Probability and Statistics Chapter 3 Notes Section 3-1 I. Probability Experiments. A. When weather forecasters say There is a 90% chance of rain tomorrow, or a doctor says There is a 35% chance of a successful surgery, they are stating the likelihood, or, that a specific event will occur. 1. Decisions such as should you plan a picnic for tomorrow, or should you proceed with surgery are almost always based on these probabilities. B. A probability experiment is an, or, through which specific results (, or ) are obtained. 1. The result of a single trial in a probability experiment is an. 2. The set of all possible outcomes of a probability experiment is the. 3. An is a of the sample space. a) An event may consist of one or more. C. Simple Example of the use of the terms probability experiment, sample space, event, and outcome. 1. Probability experiment:. 2. Sample Space: 3. Event: 4. Outcome: Section 3-1 Example 1: A probability example consists of tossing a coin and then rolling a six-sided die. Determine the number of outcomes and identify the sample space. There are possible outcomes when tossing a coin; a, or a. For each of these, there are possible outcomes when rolling a die:. One way to list all the possible outcomes (the sample space) for actions occurring in a sequence is to use a. D. Events are often represented by letters, such as and. 1. An event that consists of a single outcome is called a simple event. a) In Example 1, the event tossing heads and rolling a 3 is a simple event and can be represented as A = b) In contrast, the event tossing heads and rolling an even number is NOT simple because it consists of possible outcomes: B = Section 3-1 Example 2: Determine the number of outcomes in each event. Then decide whether each event is simple or not. 1) For quality control, you randomly select a machine part from a batch that has been manufactured that day. Event A is selecting a specific defective machine part. 2) You roll a six-sided die. Event B is rolling at least a 4. Event A event; you either pick that specific defective part or you don t Event B event; it has possible outcomes. You ask for a student s age at his or her last birthday. Decide whether each event is simple or not. 1) Event C: The student s age is between 15 and 18, inclusive. 2) Event D: The student s age is 17. Event C is ; the ages of 15, 16, 17 and 18 are all possible outcomes. Event D is : the only possible successful outcome is that the student is 17 years old.
II. The Fundamental Counting Principle A. If one event can occur in m ways and a second event can occur in n ways, the number of ways that the two events can occur in sequence is. 1) This rule can be extended for of events occurring in. B. In plain English, the number of ways that events can occur in sequence is found by the number of ways can occur by the number of ways the other event(s) can occur. Section 3-1 Example 3: 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), and green (G) How many different ways can you select one manufacturer, one car size, and one color? Use a tree diagram to check your answer. There are 3 manufacturers, 2 sizes, and 4 colors, so the number of different ways is equal to * *, or. Section 3-1 Example 4: The access code for a car s security system consists of 4 digits. Each digit can be 0 through 9. 1) How many access codes are possible if each digit can be used only once, and not repeated? 2) How many access codes are possible if each digit can be repeated? 3) How many access codes are possible if each digit can be repeated, but the first digit cannot be 0 or 1? SOLUTIONS: 1) There are possible choices for the first digit ( ) There are possible choices for the second digit. There are possible choices for the third digit. There are possible choices for the fourth digit. * * * = possible codes. 2) There are possible choices for the first digit ( ) There are possible choices for the second digit. There are possible choices for the third digit. There are possible choices for the fourth digit. * * * = possible codes. 3) There are possible choices for the first digit ( ) There are possible choices for the second digit. There are possible choices for the third digit. There are possible choices for the fourth digit. * * * = possible codes. III. Types of Probability A. Classical (or probability) is used when each outcome in a sample space is likely to occur. The classical probability for an event E is given by P(E) = Section 3-1 Example 5: 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 number of outcomes in event E Total number of outcomes in sample space.
