Statistics_4_3 - Chapter 4 Probability and counting rules 4-1 Sample Spaces and Probability 42 The Addition Rules for Probability 43 The

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Unformatted text preview: Chapter 4: Probability and counting rules 4-1 Sample Spaces and Probability 42 The Addition Rules for Probability 43 The Multiplication Rules and Conditional Probability 44 Counting Rules 45 Probability and Counting Rules Sample Spaces and Probability A probability experiment is a chance process that leads to well-defined results called outcomes. An outcome is the result of a single trial of a probability experiment. A sample space, S, is the set of all possible outcomes of a probability experiment. Examples: . Example: Find the sample space for rolling two dice. Solution: Example: Find the sample space for drawing one card from an ordinary deck of cards. Solution: . Example: Find the sample space for the gender of the children if a family has three children. Solution: BBB BBG BGB GBB GGG GGB GBG BGG A tree diagram is a device consisting of line segments emanating from a starting point and also from the outcome point. It is used to determine all possible outcomes of a probability experiment. Example, . An event consists of a set of outcomes of a probability experiment. An event with one outcome is called a simple event. a compound event consists of two or more outcomes or simple events. There are three basic interpretations of probability: Classical Probability Classical probability assumes that all outcomes in the sample space are equally likely to occur. Equally likely events are events that have the same probability of occurring. . Example: Find the probability of getting a red ace when a card is drawn at random from an ordinary deck of cards. Solution: P(red ace)= . Example: A card is drawn from an ordinary deck. Find these probabilities. a. Of getting a jack b. Of getting the 6 of clubs (i.e., a 6 and a club) c. Of getting a 3 or a diamond d. Of getting a 3 or a 6. Solution: a. P(jack)= = b. P(6 of clubs)= c. P(3 or diamond)= d. P(3 or 6)= = = Basic probability rules Probability Rule 1 The probability of any event E is a number between and including 0 and 1. This is denoted by 0 P(E) 1. Probability Rule 2 If an event E cannot occur (i.e., the event contains no members in the sample space), impossible event, , its probability is 0. Probability Rule 3 If an event E is certain, E=S, then the probability of E is 1 or P(S)=1. Probability Rule 4 The sum of the probabilities of all the outcomes in the sample space is 1. The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is denoted by Example: Find the complement of the event E= Rolling a die and getting a 4 or 6. Solution: = {1,2,3,5) Complementary Events Rule: Example: If the probability that a person lives in an industrialized country of the world is , find the probability that a person does not live in an industrialized country. Solution: P(not living in an industrialized country)= 1 P(living in an industrialized country)= 1- = . Empirical Probability Classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the likelihood of outcomes. Example: a researcher for the American Automobile Association asked 50 people who plan to travel over the Thanksgiving holiday how they will get to their destination. The results were, The probability of selecting a person who is driving is . Example: In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. a. A person has type O blood. b. A person has type A or type B blood. c. A person has neither type A nor type O blood. d. A person does not have type AB blood. Solution: . Example: Hospital records indicated that maternity patients stayed in the hospital for the number of days shown in the distribution. Find these probabilities. a. A patient stayed exactly 5 days. b. A patient stayed less than 6 days. c. A patient stayed at most 4 days. d. A patient stayed at least 5 days. Solution: . If the empirical probability of getting a head is computed by using a small number of trials, it is usually not exactly 0.5. However, as the number of trials increases, the empirical probability of getting a head will approach the theoretical probability of 0.5. This phenomenon is an example of the law of large numbers. Subjective probability A person or group makes an educated guess at the chance that an event will occur. Example, A physician might say that, on the basis of her diagnosis, there is a 30% chance the patient will need an operation. The Addition Rules for Probability Two events are mutually exclusive events if they cannot occur at the same time (i.e., they have no outcomes in common). = . Example: Determine which events are mutually exclusive and which are not, when a single die is rolled. a. Getting an odd number and getting an even number. b. Getting a 3 and getting an odd number. c. Getting an odd number and getting a number less than 4. d. Getting a number greater than 4 and getting a number less than 4. Addition Rule 1 When two events A and B are mutually exclusive, the probability that A or B will occur is = = + ( ) Example: At a political rally, there are 20 Republicans, 13 Democrats, and 6 Independents. If a person is selected at random, find the probability that he or she is either a Democrat or an Independent. Solution P(Democrat or Independent) =P(Democrat)+P(Independent) = . Addition Rule 2 If A and B are not mutually exclusive, then Example: In a hospital unit there are 8 nurses and 5 physicians; 7 nurses and 3 physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male. Solution: P(nurse or male)= P(nurse)+ P(male)- P(male nurse) = . Notice: The probability rules can be extended to three or more events. For three mutually exclusive events A, B, and C, P(A or B or C) = P(A) + P(B) + P(C) For three events that are not mutually exclusive, P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C) The Multiplication Rules and Conditional Probability The Multiplication Rules Two events A and B are independent events if the fact that A occurs does not affect the probability of B occurring. Example: Rolling a die and getting a 6, and then rolling a second die