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PROBABILITY & STATISTICS FOR P-8 TEACHERS Chapter 4 Probability PROBABILITY Probability can be defined as the chance of an event occurring. It can be used to quantify what the “odds” are that a specific event will occur. Some examples of how probability is used everyday would be weather forecasting, “75% chance of snow” or for setting insurance rates. PROBABILITY Probabilities must be between 0 and 1, inclusive. A probability of 0 indicates impossibility. A probability of 1 indicates certainty. 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 is the set of all possible outcomes of a probability experiment. An event consists of outcomes. SAMPLE SPACES Experiment Toss a coin Roll a die Answer a true/false question Toss two coins Sample Space Head, Tail 1, 2, 3, 4, 5, 6 True, False HH, HT, TH, TT SAMPLE SPACES Find the sample space for rolling two dice. 36 different possible outcomes SAMPLE SPACES Find the sample space for the outcomes of tossing 3 coins HHH HHT HTH HTT THH THT TTH TTT 8 different possible outcomes TREE DIAGRAM Use a tree diagram to find the sample space for tossing three coins. H H T H T T H HHH T HHT H HTH T HTT H THH T THT H TTH T TTT Notation P - denotes a probability. A, B, and C - denote specific events. P(A) - denotes the probability of event A occurring. SAMPLE SPACES AND PROBABILITY Probability of an event is the ratio of the outcome of the event to the sample space number of ways E can occur P (E) = number of possible outcomes PROBABILITY Determine the probability of getting exactly two heads when tossing a coin three times Sample Space HHH HHT HTH HTT THH THT THT TTT 3 ways two heads occur 8 possible outcomes The probability of exactly two heads is 3/8. PROBABILITY If two dice are rolled one time, find the probability of getting a sum of 7. P(sum of 7) = 6/36 RELATIONSHIPS AMONG EVENTS P(A or B) The event “either A or B or both occur” P(A & B) The event “both A and B occur” P(not E) The event “E does not occur” Union Intersection Complement COMPLEMENT RULE The complement of event E is the set of outcomes in the sample space that are not included in the outcomes of event P (not E) = 1 – P(E) FINDING COMPLEMENTS Find the complement of each event. Event Complement of the Event Rolling a die and getting a 4 Getting a 1, 2, 3, 5, or 6 Selecting a letter of the alphabet and getting a vowel Getting a consonant (assume y is a consonant) Selecting a month and getting a month that begins with a J Getting February, March, April, May, August, September, October, November, or December Selecting a day of the week and getting a weekday Getting Saturday or Sunday COMPLEMENT Use the complement rule to find the probability of rolling a dice and getting a number larger than 1. P(1) = 1/6 P(>1) = 5/6 Complement Rule P(>1)= 1 – P(1) = 1 – 1/6 = 5/6 ADDITION RULE The rule of addition applies to the following situation: We have two events from the same sample space, and we want to know the probability that either event occurs. P (A or B) = P(A) + P(B) – P(A and B) ADDITION RULE A card is drawn randomly from a deck of ordinary playing cards. Find the probability of picking a heart or an ace. We know the following: There are 52 cards in the deck. There are 13 hearts, so P(H) = 13/52. There are 4 aces, so P(A) = 4/52. There is 1 ace that is also a heart, so P(H and A) = 1/52. Therefore, based on the rule of addition: P(H or A) = P(H) + P(A) - P(H and A) P(H or A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13 MUTUALLY EXCLUSIVE EVENTS Two events are mutually exclusive events if they cannot occur at the same time no outcomes in common P(A and B) = 0 MUTUALLY EXCLUSIVE 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 Getting an odd number: 1, 3, or 5 Getting an even number: 2, 4, or 6 Mutually Exclusive MUTUALLY EXCLUSIVE Determine which events are mutually exclusive and which are not, when a single die is rolled. b. Getting an odd number and getting a number less than 4 Getting an odd number: 1, 3, or 5 Getting a number less than 4: 1, 2, or 3 Not Mutually Exclusive SPECIAL ADDITION RULE If two events are mutually exclusive then the addition rule can be simplified since no intersection occurs: 0 P (A or B) = P(A) + P(B) – P(A