MATH 225N MATH Stats week 4, Exams of Nursing

Download MATH 225N MATH Stats week 4 and more Exams Nursing in PDF only on Docsity! 1 3.1 Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment. A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes. An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A). The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads). Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer. To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes that meet this condition {HT, TH}, so P(A) = 24 = 0.5. Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P(E) = 26. If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 26 of the rolls would result in an outcome of "at least five". You would not expect exactly 26. The long-term relative frequency of obtaining this result would approach the theoretical probability of 26 as the number of repetitions grows larger and larger. This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.) It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would 2 be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely. "∪" Event: The Union An outcome is in the event A ∪ B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice. 5 h. P(A|B) = P(A∩B)P(B) = 36, P(B|A) = P(A∩B)P(A) = 39, No Try It 3.1 The sample space S is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)). a. S = Let event A = the sum is even and event B = the first number is prime. b. A = , B = c. P(A) = , P(B) = d. A ∩ B = , A ∪ B = e. P(A ∩ B) = , P(A ∪ B) = f. B′ = , P(B′) = g. P(A) + P(A′) = h. P(A|B) = , P(B|A) = ; are the probabilities equal? Example 3.2 A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up). a. Event T = the outcome is two. b. Event A = the outcome is an even number. c. Event B = the outcome is less than four. d. The complement of A. e. A | B f. B | A g. A ∩ B h. A ∪ B i. A ∪ B′ j. Event N = the outcome is a prime number. k. Event I = the outcome is seven. Solution 3.2 6 a. T = {2}, P(T) = 16 b. A = {2, 4, 6}, P(A) = 12 c. B = {1, 2, 3}, P(B) = 12 d. A′ = {1, 3, 5}, P(A′) = 12 e. A|B = {2}, P(A|B) = 13 f. B|A = {2}, P(B|A) = 13 g. A∩B = {2}, P(A ∩ B) = 16 h. A∪ B = {1, 2, 3, 4, 6}, P(A ∪ B) = 56 i. A ∪ B′ = {2, 4, 5, 6}, P(A ∪ B′) = 23 j. N = {2, 3, 5}, P(N) = 12 k. A six-sided die does not have seven dots. P(7) = 0. Example 3.3 Table 3.1 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left- handed. Right-handed Left-handed Males 43 9 Females 44 4 7 Table 3.1 Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left- handed. Compute the following probabilities: a. P(M) b. P(F) c. P(R) d. P(L) e. P(M ∩ R) f. P(F ∩ L) g. P(M ∪ F) h. P(M ∪ R) i. P(F ∪ L) j. P(M') k. P(R|M) l. P(F|L) m. P(L|F) Solution 3.3 a. P(M) = 0.52 b. P(F) = 0.48 c. P(R) = 0.87 d. P(L) = 0.13 e. P(M ∩ R) = 0.43 f. P(F ∩ L) = 0.04 g. P(M ∪ F) = 1 h. P(M ∪ R) = 0.96 i. P(F ∪ L) = 0.57 j. P(M') = 0.48 k. P(R|M) = 0.8269 (rounded to four decimal places) l. P(F|L) = 0.3077 (rounded to four decimal places) m. P(L|F) = 0.0833 10 Example 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. a. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD. b. Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH. Which of a. or b. did you sample with replacement and which did you sample without replacement? Solution 3.5 a. Without replacement; b. With replacement Try It 3.5 You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement. a. QS, 1D, 1C, QD b. KH, 7D, 6D, KH c. QS, 7D, 6D, KS Mutually Exclusive Events A and B are mutually exclusive events if they cannot occur at the same time. Said another way, If A occurred then B cannot occur and vise-a-versa. This means that A and B do not share any outcomes and P(A∩B)=0. For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A ∩ B = {4, 5}. P(A∩B)=210 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P(A∩C)=0. Therefore, A and C are mutually exclusive. If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms. 11 Example 3.6 Flip two fair coins. (This is an experiment.) The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads. • Let A = the event of getting at most one tail. (At most one tail means zero or one tail.) Then A can be written as {HH, HT, TH}. The outcome HH shows zero tails. HT and TH each show one tail. • Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A, so B = A′. Also, P(A) + P(B) = P(A) + P(A′) = 1. • The probabilities for A and for B are P(A) = 34 and P(B) = 14. • Let C = the event of getting all heads. C = {HH}. Since B = {TT}, P(B∩C)=0. