Probability Concepts and Counting Rules

Probability Concepts and Counting Rules Chapter 4 McGraw-Hill/Irwin Dr. Ateq Ahmed Al-Ghamedi Department of Statistics P O Box 80203 King Abdulaziz University Jeddah 21589, Saudi Arabia ateq@kau.edu.sa Copyright 2010 by The McGraw-Hill Companies, Inc. All rights reserved. GOALS 1. Define probability. 2. Describe the classical, empirical, and subjective approaches to probability. 3. Explain the terms experiment, event, outcome, permutations, and combinations. 4. Define the terms conditional probability and joint probability. 5. Calculate probabilities using the rules of addition and rules of multiplication. 6. Apply a tree diagram to organize and compute probabilities. 5-2 Probability PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. 5-3 1

Experiment, Outcome and Event An experiment is a process that leads to the occurrence of one and only one of several possible observations. An outcome is the particular result of an experiment. An event is the collection of one or more outcomes of an experiment. An event with one outcome is called a simple event and with more than one outcome is called compound event. 5-4 Tree Diagram A tree diagram is a device used to list all possibilities of a sequence of events in a systematic way. H H First Toss T T Second Toss H T 5-5 In class ex. Use a tree diagram to find the sample space for the gender of the three children Counting Rules Multiplication The multiplication formula indicates that if there are m ways of doing one thing and n ways of doing another thing, there are m x n ways of doing both. : Mr. Ahmed has 10 shirts and 8 ties. How many shirt and tie outfits does he have? (10)(8) = 80 5-6 2

5-7 Employees of a large corporation are to be issued special coded identification cards. The card consists of 4 letters of the alphabet. Each letter can be used up to 4 times in the code. How many different ID cards can be issued? Solution: Since 4 letters are to be used, there are 4 spaces to fill ( ). Since there are 26 different letters to select from and each letter can be used up to 4 times, then the total number of identification cards that can be made is 26 26 26 26 = 456,976 5-8 The digits 0, 1, 2, 3, and 4 are to be used in a 4-digit ID card. How many different cards are possible if repetitions are permitted? Solution: Since there are four spaces to fill and five choices for each space, the solution is 5 5 5 5 = 5 4 = 625 What if the repetitions were not permitted in the previous example? (=120)! Counting Rules - Permutation A permutation is any arrangement of r objects selected from n possible objects. The order of arrangement is important in permutations. 5-9 3

Permutation 5-10 Consider the possible arrangements of the letters a, b, and c. The possible arrangements are: abc, acb, bac, bca, cab, cba. If the order of the arrangement is important then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available? Solution: Number of ways = 7 P 2 = 7! / (7 2)! = 7!/!/5!! = 42. 5-11 How many different ways can four books be arranged on a shelf if they can be selected from nine books? Solution: Number of ways = 9 P 4 = 9! / (9 4)! = 9!/!/5!! = 3024. 5-12 4

Counting Rules- Combination A combination is the number of ways to choose r objects from a group of n objects without regard to order. 5-13 Combination Consider the possible arrangements of the letters a, b, and c. The possible arrangements are: abc, acb, bac, bca, cab, cba. If the order of the arrangement is not important then we say that each arrangement is the same. We say there is one combination of the three letters. 5-14 In order to survey the opinions of customers at local malls, a researcher decides to select 5 malls from a total of 12 malls in a specific geographic area. How many different ways can the selection be made? Solution: Number of combinations: 12C 5 = 12! / [(12 5)! 5!] = 12!/[!/[7!5!]!] = 792 12 5-15 5

In a work team there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there? Solution: Number of possibilities: (number of ways of selecting 3 women from 7) (number of ways of selecting 2 men from 5) = 7 C 3 5 C 2 = (35)(10) = 350. 5-16 Sample Space Ex: Find the sample space for the gender of the children if a family has three children. Use B for boy and G for girl. There are two gender and three children, so there are 8 possibilities as shown here, {BBB, BBG, BGB, GBB, GGG, GGB, GBG, BGG} 5-17 Mutually Exclusive Events Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. Events are independent if the occurrence of one event tdoes not affect tthe occurrence of another. Equally likely events are events that have the same probability of occurring. Venn diagrams are used to represent probabilities pictorially. 5-18 6