and B) P (A or B) = P(A) + P(B) BLOOD TYPES 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. Type Frequency A 22 B 5 AB 2 O 21 Total 50 f P O n 21 50 BLOOD TYPES 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. b. A person has type A or type B blood. Type Frequency A B AB O 22 5 2 21 Total 50 22 5 P A or B 50 50 27 50 BLOOD TYPES 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. c. A person does not have type AB blood. Type Frequency A B AB O 22 5 2 21 Total 50 P not AB 1 P AB 2 48 24 1 50 50 25 CONDITIONAL PROBABILITY Additional information or other events occurring may have an impact on the probability of an event. Conditional probability is the probability that the second event B occurs given that the first event A has occurred. CONDITIONAL PROBABILITY Additional information may have an impact on the probability of an event. Example: Consider rolling a dice and getting a 6 Sample space is {1, 2, 3, 4, 5, 6} P(rolling a 6) = 1/6 Suppose we know an even number was rolled New sample space is {2, 4, 6} P(rolling a 6) = 1/3 CONDITIONAL PROBABILITY Notation: P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred. (read P(B|A) as “B given A”) Conditional Probability P A and B P B A P A CONDITIONAL PROBABILITY Consider a standard 52 card deck Let A = drawing a card less than five (2, 3, or 4) B = drawing a spade Find P(B|A) Given a card less than five New sample space (12 cards) Spade includes 3 cards P(B|A) = 3/12 CONDITIONAL PROBABILITY Consider a standard 52 card deck Let A = drawing a card less than five (2, 3, or 4) B = drawing a spade Find P(B|A) using the formula P(A) = 12/52 P(A and B) = 3/52 Conditional Probability P A and B 3/52 P B A = = 3/12 12/52 P A PARKING TICKETS The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket. N= parking in a no-parking zone, T= getting a ticket P N and T 0.06 0.30 P T N 0.20 P N WOMEN IN THE MILITARY 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. WOMEN IN THE MILITARY Find the probability that the respondent answered yes (Y), given that the respondent was a female (F). P F and Y P Y F P F 8 8 100 50 50 100 4 25 MULTIPLICATION RULE We can create a multiplication rule for the intersection of two events A and B by rewriting the conditional probability formula P(A and B) P(B|A) = P(A) Solving for the intersection yields: P(A and B) = P(A) P(B|A) or P(A and B) = P(B) P(A|B) MULTIPLICATION RULE From the bag of marbles, find the probability of picking two red marbles from a bag with 4 red and 3 blue marbles. Pick a red marble from the first draw is 4/7 Since a red has already been drawn, then 3 red marbles are left from the remaining 6 marbles P(red and red) = P(red) P(red|red) = (4/7) (3/6) = 6/21 MULTIPLICATION RULE 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 burglaries to further investigate, find the probability that both will have occurred in 2004. Dependent Events P C1 and C2 P C1 P C2 C1 16 15 60 53 52 689 INDEPENDENCE Two events A and B are independent events if the fact that A occurs does not affect the probability of B occurring. Examples: Tossing a coin and rolling a die Choosing a 3 from a deck of cards, replacing it, and then choosing an ace as the second card INDEPENDENCE If two events are independent, then the events do not affect the probability outcomes One way to represent this is by: P(B|A) = P(B) since event A has no affect on event B regardless of whether A was given to us or not SPECIAL MULTIPLICATION RULE If two events are independent then the multiplication rule can be simplified since the conditional probability has no affect on the initial probability P(B) P (A and B) = P(A) P(B|A) P (A and B) = P(A) P(B) INDEPENDENT EVENTS A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die. Independent Events P Head and 4 P Head P 4 1 1 1 2 6 12 This problem could be solved using sample space. H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 INDEPENDENT EVENTS A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week. Independent Events P S and S and S P S P S P S 0.46 0.46 0.46 0.097 INDEPENDENT EVENTS Determine if the following events, based on a standard deck of cards, are independent of each other A = Drawing an ace B = Drawing a heart Show P(B|A) = P(B) P(B) = 13/52 = 1/4 