Chapter 3 Probability http www ocw cnOcw WebSloanSchoolofManagement15

Chapter 3: Probability http: //www. ocw. cn/Ocw. Web/Sloan-School-of-Management/15 -075 Applied. Statistics. Spring 2003/Course. Home/index. htm Larson/Farber Ch. 3

Useful videos/websites: Video on terminology (outcome, event, experiment, sample space): http: //screencast. com/t/ODVl. Nz. M 2 M 2 Video on 3 types of probability: http: //screencast. com/t/N 2 Vk. OTFl. Mz Website with good presentation on probability: http: //www. zweigmedia. com/Real. World/tutorialsf 15 e/frames 6_1. html Math Goodies: http: //www. mathgoodies. com/lessons/vol 6/intro_probability. html http: //www. mathgoodies. com/lessons/vol 6/independent_events. ht ml http: //www. mathgoodies. com/lessons/vol 6/conditional. html http: //www. mathgoodies. com/lessons/vol 6/complement. html Larson/Farber Ch. 3

Types of Probability Classical(equally probable outcomes) Empirical Probability blood pressure will decrease after medication Intuition Probability the line will be busy Larson/Farber Ch. 3

Important Terms Probability experiment: An action through which counts, measurements or responses are obtained Sample space: The set of all possible outcomes Event: A subset of the sample space. Outcome: The result of a single trial Larson/Farber Ch. 3

Experiment: Flip a Coin and Roll a Die Tree diagram: H 1 H 2 H 3 H 4 H 5 H 6 T 1 T 2 T 3 T 4 T 5 T 6 The sample space has 12 outcomes: {H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6} Larson/Farber Ch. 3

Test yourself: What’s the probability of the following events: P(getting heads) = b. P(rolling a 3) = c. P(H 3 or H 4) = d. P(rolling a Head AND getting a 3 or 4) = a. The sample space has 12 outcomes: {H 1, H 2, H 3, H 4, H 5, H 6, T 1, T 2, T 3, T 4, T 5, T 6} Larson/Farber Ch. 3

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. Larson/Farber Ch. 3

Example How many license plates can you make if a license number consists of four alphabetic characters followed by two numbers? __ __ __ Larson/Farber Ch. 3

Example How many license plates can you make if a license number consists of four unique alphabetic characters followed by two unique digits? __ __ __ Larson/Farber Ch. 3

Definition: Complementary Events Complement of event E [Denoted E ′ (E prime)] The set of all outcomes in a sample space that are not included in event E. • P(E ′) + P(E) = 1 • P(E) = 1 – P(E ′) • P(E ′) = 1 – P(E) Larson/Farber Ch. 3 E E′

Example 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 24 54 25 to 34 366 35 to 44 233 45 to 54 180 55 to 64 125 65 and over 42 Σf = 1000 Larson/Farber 4 th ed Larson/Farber Ch. 3 11

Solution – Complementary Events Event A = Age is between 25 and 34 Event A’ = Age is NOT between 25 and 34 Employee ages Frequency, f 15 to 24 54 25 to 34 366 35 to 44 233 45 to 54 180 55 to 64 125 65 and over 42 Σf = 1000 Larson/Farber 4 th ed Larson/Farber Ch. 3 12

Section 3. 2 Conditional Probability and the Multiplication Rule Larson/Farber Ch. 3

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”) For Example: There are 6 frosted donuts and 6 plain donuts in a box. If you select one at random, what’s the probability that you get a frosted donut? [Simple probability. Answer: 1/2] Your friend takes a frosted donut, now what’s the probability that you get a frosted donut? [Conditional probability. Answer: 5/11] Larson/Farber Ch. 3

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. ) Event you are looking for: Second card is a queen. Condition: You’ve already taken a king out of the deck. Solution: Because the first card is a king and is not replaced, the remaining deck has 51 cards, 4 of which are queens. 15 Larson/Farber Ch. 3

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. 16 Gene Present Gene not present Total High IQ 33 19 52 Normal IQ 39 11 50 Total 72 30 102 Larson/Farber 4 th ed Larson/Farber Ch. 3

Solution: Finding Conditional Probabilities There are 72 children who have the gene. So, the sample space consists of these 72 children. Gene Present Gene not present Total High IQ 33 19 52 Normal IQ 39 11 50 Total 72 30 102 Of these, 33 have a high IQ. 17 Larson/Farber Ch. 3

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 18 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Independent Events Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence (or non-occurrence) of event A. For example: A = Being female B = Having type O blood A = 1 st child is a boy B = 2 nd child is a boy Larson/Farber Ch. 3

Dependent Events Two events that are not independent are dependent. For example: A = Living in Houston B = Living in Texas A = Selecting a red ball from (3 red, 3 blue) B = Selecting a red ball, then a blue ball Larson/Farber Ch. 3

Contingency Table The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Omaha Yes 100 No 125 Undecided 75 Total 300 Seattle 150 130 170 450 Miami 150 95 5 250 Total 400 350 250 1000 One of the responses is selected at random. Find: 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No | Miami) = probability the answer is “no” given that adult is a resident of Miami Larson/Farber Ch. 3

Solutions Yes No Undecided Total Omaha 100 125 75 300 1. P(Yes) Seattle 150 130 170 450 Miami 150 95 5 250 = 400 / 1000 = 0. 4 2. P(Seattle) = 450 / 1000 = 0. 45 3. P(Miami) = 250 / 1000 = 0. 25 4. P(No, given Miami) = 95 / 250 = 0. 38 Answers: 1) 0. 4 Larson/Farber Ch. 3 Total 400 350 250 1000 2) 0. 45 3) 0. 25 4) 0. 38

Imagine this Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective. x x Larson/Farber Ch. 3 x

Multiplication Rule To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has already occurred. A = first car is defective B = second car is defective. P(A and B) = P(A) x P(B|A) P(A) = 5/12 P(B|A) = 4/11 P(A and B) = 5/12 x 4/11 = 5/33 = 0. 1515 Larson/Farber Ch. 3

Multiplication Rule Two dice are rolled. Find the probability both are 4’s. A = first die is a 4 and B = second die is a 4. P(A) = 1/6 P(B|A) = 1/6 P(A and B) = 1/6 x 1/6 = 1/36 = 0. 028 When two events A and B are independent, then P (A and B) = P(A) x P(B) (because for independent events P(B) = P(B|A) ) Larson/Farber Ch. 3 .