and getting a 3. Drawing a card from a deck and getting a queen, replacing it, and drawing a second card and getting a queen. Multiplication Rule 1 When two events are independent, the probability of both occurring is = = . ( ) Example: A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace. Solution: P(queen and ace) = P(queen). P(ace) = = . Example: An urn contains 3 red balls, 2 blue balls, and 5 white balls. A ball is selected and its color noted. Then it is replaced. A second ball is selected and its color noted. Find the probability of each of these. a. Selecting 2 blue balls. b. Selecting 1 blue ball and then 1 white ball. c. Selecting 1 red ball and then 1 blue ball. Solution: a. P(blue and blue) = P(blue) . P(blue) = b. P(blue and white) = P(blue) . P(white) = c. P(red and blue) = P(red) . P(blue) = = = = Notice: Multiplication rule 1 can be extended to three or more independent events by using the formula P(A and B and C and . . . and K)= P(A) . P(B) . P(C) . . . P(K) When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent events. Example, drawing a card from a deck, not replacing it, and then drawing a second card. Multiplication Rule 2 When two events are dependent, the probability of both occurring is P(A and B) = P(A) . P Where P means that the probability that event B occurs given that event A has already occurred. Example: At a university in western Pennsylvania, there were 5 burglaries reported in 2003, 16 in 2004, and 32 in 2005. If a researcher wishes to select at random two different burglaries to further investigate, find the probability that both will have occurred in 2004. Solution: 16 15 60 = = = 53 52 689 Example: Statistical section consists of 15 female students and 11 male students. If two students are to be selected without replacement. a. If the first selected student is a female what is the probability that the second student will be a male student. b. What is the probability that both the two selected students will be female. c. What is the probability that both selected students will be of the same gender. d. What is the probability that the two selected students will of a different gender. Solution: Let F=the selected student will be a female and M=the selected student will be a male Then 14 15 15 P( F2 F1 ) = . P(F ) = . . P ( F2 M 1 ) = 25 26 25 11 11 10 P (M ) = P( M 2 F1 ) = P(M 2 M 1 ) = 26 25 25 a. If the first is female, the probability that the second will be male is P ( M F ) = 11 25 b. The probability that both are females is 2 1 P ( F 2 I F 1 ) = P ( F 2 F 1 ). P ( F 1 ) = 14 15 21 . = = 0 . 3231 25 26 65 c. The probability that both are of same gender is P ( same gender ) = P ( F2 I F1 ) + P ( M 2 I M 1 ) 14 15 21 . = = 0 . 3231 25 26 65 10 11 11 P ( M 2 I M 1 ) = P ( M 2 M 1 ). P ( M 1 ) = . = = 0 . 1692 25 26 65 P ( same gender ) = 0.3231 + 0.1692 = 0.4923 P ( F2 I F1 ) = P ( F2 F1 ). P ( F1 ) = d. P ( different gender ) = 1 - P ( same gender ) = 1 - 0 .4923 = 0 .5077 Conditional Probability Formula for Conditional Probability Example: A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown. Find these probabilities. a. The respondent answered yes, given that the respondent was a female. b. The respondent was a male, given that the respondent answered no. Solution: Let M = respondent was a male , Y = respondent answered yes F = respondent was a female, N = respondent answered no a. b. Notice: Example: A coin is tossed 5 times. Find the probability of getting at least 1 tail. Solution: . Counting Rules Fundamental Counting Rule In a sequence of n events in which the first one has possibilities and the second event has and the third has , and so forth, the total number of possibilities of the sequence will be Example: A paint manufacturer wishes to manufacture several different paints. The categories include Color: Red, blue, white, black, green, brown, yellow Type: Latex, oil Texture: Flat, semigloss, high gloss Use: Outdoor, indoor How many different kinds of paint can be made if you can select one color, one type, one texture, and one use? Solution: number of paints=(7)(2)(4)(2)=84. Factorial Formulas For any counting n, n!=n(n-1)(n-2)...1 Where 0!=1. Permutations A permutation is an arrangement of n objects in a specific order. Example: Suppose a business owner has a choice of 5 locations in which to establish her business. She decides to rank each location according to certain criteria, such as price of the store and parking facilities. a. How many different ways can she rank the 5 locations? b. How many different ways can she rank the top 3 out of the 5 locations? Solution: a. There are 5!=5.4. 3. 2. 1= 120 different possible rankings. b. There are 5.4. 3= 60 different possible rankings. Permutation Rule The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taking r objects at a time. It is written as , and the ! formula is ! Example: A school musical director can select 2 musical plays to present next year. One will be presented in the fall, and one will be presented in the spring. If she has 9 to pick from, how many different possibilities are there? Solution: . Combinations A selection of distinct objects without regard to order is called a combination. Combination Rule The number of combinations of r objects selected from n objects is denoted by and is given by the ! formula = . ! ! Example: There are 12 players on a High School basketball team. The Coach must pick five players among the twelve on the team to comprise the starting lineup. How many different groups are possible? 12! Solution: 12C 5 = = 792 5!(12 - 5)! Probability and Counting Rules Example: Find the probability of getting 4 aces when 5 cards are drawn from an ordinary deck of cards. Solution: Example: A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities. a. Exactly 2 are defective. b. None is defective. c. All are defective. d. At least 1 is defective. Solution: a. b. c. d. ...
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This note was uploaded on 03/10/2012 for the course STAT 101 taught by Professor Johnanderson during the Spring '12 term at Amity University.

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