SOLUTIONS: 1) The sample space consists of outcomes: There is outcome in Event A: A = So, P (rolling a 3) = 2) The sample space consists of 6 outcomes: {1, 2, 3, 4, 5, 6} There are outcomes in Event B: So, P (rolling a 7) = 2) The sample space consists of 6 outcomes: {1, 2, 3, 4, 5, 6} There are outcomes in Event B: So, P (rolling < 5) = B. Empirical (or statistical) probability is used when each outcome of an event is likely to occur. When an experiment is repeated times, regular patterns are formed. These patterns make it possible to find probability. Empirical (or statistical) probability) is based on obtained from. The empirical probability of an event E is the frequency of event E. Frequency of event E P(E) = Total frequency n Section 3-1 Example 6: A company is conducting an online survey of randomly selected individuals to determine if traffic congestion is a problem in their community. So far, 320 people have responded to the survey. The frequency distribution shows the results. What is the probability that the next person that responds to the survey says that traffic congestion is a serious problem in their community? The event is a response of It is a serious problem. The frequency of this event is. Because the total of the frequencies is, the empirical probability of the next person saying that traffic congestion is a serious problem in their community is: P(Serious problem) = Frequency of event "It is a serious problem" Total frequency = C. probability: these result from intuition, educated guesses and estimates. They are not as trusted in statistical studies as theoretical or empirical probabilities are. D. Law of Large Numbers A. As an experiment is repeated over and over, the empirical probability of an event the ( ) probability of an event. If you toss a fair coin times, you may only get heads. If you toss a fair coin times, you will get very close to heads. Section 3-1 Example 7: You survey a sample of 1000 employees at a large company and record the age of each. The results are shown in the frequency distribution. If you randomly select another employee, what is the probability that the employee will be between 25 and 34 years old? Ages 15 to 24 25 to 34 35 to 44 45 to 54 55 to 64 65 and over Total Frequencies 54 366 233 180 125 42 1000
The event is selecting an employee who is between 25 and 34 years old. In your survey, the frequency of this event is 366. Because the total of the frequencies is 1000, the probability of selecting an employee between the ages of 25 and 34 years old is: P(25 to 34 years old) = Frequency of event "25 to 34 years old" Total frequency = Section 3-1 Example 8: Classify each statement as an example of classical probability, empirical probability, or subjective probability 1. The probability that you will be married by age 30 is 0.50. 2. The probability that a voter chosen at random will vote Republican is 0.45. 3. The probability of winning a 1000-ticket raffle with one ticket is SOLUTIONS 1. This is a probability, based on a or a. 2. This is an example of an probability, since it most likely resulted from a or and the outcomes are likely. 3. This is an example of ( ) probability. You know the number of outcomes, and each one is likely to occur. E. Range of Probabilities Rule 1. Very important to remember this AT ALL TIMES!! ALL probabilities have a value between and. An impossible event has a probability of. An event that is guaranteed to occur has a probability of. An event that has an equal chance of occurring or not occurring has a probability of. Any event with a probability of less than or greater than is considered to be unusual when it occurs. F. Complementary Events 1. Complementary events have probabilities that add up to. If there is a 76% chance of rain, and a 24% chance that it doesn t rain, raining or not raining are events. 2. Definition: The complement of event E is the set of all in a sample space that are not in event E. The complement of event E is denoted by E and is read as E prime. Section 3-1 Example 9: Use the frequency distribution in Example 7 to find the probability of randomly choosing an employee who is not between 25 and 34 years old. We already know that the probability of an employee being between 25 and 34 years old is. We also know that the event not between 25 and 34 is the of between 25 and 34. So, we can subtract from to get the probability that an employee is not between 25 and 34 years old. =. Section 3-2 I. Conditional Probability A conditional probability is the probability of an event occurring, given that another event has already occurred. The conditional probability of event B occurring, given that event A has occurred, is denoted P(B A) and is read as probability of B, given A. II. Independent and Dependent Events Two events are if the occurrence of one of the events affect the probability of the occurrence of the other event. Two events A and B are if P(B A) = P(B), or if P(A B) = P(A). 1. 1000