Ways of Assigning Probability 5-19 There are three ways of assigning probability: 1. CLASSICAL PROBABILITY Based on the assumption that the outcomes of an experiment are equally likely. 2. EMPIRICAL PROBABILITY The probability of an event happening is the fraction of the time similar events happened in the past. 3. SUBJECTIVE CONCEPT OF PROBABILITY The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available. Classical Probability Consider an experiment of rolling a six-sided die. What is the probability of the event an even number of spots appear face up? The possible outcomes are: There are three favorable outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes. =(3/6)=(1/2) 5-20 Empirical Probability The empirical approach to probability is based on what is called the law of large numbers. The key to establishing probabilities empirically is that more observations will provide a more accurate estimate of the probability. 5-21 7

Empirical Probability - On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 113 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed? Probability of a successfulflight = Number of successfulflights Total number of flights 111 = = 0.98 113 5-22 Subjective Probability - If there is little or no past experience or information on which to base a probability, it may be arrived at subjectively. Illustrations of subjective probability are: 1. Estimating the likelihood the New England Patriots will play in the Super Bowl next year. 2. Estimating the likelihood you will be married before the age of 30. 3. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10 years. 5-23 Summary of Types of Probability 5-24 8

The Complement Rule The complement rule is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. 5-25 s Find the complement of each event. 1-Rolling a die and getting a 4. Solution: Getting a 1, 2, 3, 5, or 6. 2-Selecting a letter of the alphabet and getting a vowel. Solution: Getting a consonant 3-Selecting a day of the week and getting a weekday. Solution: Getting a day of the weekend. 4-Selecting a one-child family and getting a boy. Solution: Getting a girl. 5-26 Rules for Computing Probabilities Rules of Addition Special Rule of Addition - If two events A and B are mutually exclusive, the probability of one or the other event s occurring equals the sum of their probabilities. P(A or B) = P(A) + P(B) The General Rule of Addition - If A and B are two events that are not mutually exclusive, then P(A or B) is given by the following formula: P(A or B) = P(A) + P(B) - P(A and B) 5-27 9

5-28 For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond. Solution: (a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13. (b) Since there is only one 6 of clubs, then P(6 of clubs) = 1/52. (c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond) = 16/52 = 4/13. In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution. Type Frequency A B AB O 22 5 2 21 50 = n 5-29 5-30 Find the following probabilities for the previous example. 1-A person has type O blood. Solution: P(O) = f /n = 21/50. 2- A person has type A or type B blood. Solution: P(A or B) = 22/50+ 5/50 = 27/50. 3- A person has neither type A nor type O blood. Solution: P(B or AB) = 5/50+2/50=7/50 4- A person does not have type AB blood. Solution: P(not AB) =1- P(AB) =1-2/50=48/50 10

5-31 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. The events are mutually exclusive, since the first event can be 1, 3 or 5 and the second event can be 2, 4 or 6. b. Getting a 3 and getting an odd number. The events are not mutually exclusive, since the first event is a 3 and then second event can be 1, 3 or 5. Hence, 3 is contained in both events. c. Getting an odd number and getting a number less than 4. The events are not mutually exclusive, since the first event can be 1, 3 or 5 and the second event can be 1, 2 or 3. Hence, 1 and 3 are contained in both events. d. Getting a number greater than 4 and getting a number less than 4. The events are mutually exclusive, since the first event can be 5 or 6 and the second event can be 1, 2 or 3. In a hospital unit there are 8 nurses and 5 physicians; 7 nurses and 3 physicians are females. If a staff is selected, find the probability that the subject is a nurse or a male. The events are not mutually exclusive and the sample space is Staff Female Males Total Nurses 7 1 8 Physicians 3 2 5 Total 10 3 13 Solution: P(nurse or male) = P(nurse) + P(male) P(male nurse) = 8/13 + 3/13 1/13 = 10/13 5-32 What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart? 5-33 P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52-1/52 = 16/52, or.3077 11

