Chapter. Probability

Transcription

1 Chapter 3 Probability

2 Section 3.1 Basic Concepts of Probability

3 Section 3.1 Objectives Identify the sample space of a probability experiment Identify simple events Use the Fundamental Counting Principle Distinguish among classical probability, empirical probability, and subjective probability Determine the probability of the complement of an event Use a tree diagram and the Fundamental Counting Principle to find probabilities

4 Probability Experiments Probability experiment An action, or trial, through which specific results (counts, measurements, or responses) are obtained. Outcome The result of a single trial in a probability experiment. Sample Space The set of all possible outcomes of a probability experiment. Event Consists of one or more outcomes and is a subset of the sample space.

5 Probability Experiments Probability experiment: Roll a die Outcome: {3} Sample space: {1, 2, 3, 4, 5, 6} Event: {Die is even}={2, 4, 6}

6 Example: Identifying the Sample Space A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space. Solution: There are two possible outcomes when tossing a coin: a head (H) or a tail (T). For each of these, there are six possible outcomes when rolling a die: 1, 2, 3, 4, 5, or 6. One way to list outcomes for actions occurring in a sequence is to use a tree diagram.

7 Solution: Identifying the Sample Tree diagram: Space H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 The sample space has 12 outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

8 Simple Events Simple event An event that consists of a single outcome. e.g. Tossing heads and rolling a 3 {H3} An event that consists of more than one outcome is not a simple event. e.g. Tossing heads and rolling an even number {H2, H4, H6}

9 Fundamental Counting Principle Fundamental Counting Principle If one event can occur in m ways and a second event can occur in n ways, the number of ways the two events can occur in sequence is m n. Can be extended for any number of events occurring in sequence.

10 Example: Fundamental Counting Principle You are purchasing a new car. The possible manufacturers, car sizes, and colors are listed. Manufacturer: Ford, GM, Honda Car size: compact, midsize Color: white (W), red (R), black (B), green (G) How many different ways can you select one manufacturer, one car size, and one color? Use a tree diagram to check your result.

11 Solution: Fundamental Counting Principle There are three choices of manufacturers, two car sizes, and four colors. Using the Fundamental Counting Principle: = 24 ways

12 Exercise In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts (a) through (j) below. (Note that you can't find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) Let F be the event that a student is female. Let M be the event that a student is male. Let S be the event that a student has short hair. Let L be the event that a student has long hair.

13 a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female.

14 a. P(L')=P(S) b. P(M or S) c. P(F and L) d. P(M L) e. P(L M) f. P(S F) g. P(F L) h. P(F or L) i. P(M and S) j. P(F) Solutions

15 Types of Probability Classical (theoretical) Probability Each outcome in a sample space is equally likely. PE ( ) Number of outcomes in event E Number of outcomes in sample space

16 Example: Finding Classical Probabilities You roll a six-sided die. Find the probability of each event. 1. Event A: rolling a 3 2. Event B: rolling a 7 3. Event C: rolling a number less than 5 Solution: Sample space: {1, 2, 3, 4, 5, 6}

17 Solution: Finding Classical Probabilities 1. Event A: rolling a 3 Event A = {3} 1 P( rolling a 3) Event B: rolling a 7 Event B= { } (7 is not in the sample space) 0 P( rolling a 7) Event C: rolling a number less than 5 Event C = {1, 2, 3, 4} 4 P( rolling a number less than 5)

18 Types of Probability Empirical (statistical) Probability Based on observations obtained from probability experiments. Relative frequency of an event. PE ( ) Frequency of event E Total frequency f n

19 Example: Finding Empirical Probabilities 1. A company is conducting a telephone survey of randomly selected individuals to get their overall impressions of the past decade (2000s). So far, 1504 people have been surveyed. What is the probability that the next person surveyed has a positive overall impression of the 2000s? (Source: Princeton Survey Research Associates International) Response Number of times, f Positive 406 Negative 752 Neither 316 Don t know 30 Σf = 1504

20 Solution: Finding Empirical Probabilities event Response Number of times, f Positive 406 Negative 752 Neither 316 Don t know 30 Σf = 320 frequency f 406 P( positive) n 1504

21 Law of Large Numbers Law of Large Numbers As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event. pp/index.php?dataid=205337

22 Range of Probabilities Rule Range of probabilities rule The probability of an event E is between 0 and 1, inclusive. 0 P(E) 1 Impossible Unlikely Even chance Likely Certain [ ]

23 Complementary Events Complement of event E The set of all outcomes in a sample space that are not included in event E. Denoted E (E prime) P(E) + P(E ) = 1 P(E) = 1 P(E ) P(E ) = 1 P(E)

24 Example: Probability of the Complement of an Event You survey a sample of 1000 employees at a company and record the age of each. Find the probability of randomly choosing an employee who is not between 25 and 34 years old. Employee ages Frequency, f 15 to to to to to and over 42 Σf = 1000

25 Solution: Probability of the Complement of an Event Use empirical probability to find P(age 25 to 34) P(age 25 to 34) f n Use the complement rule P(age is not 25 to 34) Employee ages Frequency, f 15 to to to to to and over 42 Σf = 1000

26 Example: Probability Using a Tree Diagram A probability experiment consists of tossing a coin and spinning the spinner shown. The spinner is equally likely to land on each number. Use a tree diagram to find the probability of tossing a tail and spinning an odd number.