In other words, event B is likely to occur whether event A has or not. Events that are not independent are. To determine if A and B are independent, first calculate, the probability of event. Then calculate, the probability of, given. If the two probabilities are, the events are. If the two probabilities are not, the events are. III. The Multiplication Rule P(A and B) = P(A) P(B A) If Events A and B are dependent; 1) Find the probability the first event occurs 2) Find the probability the second event occurs given the first event has occurred. 3) Multiply these two probabilities to find the probability that both events will occur in sequence. If Events A and B are independent, P(A and B) = P(A) P(B). This simplified rule can be extended for any number of independent events, just like the Fundamental Counting Principle could be extended. Example 1 (Page 149) 1) 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 (it s not put back into the deck before the queen is drawn. 2) 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 33 19 52 Normal IQ 39 11 50 Total 72 30 102 SOLUTIONS: 1) The probability of drawing a queen after a king (or any other card) has been taken out of the deck is, or about. 2) There are children who have the gene. They are our. Of these, have a high IQ. So, P(B A) =. Example 2 (Page 150) Decide whether the events are independent or dependent. 1) Selecting king from a standard deck (A), not replacing it, and then selecting a queen from the deck (B). 2) Tossing a coin and getting a head (A), and then rolling a six-sided die and obtaining a 6 (B). SOLUTIONS: 1) The probability of pulling a queen out of the deck is. The probability of pulling a queen out of the deck after a king has already been removed is. Since these probabilities are, the events are. 2) The probability of obtaining a 6 is. The probability of obtaining a 6 given that the coin came up heads is. Since these probabilities are, the events are. Example 3 (Page 151) 1) 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. 2) A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6. SOLUTIONS: 1) Because the first card is not replaced, the events are. P(K and Q) = * * = 2) The two events are. P(H and 6) = P(H) P(6) * = Example 4 (Page 152) 1) A coin is tossed and a die is rolled. Find the probability of getting a tail and then rolling a 2.
2) The probability that a particular knee surgery is successful is 0.85. Find the probability that three surgeries in a row are all successful. 3) Find the probability that none of the three knee surgeries is successful. 4) Find the probability that at least one of the three knee surgeries is successful. SOLUTIONS 1) The probability of getting a tail and then rolling a 2 is P(T and 2) = P(T) P(2) * = 2) The probability that each knee surgery is successful is 0.85. Since these events are, the probability that all three are successful is found by their probabilities together. The probability that all three surgeries are successful is about. 3) Because the probability of success for one surgery is.85, the probability of failure for one surgery is. This is true because failure is the of success. The probability that each knee surgery is not successful is. Since these events are, the probability that all three are not successful is found by their probabilities together. The probability that none of the surgeries are successful is about. 4) The phrase at least one means. The complement to the event at least one is successful is the event. Using the complement rule, we can simply the probability that none were successful from to find the probability that at least one was successful. =. The probability that at least one of the surgeries is successful is about. Example 5 (Page 153) More than 15,000 medical school seniors applied to residency programs in 2007. Of those, 93% were matched to a residency position. 74% percent of the seniors matched to a residency position were matched to one of their top two choices. 1) Find the probability that a randomly selected senior was matched to a residency program and it was one of the senior s top two choices. 2) Find the probability that a randomly selected senior that was matched to a residency program did not get matched with one of the senior s top two choices. 3) Would it be unusual for a randomly selected senior to result in a senior that was matched to a residency position and it was one of the senior s top two choices? 1) These two events are : P(A and B) = P(A) P(B A) = *. 2) To find this probability, use the complement: P(B A) = 1 P(B A) = 1 - =. 3) : This event occurs around % of the time. Section 3-3 Mutually exclusive events are events that both happen at the time. The Addition Rule (For OR probabilities) Or can mean one of three things: A occurs and B does not occur. B occurs and A does not occur. Both A and B occur. If A and B are mutually exclusive, simply their probabilities of occurring together. If A and B are NOT mutually exclusive, add their probabilities of occurring together and then subtract the probability that.