P(B|A) = 1/4 Since P(B|A) = P(B) = 1/4 then the events are independent BASIC COUNTING PRINCIPLES Counting problems are of the following kind: “How many different 8-letter passwords are there?” “How many possible ways are there to pick 11 soccer players out of a 20player team?” “What is the probability of winning the lottery?” FUNDAMENTAL COUNTING RULE Stan is about to order dinner at a restaurant. He has a choice of: • two appetizers (soup or salad) • three main courses (pasta, steak, or fish) • two desserts (cake or pie) How many different meal combinations can Stan choose? Using a Tree Diagram to Determine the Number of Combinations Appetizers Main Courses pasta soup steak fish pasta salad steak fish Desserts cake pie cake pie cake pie 12 Combinations cake pie cake pie cake pie COUNTING RULES The fundamental counting rule is also called the multiplication of choices. If a task is made up of many stages, the total number of possibilities for the task is given by m×n×p× ... where m is the number of choices for the first stage, and n is the number of choices for the second stage, p is the number of choices for the third stage, and so on. FUNDAMENTAL COUNTING RULE 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? # of × # of × # of × # of Colors Types Textures Uses 7 × 2 × 3 84 different kinds of paint × 2 = 84 FUNDAMENTAL COUNTING RULE Colleen has six blouses, four skirts and four sweaters. How many different outfits can she choose from, assuming she wears three different items at once? Blouses Skirts Sweaters = 96 outfits 6 × 4 × 4 The final score in a soccer game is 5 to 4 for Team A. How many different half-time scores are possible? Team A Team B (0 - 4) (0 - 5) 6 × 5 = 30 different possible half-time scores. FUNDAMENTAL COUNTING RULE How many four-letter arrangements are possible? use any of the 26 letters: A-Z 26 × 26 × 26 × 26 1st 2nd 3rd 4th = 456,976 How many four-letter arrangements are possible if a letter cannot be repeated? use any of the 26 letters: A-Z 26 × 25 × 24 × 23 1st 2nd 3rd 4th only 25 only 24 only 23 letters letters letters left left left = 358,800 PERMUTATIONS A permutation is an arrangement of a set of objects for which the order of the objects is important. Show the number of ways of arranging the letters of the word CAT: CAT CTA ACT ATC TCA TAC There are six arrangements or permutations of the word CAT. By changing the order of the letters, you have a different permutation. Using the Fundamental Counting Principle, we would have 3 x 2 x 1 number of distinct arrangements or permutations. FACTORIAL NOTATION In calculating permutations, we often encounter expressions such as 3 x 2 x 1. The product of consecutive natural numbers in decreasing order down to the number one can be represented using factorial notation: Factorial is the product of all the positive numbers from 1 to a number. n ! n n 1 n 2 3 2 1 0! 1 3 x 2 x 1 = 3! This is read as “three factorial”. FACTORIAL NOTATION 4! = 4 × 3 × 2 × 1 = 24 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800 10! 10 9 8 7 .... 3 2 1 7! 7 6 5 ... 3 2 1 10 9 8 = 720 PERMUTATIONS Permutation is an arrangement of objects in a specific order. Order matters. n! n n 1 n 2 n r 1 n Pr n r ! r items In general, if we have n objects but only want to select r objects at a time PERMUTATIONS 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. How many different ways can she rank the 5 locations? first second third fourth fifth choice choice choice choice choice 5 4 3 2 1 120 different ways to rank the locations Using factorials, 5! = 120. Using permutations, 5P5 = 120. PERMUTATIONS Suppose the business owner wishes to rank only the top 3 of the 5 locations. How many different ways can she rank them? first second third choice choice choice 5 4 3 60 different ways to rank the locations Using permutations, 5P3 = 60. PERMUTATIONS How many three-letter permutations can be formed from the letters of the word DINOSAUR? the number of permutations of eight objects taken three at a time. This may be written as 8P3. 8! = 8P3 = (8–3)! 8! = 5! 8×7×6×5×4×3×2×1 = 336 5×4×3×2×1 PERMUTATIONS Find the following permutations 6! = 6 × 5 = 30 6P2 = 4! 7P4 = 7! = 7 × 6 × 5 × 4 = 840 3! 