B) c = A c B c. 3. Doing Venn diagrams, prove that (A B) c = A c B c. 4. Doing Venn diagrams, prove that P[A B C] = P[A] + P[B) + P [C] P (A B) P [A C] P [B C] + P [A B C]. 5. Let P [A] = 1/2, P [B] = 1/3 and P [A B] = 1/6. Find P [A B c ]. 6. Prove that the probability that either A or B occur but not both is P [A] + P [A] 2P [A B]. 7. Let P [A B C] = 1/2, let P [A] = P [B] = P [C] = 1/3 and let P [A B] = P [A C] = P [B C] = 1/4. Find P [A B C]. 8. A insurer classifies insurance applicants according sex and whether they are homeowners or not. From its insurance pool the insurer has the following information: 45 % of applicants are female, 35 % of applicants are homeowners and 20 % of applicants are male who do not own a house. Find the percentage of applicants who are female who own a home.

10 12.3. PROBLEMS A total of 8 percent of American males smoke cigarettes, 1 percent smoke cigars, and 0.3 percent smoke both cigars and cigarettes. What percentage of males smoke neither cigars nor cigarettes? 10. Suppose that a deck of 52 cards is shuffled and the top two cards are dealt. Find the probability that at least one ace is among the two cards. 11. In testing the water supply for various cities for two kinds of impurities commonly found in water, it was found that 20 % of the water supplies had neither sort of impurity, 40 % had an impurity of type A, and 50 % had an impurity of type B. If a city is chosen at random, what is the probability that is water supply has exactly one type of impurity? 12. A high school is offering 3 language classes: one in Spanish, one in French, and one in German. There are 258 students in this high school. There are 30 students in the Spanish class, 20 in the French class and 18 in the German class. There are 6 students in that are both Spanish and French, 10 that are in both Spanish and German, and 4 that are in both French and German. Four student taking all 3 classes. If a student is chosen randomly, what is the probability that he/she is in at least one of these classes? 13. Eighty percent of the students at a certain school do not have a tattoo and they are not pierced. Twenty five percent have a tattoo and they 60 are pierced. If one of the students is chosen randomly, what is the probability that this student has a tattoo and it is pierced.