Type of Probability & Rules Classical Probability Empirical Probability Range of Probabilities Rule Complementary Events Multiplication Rule Addition Rule In Words The number of outcomes in the sample space is known and each outcome is equally likely to occur The frequency of outcomes in the sample space is estimated from experimentation (you have data) ALL probabilities are between zero and 1, inclusive The complement of event E is the set of all outcomes in a sample space that are NOT included in E, denoted E The Multiplication Rule is used to find the probability of two events occurring in a sequence (AND) The Addition Rule is used to find the probability of at least one of two events In Symbols Number of outcomes in E P(E) = Number of outcomes in sample space Frequency of E Total Frequency = Frequency of E Total Frequency = f n 0 P(E) 1 P(E ) = 1 P(E); P(E) = 1 P(E ) P(A and B) = P(A) P(B A) (dependent) P(A and B) = P(A) P(B) (independent) P(A or B) = P(A) + P(B) P(A and B) P(A or B) = P(A) + P(B) (mutually exclusive) occurring (OR) Example 1: Determine whether these two events are mutually exclusive or not. 1) Roll 3 on a die AND roll 4 on a die. 2) Randomly select a male student AND randomly select a nursing major. 3) Randomly select a blood donor with type O blood AND randomly select a female blood donor. 4) Randomly select a jack from a deck of cards AND randomly select a face card from a deck of cards. 5) Randomly select a 20-year-old student AND randomly select a student with blue eyes. 6) Randomly select a vehicle that is a Ford AND randomly select a vehicle that is a Toyota. SOLUTIONS 1) You roll and 3 AND a 4 on the same roll. These are. 2) You be both a male and a nursing major. These are. 3) You be both a female and have type 0 blood. These are. 4) Jacks are face cards, so any jack is also a face card. These are. 5) You be a 20-year-old student and have blue eyes. These are. 6) A vehicle be a Ford and a Toyota at the same time. These are. Example 2: 1) You select a card from a standard deck. Find the probability that the card is a 4 or an ace. 2) You select a card from a standard deck. Find the probability that the card is a queen or a red card. SOLUTIONS 1) Ace and 4 are. the probability of getting an ace to the probability of getting a 4. The probability of getting an ace is ; the probability of getting a four is also. The probability of getting an ace or a four is 4 52 + 4 52 = 8 52 = 2 13. 2) The events are (there are red queens). the probability of getting a to the probability of getting a, then the probability of getting a. P(queen) = ; P(red) = ; P(red queen) = ( 4 + 26 52 52 52 52 52 52 Example 3 The frequency distribution shows the volume of sales (in dollars) and the number of months 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 0-24,999 25,000-49,999 50,000-74,999 75,000-99,999 100,000-124,999 125,000-149,999 150,000-174,999 175,000-199,999 Months 3 5 6 7 9 2 3 1
Define Event A as monthly sales between and Define Event B as monthly sales between and Because events A and B are, the probability that the sales representative will sell between $75,000 and $124,999 next month is: P(A or B) = 0.444 Example 4 A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last five days. The number of donors who gave each blood type is shown in the table. A donor is selected at random. 1. Find the probability that the donor has type O or type A blood. 2. Find the probability that the donor has type B or is Rh-negative. Blood Type Rh-Factor SOLUTIONS: 1. These events are. P(O or A) = P(O) + P(A) 184 + 164 = 348 0.851 409 409 409 O A B AB Total Positive 156 139 37 12 344 Negative 28 25 8 4 65 Total 184 164 45 16 409 2. These events are. P(B or Rh neg) = P(B) + P(Rh neg) P(B and Rh neg) 45 + 65 8 = 102 0.249 409 409 409 409 Example 5 Use the graph to find the probability that a randomly selected draft pick is not a running back or a wide receiver. Define Events A and B Event A: Draft pick is a running back. Event B: Draft pick is a wide receiver. These events are, so the probability that the draft pick is a running pick or a wide receiver is: P(A or B) = P(A) + P(B) 25 + 34 = 59 0.231 255 255 255 To find the probability that the draft is NOT a running back or wide receiver, simply subtract the probability that he was one of those positions from 1 ( rule) 1 0.231 = 0.769 Section 3-4 Permutations A permutation is an arrangement of objects. The number of different permutations of n distinct object is n! The! Symbol means factorial and indicates that you start with the number given and multiply that times every number between that number and zero. 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880 The calculator will do factorials for you; Math Prb 4 accesses that feature. To find the number of ways that a permutation can occur, use the TI-84. Math PRB 2 Enter the total number of items in your list.