3! =3×2×1=6 3P3 = 0! Note: nPn = n! PERMUTATIONS 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? Order matters, so we will use permutations. 9! 72 9 P2 7! or P 9 8 72 9 2 2 PERMUTATIONS WITH REPETITION How many four-letter words can be made using the letters of PEER? 4P4 = 4! Note that PEER and PEER are not distinguishable, therefore, they count as one permutation. When this happens, you must divide by the factorial of the number of repeated terms. For PEER, the number of arrangements is: “E” occurs twice 4! = 4 × 3 = 12 = 2! 2! 4P4 COUNTING RULES Consider selecting two people out of three individuals (A, B, C) for President and Vice-President of a committee 3P2 =6 AB BA AC CA BC CB Now consider selecting two people out of three individuals (A, B, C) for AB committee membership Here choice A,B is the same committee as choice B,A AC BC 6 choices 3 choices COMBINATIONS Combination is a grouping of objects. Order does not matter. n! n Cr n r !r ! Pr r! n When order matters, you have permutations. When order does not matter, you have combinations COMBINATIONS A committee of four students is to be selected from a group of ten students. In how many ways can this be done? 10C4 10! (10 4)! 4! = 210 The committee of four can be selected in 210 ways. COMBINATIONS A newspaper editor has received 8 books to review. He decides that he can use 3 reviews in his newspaper. How many different ways can these 3 reviews be selected? The placement in the newspaper is not mentioned, so order does not matter. We will use combinations. 8! 8!/ 5!3! 56 8 C3 5!3! P3 or 8C3 56 3! 8 COMBINATIONS A company is hiring people to fill five identical positions. There are 12 applicants. How many ways can the five positions be filled? 12C5 12! (12 5)! 5! The company can fill the five positions 792 ways. = 792 The company wants to hire Applicant A and any four of the others. How many ways can the five positions be filled now? 1! 11! With Applicant A, 1C1 x 11C4 1!0! 7!4! there are 330 ways to = 330 fill the position COMBINATIONS There are seven books to choose from. How many ways can five or more books be selected? Select 5 or 6 or 7: 7C5 + 7C6 + 7C7 = 21 + 7 + 1 = 29 There are 29 ways to select 5 or more books. COMBINATIONS In a club 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? There are not separate roles listed for each committee member, so order does not matter. We will use combinations. 7! 5! Women: 7C3 35, Men: 5C2 10 4!3! 3!2! There are 35·10=350 different possibilities. COMBINATIONS How many five-card hands can be dealt from a standard deck of 52 cards if: a) each hand must contain two aces? 4C2 x 48C3 = 103 776 b) each hand must contain three red cards? 26C3 x 26C2 = 845 000 PROBABILITY AND COUNTING RULES The counting rules can be combined with the probability rules in this chapter to solve many types of probability problems. By using the fundamental counting rule, the permutation rules, and the combination rule, you can compute the probability of outcomes of many experiments, such as getting a full house when 5 cards are dealt or selecting a committee of 3 women and 2 men from a club consisting of 10 women and 10 men. PLAYING CARDS Two cards are picked at random from a deck of playing cards. What is the probability that they are both kings? Method 1: Using the fundamental counting principle 1 4 3 P(2 kings) = 221 52 51 Method 2: Using combinations Number of ways of choosing 2 kings from 4 kings is 4C2. Number of ways of choosing 2 cards from 52 cards is 52C2. C2 P(2 kings) = 52 C 2 4 4! (4 2)!2! 52! (52 2)!2! 1 221 MAGAZINE SELECTION A store has 6 TV Graphic magazines and 8 Newstime magazines on the counter. If two customers purchased a magazine, find the probability that one of each magazine was purchased. TV Graphic: One magazine of the 6 magazines Newstime: One magazine of the 8 magazines Total: Two magazines of the 14 magazines 6 C1 8 C1 6 8 48 91 91 14 C2 COMBINATION LOCKS A combination lock consists of 50 numbers (1 through 50). If a 3-number combination is needed, find the probability that the combination is the number 1-2-3 in that order. The same number can be used more than once. (Note: A combination lock is really a permutation lock.) There are 50·50·50 = 125,000 possible combinations. The number 1-2-3 is one combination. P(1-2-3) = 1/125,000