Larson/Farber Ch. 3 Chapter 3: Probability StatisticsSpring2003/CourseHome/index.htm

Larson/Farber Ch. 3 Useful videos/websites: Video on terminology (outcome, event, experiment, sample space): Video on 3 types of probability: Website with good presentation on probability: Math Goodies: ml ml

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

Larson/Farber Ch. 3 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: Important Terms The result of a single trial

Larson/Farber Ch. 3 Experiment: Flip a Coin and Roll a Die Tree diagram: 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}

Larson/Farber Ch. 3 What’s the probability of the following events: a. P(getting heads) = b. P(rolling a 3) = c. P(H3 or H4) = d. P(rolling a Head AND getting a 3 or 4) = Test yourself: The sample space has 12 outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

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) E ′ E

Larson/Farber Ch. 3 Example Larson/Farber 4th ed11 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 agesFrequency, f 15 to to to to to and over42 Σf = 1000

Larson/Farber Ch. 3 Solution – Complementary Events Larson/Farber 4th ed12 Event A = Age is between 25 and 34 Event A’ = Age is NOT between 25 and 34 Employee agesFrequency, f 15 to to to to to and over42 Σf = 1000

Larson/Farber Ch. 3 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 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.

Larson/Farber Ch. 3 Larson/Farber 4th ed 16 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 presentTotal High IQ Normal IQ Total Example: Finding Conditional Probabilities

Larson/Farber Ch Solution: Finding Conditional Probabilities There are 72 children who have the gene. So, the sample space consists of these 72 children. Of these, 33 have a high IQ. Gene Present Gene not presentTotal High IQ Normal IQ Total

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 Larson/Farber 4th ed 18

Larson/Farber Ch. 3 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. A = Being female B = Having type O blood A = 1st child is a boy B = 2nd child is a boy Independent Events For example:

Larson/Farber Ch. 3 Two events that are not independent are dependent. Dependent Events 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 For example:

Larson/Farber Ch. 3 The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: 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 OmahaSeattleMiamiTotal Yes No Undecided Total One of the responses is selected at random. Find: Contingency Table

Larson/Farber Ch P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) OmahaSeattleMiamiTotal Yes No Undecided Total = 95 / 250 = 0.38 = 250 / 1000 = 0.25 Answers: 1) 0.4 2) ) ) 0.38 = 450 / 1000 = 0.45 Solutions = 400 / 1000 = 0.4

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

Larson/Farber Ch. 3 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. P(A and B) = P(A) x P(B|A) A = first car is defective B = second car is defective. P(A) = 5/12 P(B|A) = 4/11 P(A and B) = 5/12 x 4/11 = 5/33 = Multiplication Rule

Larson/Farber Ch. 3 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) = 1/6P(B|A) = 1/6 P(A and B) = 1/6 x 1/6 = 1/36 = 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) ). Multiplication Rule

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). A B A or B A and B A B

Larson/Farber Ch. 3 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 B Non-mutually exclusive P(A and B) ≠ 0 A and B

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.

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) = 4/ /52 – 2/52 = 28/52 = 0.538

Larson/Farber Ch. 3 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. Larson/Farber 4th ed 32 Type OType AType BType ABTotal Rh-Positive Rh-Negative Total

Larson/Farber Ch. 3 Solution: Using the Addition Rule 33 The events are mutually exclusive (a donor cannot have type O blood and type A blood) Type OType AType BType ABTotal Rh-Positive Rh-Negative Total

Larson/Farber Ch. 3 Example: Using the Addition Rule Find the probability the donor has type B or is Rh-negative. Larson/Farber 4th ed 34 Solution: The events are not mutually exclusive (a donor can have type B blood and be Rh-negative) Type OType AType BType ABTotal Rh-Positive Rh-Negative Total

Larson/Farber Ch. 3 Solution: Using the Addition Rule Larson/Farber 4th ed 35 Type OType AType BType ABTotal Rh-Positive Rh-Negative Total

Larson/Farber Ch. 3 Section 3.4 Counting Principles

Larson/Farber Ch. 3 Permutations Larson/Farber 4th ed 37 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

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

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

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

Larson/Farber Ch. 3 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? Larson/Farber 4th ed 41 Method: You need to select 4 companies from a group of 16 n = 16, r = 4 Order is not important

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

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 2nd car is defective, given the first car was defective?

Larson/Farber Ch. 3 The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Contingency Table 3. P(Miami or Yes) 4. P(Miami or Seattle) OmahaSeattleMiamiTotal Yes No Undecided Total One of the responses is selected at random. Find : 1. P(Miami and Yes) 2. P(Miami and Seattle)

Larson/Farber Ch 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 B C D

Larson/Farber Ch. 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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. A B C D Slide CheesePepperoniSausageTotal Male85215 Female2439 Total109524

Larson/Farber Ch. 3 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 – 150/1000 = 0.5 P(Miami or Seattle ) = 250/ /1000 = (B) (A).458