Press Math - PRB 2 and Enter Enter the number of items you wish to order and press enter. Distinguishable Permutations A distinguishable permutation must be done by hand. n! n 1!n 2!n 3! n k! Alpha and y= on your calculator will allow you to enter this as a fraction, making it easier. Again, the calculator will also let you enter the factorial symbol, saving time and reducing the chance of making a simple error in entering the numbers. Combinations A combination is a of r objects from a group of n objects regard to. To find the number of ways that a combination can occur, use the TI-84. Math PRB 3 Enter the total number of items in your list. Press Math - PRB 3 and Enter Enter the number of items you wish to select and press enter. Example 1 The objective of a 9 x 9 Sudoku number puzzle is to fill the grid so that each row, each column, and each 3 x 3 grid contains the digits 1 to 9, with no repeats. How many different ways can the first row of a blank 9 x 9 Sudoku grid be filled. There are digits that could fill the first spot, for the second spot, and so forth. This can be expressed as either, or as. 9! = 9P 9 also equals. So, there are ways to fill the first row of a 9 x 9 Sudoku number puzzle. Example 2 Find the number of ways of forming three-digit codes in which no digit is repeated. To form a three-digit code with no repeating digits, you need to select digits from a group of, so n = and r =. 10P 3 =. You could also simply say that you had choices for the first digit, choices for the second digit, and choices for the third digit. * * =. Example 3 Forty-three race cars started the 2007 Daytona 500. How many ways can the cars finish first, second, and third? You want to order 3 cars out of 43. This is a. = * * = cars could win the race. Once the winner has won, there are left to fight for 2 nd. Once 2 nd has been determined, there are left to battle it out for 3 rd. Example 4 A building contractor is planning to develop a sub-division. The sub-division is to consist of 6 one-story houses, 4 twostory houses and 2 split-level houses. In how many distinguishable ways can the houses be arranged? 12! 6!4!2! = different distinguishable ways to arrange the houses. Example 5 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? =. There are different combinations of 4 companies that can be selected from the 16 bidding companies.
Example 6 A student advisory board consists of 17 members. Three members serve as the board s chair, secretary, and webmaster. Each member is equally likely to serve any of the positions. What is the probability of selecting at random the three members that hold each position? There are different ways that the three positions can be filled. There are different ways that the three positions can be filled. The chance that you randomly pick the one correct outcome is 1 4080 Example 7 You have 11 letters consisting of one M, four I s, four S s and two P s. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word Mississippi? This is a distinguishable permutation question. There are 11! 1!4!4!2! different distinguishable ways to arrange those 11 letters. There are different distinguishable ways to arrange those 11 letters. 1 The probability of randomly picking the one arrangement that spells Mississippi is. 34650 Example 8 Find the probability of being dealt five diamonds from a standard deck of playing cards. You need to determine how many ways you can get 5 diamonds, and then how many different 5 card hands are possible. Once you have these values, the number of ways to get 5 diamonds by the number of possible hands to find out how likely it is that you get 5 diamonds. 13C 5 = and 52C 5 = 1287 P(5 diamonds) =. 2,598,960 The probability of getting a diamond flush is approximately. Example 9 A food manufacturer is analyzing a sample of 400 corn kernels for the presence of a toxin. In this sample, three kernels have dangerously high levels of the toxin. If four kernels are randomly selected from the sample, what is the probability that exactly one kernel contains a dangerously high level of the toxin? Find the number of possible ways to choose one toxic kernel and three non-toxic kernels, those together and the answer by the number of ways to choose 4 kernels. This tells you how likely it is that you will randomly choose exactly one toxic kernel out of four. = and = 3 * 10,349,790 = There are different ways to select one toxic kernel out of 400. = There are different ways to select 4 kernels out of 400. 31,049,370 P(exactly 1 toxic kernel) =. 1,050,739,900 The probability of getting exactly one toxic kernel is approximately.