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.) • Let D = event of getting more than one tail. D = {TT}. P(D) = 14 • Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = {HT, HH}. P(E) = 24 • Find the probability of getting at least one (one or two) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = 34 12 Try It 3.6 Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card. Example 3.7 Flip two fair coins. Find the probabilities of the events. a. Let F = the event of getting at most one tail (zero or one tail). b. Let G = the event of getting two faces that are the same. c. Let H = the event of getting a head on the first flip followed by a head or tail on the second flip. d. Are F and G mutually exclusive? e. Let J = the event of getting all tails. Are J and H mutually exclusive? Solution 3.7 Look at the sample space in Example 3.6. a. Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = 34 b. Two faces are the same if HH or TT show up. P(G) = 24 c. A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = 24 d. F and G share HH so P(F∩G) is not equal to zero (0). F and G are not mutually exclusive. e. Getting all tails occurs when tails shows up on both coins (TT). H’s outcomes are HH and HT. J and H have nothing in common so P(J∩H) = 0. J and H are mutually exclusive. Try It 3.7 A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events: a. Let F = the event of getting the white ball twice. b. Let G = the event of getting two balls of different colors. c. Let H = the event of getting white on the first pick. d. Are F and G mutually exclusive? e. Are G and H mutually exclusive? 15 P(G)P(H)=(0.6)(0.5)=0.3=P(G∩H) Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that P(H|G)=P(H) to show that G and H are independent events. Try It 3.9 In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4. • R = a red marble • G = a green marble • O = an odd-numbered marble • The sample space is S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}. S has ten outcomes. What is P(G∩O)? Example 3.10 Let event C = taking an English class. Let event D = taking a speech class. 16 Suppose P(C)=0.75, P(D)=0.3, P(C|D)=0.75 and P(C∩D)=0.225. Justify your answers to the following questions numerically. a. Are C and D independent? b. Are C and D mutually exclusive? c. What is P(D|C)? Solution 3.10 a. Yes, because P(C|D)=P(C). b. No, because P(C∩D) is not equal to zero. c. P(D∣∣∣C)=P(C∩D)P(C)=0.2250.75=0.3 Try It 3.10 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B)=0.40, P(D)=0.30 and P(B∩D)=0.20. a. Find P(B|D). b. Find P(D|B). c. Are B and D independent? d. Are B and D mutually exclusive? Example 3.11 In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card. Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn. The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes. • P(R)=38.P(B)=58.P(R∩B)=0. (You cannot draw one card that is both red and blue.) • P(E)=38. (There are three even-numbered cards, R2, B2, and B4.) • P(E∣∣B)=25. (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 and B4.) 17 • P(B∣∣E)=23. (There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 and B4.) • The events R and B are mutually exclusive because P(R∩B)=0. • Let G = card with a number greater than 3. G = {B4, B5}. P(G)=28. Let H = blue card numbered between one and four, inclusive. H = {B1, B2, B3, B4}. P(G∣∣H)=14. (The only card in H that has a number greater than three is B4.) Since 28 = 14, P(G)=P(G|H), which means that G and H are independent. Try It 3.11 In a basketball arena, • 70% of the fans are rooting for the home team. • 25% of the fans are wearing blue. • 20% of the fans are wearing blue and are rooting for the away team. • Of the fans rooting for the away team, 67% are wearing blue. Let A be the event that a fan is rooting for the away team. Let B be the event that a fan is wearing blue. Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive? Example 3.12 20 Try It 3.13 A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing. a. Compute P(T). b. Compute P(T|F). c. Are T and F independent?. d. Are F and S mutually exclusive? e. Are F and S independent? 21 3.4 Contingency Tables A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner. Example 3.20 Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data: Speeding violation in the last year No speeding violation in the last year Total Uses cell phone while driving 25 280 305 Does not use cell phone while driving 45 405 450 Total Table 3.2 70 685 755 The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755. Calculate the following probabilities using the table. a. Find P(Driver is a cell phone user). Solution 3.20 22 a. number of cell phone userstotal number in study = 305755 b. Find P(Driver had no violation in the last year). Solution 3.20 b. number that had no violationtotal number in study = 685755 c. Find P(Driver had no violation in the last year ∩ was a cell phone user). Solution 3.20 c. 280755 d. Find P(Driver is a cell phone user ∪ driver had no violation in the last year). Solution 3.20 25 Sex The coastline Near lakes and streams On mountain peaks Total Male 16 25 14 55 Total 34 41 25 100 Table 3.5 Hiking Area Preference b. Are the events "being female" and "preferring the coastline" independent events? Let F = being female and let C = preferring the coastline. 