27 Solution: Probability Using a Tree Diagram Tree Diagram: H T H1 H2 H3 H4 H5 H6 H7 H8 T1 T2 T3 T4 T5 T6 T7 T8 P(tossing a tail and spinning an odd number) =

28 Example: Probability Using the Fundamental Counting Principle Your college identification number consists of 8 digits. Each digit can be 0 through 9 and each digit can be repeated. What is the probability of getting your college identification number when randomly generating eight digits?

29 Solution: Probability Using the Fundamental Counting Principle Each digit can be repeated There are 10 choices for each of the 8 digits Using the Fundamental Counting Principle, there are = 10 8 = 100,000,000 possible identification numbers Only one of those numbers corresponds to your ID number 1 P(your ID number) = 100, 000, 000

30 Section 3.2 Conditional Probability and the Multiplication Rule

31 Section 3.2 Objectives Determine conditional probabilities Distinguish between independent and dependent events Use the Multiplication Rule to find the probability of two events occurring in sequence Use the Multiplication Rule to find conditional probabilities

32 Conditional Probability Conditional Probability The probability of an event occurring, given that another event has already occurred Denoted P(B A) (read probability of B, given A )

33 Example: Finding Conditional Probabilities Two cards are selected in sequence from a standard deck. Find the probability that the second card is a queen, given that the first card is a king. (Assume that the king is not replaced.) Solution: Because the first card is a king and is not replaced, the remaining deck has 51 cards, 4 of which are queens. nd st 4 P( B A) P(2 card is a Queen 1 card is a King)

34 Example: Finding Conditional Probabilities The table shows the results of a study in which researchers examined a child s IQ and the presence of a specific gene in the child. Find the probability that a child has a high IQ, given that the child has the gene. Gene Present Gene not present Total High IQ Normal IQ Total

35 Solution: Finding Conditional Probabilities There are 72 children who have the gene. So, the sample space consists of these 72 children. Gene Present Of these, 33 have a high IQ. Gene not present Total High IQ Normal IQ Total P(B A) P(high IQ gene present )

36 Independent and Dependent Events Independent events The occurrence of one of the events does not affect the probability of the occurrence of the other event P(B A) = P(B) or P(A B) = P(A) Events that are not independent are dependent

37 Example: Independent and Dependent Events Decide whether the events are independent or dependent. 1. Selecting a king from a standard deck (A), not replacing it, and then selecting a queen from the deck (B). Solution: nd st P( B A) P(2 card is a Queen 1 card is a King) P( B) P( Queen) 4 52 Dependent (the occurrence of A changes the probability of the occurrence of B) 4 51

38 Example: Independent and Dependent Events Decide whether the events are independent or dependent. 2. Tossing a coin and getting a head (A), and then rolling a sixsided die and obtaining a 6 (B). Solution: P( B A) P( rolling a 6 head on coin) P( B) P( rolling a 6) Independent (the occurrence of A does not change the probability of the occurrence of B)

39 The Multiplication Rule Multiplication rule for the probability of A and B The probability that two events A and B will occur in sequence is P(A and B) = P(A) P(B A) For independent events the rule can be simplified to P(A and B) = P(A) P(B) Can be extended for any number of independent events

40 Example: Using the Multiplication Rule Two cards are selected, without replacing the first card, from a standard deck. Find the probability of selecting a king and then selecting a queen. Solution: Because the first card is not replaced, the events are dependent. P( K and Q) P( K) P( Q K)

41 Example: Using the Multiplication Rule A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6. Solution: The outcome of the coin does not affect the probability of rolling a 6 on the die. These two events are independent. P( H and 6) P( H ) P(6)

42 Example: Using the Multiplication Rule The probability that a particular knee surgery is successful is Find the probability that three knee surgeries are successful. Solution: The probability that each knee surgery is successful is The chance for success for one surgery is independent of the chances for the other surgeries. P(3 surgeries are successful) = (0.85)(0.85)(0.85) 0.614

43 Example: Using the Multiplication Rule Find the probability that none of the three knee surgeries is successful. Solution: Because the probability of success for one surgery is The probability of failure for one surgery is = 0.15 P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15) 0.003

44 Example: Using the Multiplication Rule Find the probability that at least one of the three knee surgeries is successful. Solution: At least one means one or more. The complement to the event at least one is successful is the event none are successful. Using the complement rule P(at least 1 is successful) = 1 P(none are successful) = 0.997

45 More on Independent Events Two events are independent if the following are true: P(A B) = P(A) P(B A) = P(B) P(A AND B) = P(A) P(B) Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.