The Venn Diagram shows the result of a survey of 200 tourists who visited Florida during the year. The survey revealed that 120 went to Disney World, 100 went to Busch Gardens and 60 visited both. What is the probability a selected person visited either Disney World or Busch Gardens? P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch) = 120/200 + 100/200 60/200 =.60 +.50.80 5-34 Joint Probability Venn Diagram JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently. 5-35 Independent and Dependent Events Two events A and B are independent if the occurrence of one has no effect on the probability of the occurrence of the other. This rule is written: P(A and B) = P(A)P(B) 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 5-36 12

5-37 1- 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: Because these two events are independent (why?), P(queen and ace) = (4/52) (4/52) = 16/2704 = 1/169. 2- The probability that a specific medical test will show positive is 0.32. If four people are tested, find the probability that all four will show positive. Solution: Let T denote a positive test result. Then P(T and T and T and T) = (0.32) 4 = 0.010. Conditional Probability A conditional probability is the probability of a particular event occurring, given that another event has occurred. The probability of the event A given that the event B has occurred is written P(A B). 5-38 In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested. Solution: Since the events are dependent, P(D 1 and D 2 ) = P(D 1 ) P(D 2 D 1 ) = (2/25)(1/24) = 2/600 = 1/300. 5-39 13

Insurance Company found that 53% of the residents of a city had homeowner s insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner s and automobile insurance. Solution: Since the events are dependent, P(H and A) = P(H) P(A H) = (0.53)(0.27) = 0.1431. 5-40 Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball. Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24 see the tree diagram 5-41 Tree Diagram P(R B 1 ) 2/3 Red (1/2)(2/3) P(B 1 ) 1/2 Box 1 P(B B 1 ) 1/3 Blue (1/2)(1/3) P(B 2 ) 1/2 P(R B 2 ) 1/4 Box 2 Red (1/2)(1/4) 5-42 P(B B 2 ) 3/4 Blue (1/2)(3/4) 14

A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. What is the likelihood both shirts selected are white? 5-43 The event that the first shirt selected is white is W 1. The probability is P(W 1 ) = 9/12 The event that the second shirt (W 2 )selected is also white. The conditional probability that the second shirt selected is white, given that the first shirt selected is also white, is P(W 2 W 1 ) = 8/11. To determine the probability of 2 white shirts being selected we use formula: P(AB) = P(A) P(B A) P(W 1 and W 2 ) = P(W 1 )P(W 2 W 1 ) = (9/12)(8/11) = 0.55 Contingency Tables A CONTINGENCY TABLE is a table used to classify sample observations according to two or more identifiable characteristics A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table Gender Yes No Total Male 32 32 18 18 50 50 Female 8 42 42 50 50 Total 40 40 60 60 100 5-44 1- Find the probability that the respondent answered yes given that the respondent was a female. Solution: Let M = respondent was a male; F = respondent was a female; Y = respondent answered yes ; N = respondent answered no. P(Y F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25. 5-45 2- Find the probability that the respondent was a male, given that the respondent answered no. Solution: P(M N) = [P(N and M)]/P(N) = [18/100]/[60/100] = 3/10. 15

Using Tree Diagrams Using a tree diagram is useful for portraying conditional and joint probabilities. It is particularly useful for analyzing business decisions involving several stages. 5-46 5-47 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. (# of selecting 2 defectives) (# of selecting 2 nondefectives) P ( exactly 2 defectives ) = # of selecting 4 transistors 4! 20! 4C 2 20C 2 2! 2! 18! 2! 1140 190 = = = = 24! 24C 4 10626 1771 20! 4! # of selecting no defective 20C 4 1615 P( no defective ) = = = # of selecting 4 transistors 24C 4 3542 # of selecting defective 4C 4 1 P ( all defective) = = = # of selecting 4 transistors 24C 4 10626 20C 4 1615 1927 P ( at least 1 defective) = 1 P ( no defective) = 1 = 1 = 24C 4 3542 3542 b. None is defective. c. All are defective. d. At least 1 is defective. 16