Section 3. 3 The Addition Rule Larson/Farber Ch. 3

Compare “A and B” to “A or B” The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B). The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B). B A A and B Larson/Farber Ch. 3 B A A or B

Mutually Exclusive Events Two events, A and B, are mutually exclusive if they cannot occur in the same trial. A = A person is under 21 years old B = A person is running for the U. S. Senate A = A person was born in Philadelphia B = A person was born in Houston A B Mutually exclusive P(A and B) = 0 When event A occurs it excludes event B in the same trial. Larson/Farber Ch. 3

Non-Mutually Exclusive Events If two events can occur in the same trial, they are NOT mutually exclusive. A = A person is under 25 years old B = A person is a lawyer A = A person was born in Philadelphia B = A person watches “Jeopardy” on TV A and B Non-mutually exclusive P(A and B) ≠ 0 Larson/Farber Ch. 3 A B

The Addition Rule The probability that one or the other of two events will occur is: P(A) + P(B) – P(A and B) A card is drawn from a deck. Find the probability it is a king or it is red. Larson/Farber Ch. 3

The Addition Rule The probability that one or the other of two events will occur is: P(A) + P(B) – P(A and B) A card is drawn from a deck. Find the probability it is a king or it is red. A = the card is a king B = the card is red. P(A) = 4/52 and P(B) = 26/52 but P(A and B) = 2/52 P(A or B) Larson/Farber Ch. 3 = 4/52 + 26/52 – 2/52 = 28/52 = 0. 538

Example: Using the Addition Rule A blood bank catalogs the types of blood given by donors during the last five days. A donor is selected at random. Find the probability the donor has type O or type A blood. Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 184 164 45 16 409 Total 32 Larson/Farber Ch. 3 Larson/Farber 4 th ed

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 156 139 37 12 344 Rh-Negative 28 25 8 4 65 184 164 45 16 409 Total 33 Larson/Farber Ch. 3

Example: Using the Addition Rule Find the probability the donor has type B or is Rh-negative. Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 184 164 45 16 409 Total Solution: The events are not mutually exclusive (a donor can have type B blood and be Rh-negative) 34 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Solution: Using the Addition Rule Type O Type A Type B Type AB Total Rh-Positive 156 139 37 12 344 Rh-Negative 28 25 8 4 65 184 164 45 16 409 Total 35 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Section 3. 4 Counting Principles Larson/Farber Ch. 3

Permutations Permutation An ordered arrangement of objects The number of different permutations of n distinct objects is n! (n factorial) n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙ 3∙ 2 ∙ 1 0! = 1 Examples: 6! = 6∙ 5∙ 4∙ 3∙ 2∙ 1 = 720 4! = 4∙ 3∙ 2∙ 1 = 24 37 Larson/Farber 4 th ed Larson/Farber Ch. 3

Permutations Permutation ofn objects takenr at a time The number of different permutations of n distinct objects taken r at a time ■ wherer ≤ n 38 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Example: Finding n. Pr Find the number of ways of forming three-digit codes in which no digit is repeated. Solution: You need select 3 digits from a group of 10 n = 10, r = 3 39 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Combinations Combination ofn objects takenr at a time A selection of r objects from a group of n objects without regard to order ■ 40 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Example: Combinations A state’s department of transportation plans to develop a new section of interstate highway and receives 16 bids for the project. The state plans to hire four of the bidding companies. How many different combinations of four companies can be selected from the 16 bidding companies? Method: • You need to select 4 companies from a group of 16 • n = 16, r = 4 • Order is not important 41 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Solution: Combinations 42 Larson/Farber Ch. 3 Larson/Farber 4 th ed

Larson/Farber Ch. 3

1. Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2 nd car is defective, given the first car was defective? Larson/Farber Ch. 3

Contingency Table The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Omaha Yes 100 No 125 Undecided 75 Total 300 Seattle 150 130 170 450 Miami 150 95 5 250 Total 400 350 250 1000 One of the responses is selected at random. Find: 1. P(Miami and Yes) 3. P(Miami or Yes) 2. P(Miami and Seattle) 4. P(Miami or Seattle) Larson/Farber Ch. 3

3. One card is selected at random from a standard deck, then replaced, and a second card is drawn. Find the probability of selecting two face cards. A. 0. 050 B. 0. 053 C. 0. 038 D. 0. 462 Larson/Farber Ch. 3

4. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage. Cheese Pepperoni Sausage Total Male 8 5 2 15 A. 0. 458 B. 0. 583 C. 0. 125 Female Total 2 10 4 9 3 5 9 24 D. 0. 556 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Larson/Farber Ch. 3 Slide 3 - 47

Answers 1. Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11 2. P(Miami and Yes) = 0. 15 P(Miami and Seattle) = 0 P(Miami or Yes) = 250/1000 + 400/1000 – 150/1000 = 0. 5 P(Miami or Seattle ) = 250/1000 + 450/1000 = 0. 7 3. (B) 0. 053 4. (A). 458 Larson/Farber Ch. 3