46 With and without replacement Sampling may be done with replacement or without replacement. With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick. Without replacement: When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent. If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise

47 Section 3.3 Addition Rule

48 Section 3.3 Objectives Determine if two events are mutually exclusive Use the Addition Rule to find the probability of two events

49 Example: Mutually Exclusive Events Decide if the events are mutually exclusive. Event A: Roll a 3 on a die. Event B: Roll a 4 on a die. Solution: Mutually exclusive (The first event has one outcome, a 3. The second event also has one outcome, a 4. These outcomes cannot occur at the same time.)

50 Example: Mutually Exclusive Events Decide if the events are mutually exclusive. Event A: Randomly select a male student. Event B: Randomly select a nursing major. Solution: Not mutually exclusive (The student can be a male nursing major.)

51 The Addition Rule Addition rule for the probability of A or B The probability that events A or B will occur is P(A or B) = P(A) + P(B) P(A and B) For mutually exclusive events A and B, the rule can be simplified to P(A or B) = P(A) + P(B) Can be extended to any number of mutually exclusive events

52 Example: Using the Addition Rule You select a card from a standard deck. Find the probability that the card is a 4 or an ace. Solution: The events are mutually exclusive (if the card is a 4, it cannot be an ace) P(4 or ace) P(4) P( ace)

53 Example: Using the Addition Rule You roll a die. Find the probability of rolling a number less than 3 or rolling an odd number. Solution: The events are not mutually exclusive (1 is an outcome of both events)

54 Solution: Using the Addition Rule P( less than 3 or odd) P( less than 3) P( odd) P( less than 3 and odd)

55 Example: Using the Addition Rule The frequency distribution shows the volume of sales (in dollars) and the number of months in which a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between $75,000 and $124,999 next month? Sales volume ($) Months 0 24, ,000 49, ,000 74, ,000 99, , , , , , , , ,999 1

56 Solution: Using the Addition Rule A = monthly sales between $75,000 and $99,999 B = monthly sales between $100,000 and $124,999 A and B are mutually exclusive P( A or B) P( A) P( B) Sales volume ($) Months 0 24, ,000 49, ,000 74, ,000 99, , , , , , , , ,999 1

57 Example: Using the Addition Rule A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last five days. A donor is selected at random. Find the probability that the donor has type O or type A blood. Type O Type A Type B Type AB Total Rh-Positive Rh-Negative Total

58 Solution: Using the Addition Rule The events are mutually exclusive (a donor cannot have type O blood and type A blood) Type O Type A Type B Type AB Total Rh-Positive Rh-Negative Total P(type O or type A) P(type O) P(type A)

59 Example: Using the Addition Rule Find the probability that the donor has type B blood or is Rhnegative. Solution: Type O Type A Type B Type AB Total Rh-Positive Rh-Negative Total The events are not mutually exclusive (a donor can have type B blood and be Rh-negative)

60 Solution: Using the Addition Rule Type O Type A Type B Type AB Total Rh-Positive Rh-Negative Total P(type B or Rh neg) P(type B) P(Rh neg) P(type B and Rh neg)

61 Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

62 EXAMPLE In an urn, there are 11 balls. Three balls are red (R) and 8 balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. Larson/Farber 5th edition 62

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64 Calculate 1. List the 24 BR outcomes: B1R1, B1R2, B1R3, Using the tree diagram, calculate P(RR). 3. Using the tree diagram, calculate P(RB OR BR). 4. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw). 5. Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw). 6. Using the tree diagram, calculate P(BB). 7. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw). Larson/Farber 5th edition 64

65 Answers 1. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B 4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B 8R2; B8R3 2. P(RR)=(3/11) (3/11)=9/ P(RB OR BR)=(3/11) (8/11) + (8/11) (3/11) = 48/ P(R on 1st draw AND B on 2nd draw)=p(rb)=(3/11) (8/11)=24/ P(R on 2nd draw given B on 1st draw)=p(r on 2nd B on 1st)=24/88=3/11 6. P(BB) = 64/ P(B on 2nd draw R on 1st draw) = 8/11 There are outcomes that have R on the first draw (9 RR and 24 RB).

66 Example An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time, this time without replacement from the urn. "Without replacement" means that you do not put the first ball back before you select the second ball. Below is a tree diagram. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (3/11) (2/10)=6/110.

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68 Calculate 1. P(RR) = 2. P(RB OR BR)= 3. P(R on 2d B on 1st) = 4. P(R on 1st and B on 2nd) = 5. P(BB) = 6. P(B on 2nd R on 1st) =

69 Answers 1. P(RR)=(3/11) (2/10)=6/ P(RB or BR)=(3/11) (8/10) + (8/11)(3/10) =48/ P(R on 2d B on 1st) = P(R on 1st and B on 2nd) = P(RB) = (3/11)(8/10) = 24/ P(BB) = (8/11) (7/10) 6. The probability is 24/30