1. Find P(F∩C). 2. Find P(F)P(C) Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent. Solution 3.21 b. 1. P(F∩C)=18100 = 0.18 2. P(F)P(C) = (45100)(34100) = (0.45)(0.34) = 0.153 P(F∩C) ≠ P(F)P(C), so the events F and C are not independent. c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams. 1. What word tells you this is a conditional? 2. Fill in the blanks and calculate the probability: P( | ) = . 3. Is the sample space for this problem all 100 hikers? If not, what is it? 26 Solution 3.21 c. 1. The word 'given' tells you that this is a conditional. 2. P(M|L) = 2541 3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams. d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks. 1. Find P(F). 2. Find P(P). 3. Find P(F∩P). 4. Find P(F∪P). Solution 3.21 d. 27 1. P(F) = 45100 2. P(P) = 25100 3. P(F∩P) = 11100 4. P(F∪P) = 45100 + 25100 - 11100 = 59100 Try It 3.21 Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path. Gender Lake path Hilly path Wooded path Total Female 45 38 27 110 Male 26 52 12 90 Total 71 90 39 200 Table 3.6 a. Out of the males, what is the probability that the cyclist prefers a hilly path? b. Are the events “being male” and “preferring the hilly path” independent events? Example 3.22 Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 15 and the probability he is not caught is 45. If he goes out the second door, the probability he gets caught by Alissa is 14 and the probability he is not caught is 34. The probability that Alissa catches Muddy coming out of the third door is 12 and the probability she does not catch Muddy is 12. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 13. Caught or not Door one Door two Door three Total Caught 115 112 16 Not caught 415 312 16 Total 1 Table 3.7 Door Choice 30 a. Find P(2009∩Robbery). b. Find P(2010∩Burglary). c. Find P(2010∪Burglary). d. Find P(2011|Rape). e. Find P(Vehicle|2008). Solution 3.23 a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575 Try It 3.23 Table 3.10 relates the weights and heights of a group of individuals participating in an observational study. Weight/height Tall Medium Short Totals Obese 18 28 14 Normal 20 51 28 Underweight 12 25 9 Totals 31 Table 3.10 a. Find the total for each row and column b. Find the probability that a randomly chosen individual from this group is Tall. c. Find the probability that a randomly chosen individual from this group is Obese and Tall. d. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese. e. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall. f. Find the probability a randomly chosen individual from this group is Tall and Underweight. g. Are the events Obese and Tall independent? Tree Diagrams Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities. Tree Diagrams A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram. Example 3.24 In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows. 32 Figure 3.2 Total = 64 + 24 + 24 + 9 = 121 The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as: R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3 The other outcomes are similar. There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space. a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ... 35 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 2433. Try It 3.24 In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF). Figure 3.3 Example 3.25 An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (311)(210)=6110. B RB a 3 8 10 10 © ©—10 56. 24 24 110 110 110 BB BR RB Figure 3.4 Total = 56+24+24+6110=110110=1 NOTE Blya BE 2nd Draw  37 If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn. Calculate the following probabilities using the tree diagram. a. P(RR) = Solution 3.25 a. P(RR) = (311)(210)=6110 b. Fill in the blanks: P(RB∪BR) = (311)(810) + ( )( ) = 48110 Solution 3.25 b. P(RB∪BR) = (311)(810) + (811)(310) = 48110 c. P(R on 2nd|B on 1st) = Solution 3.25 c. P(R on 2nd|B on 1st) = 310 40 a. Find P(FN∪NF). b. Find P(N|F). c. Find P(at most one face card). Hint: "At most one face card" means zero or one face card. d. Find P(at least on face card). Hint: "At least one face card" means one or two face cards. Example 3.26 A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption. 41 a. What is the probability that both kittens are tabby? a.(12)(12) b.(49)(49) c.(49)(38) d.(49)(59) b. What is the probability that one kitten of each coloring is selected? a.(49)(59) b.(49)(58) c.(49)(59)+(59)(49) d.(49)(58)+(59)(48) c. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first? d. What is the probability of choosing two kittens of the same color? Solution 3.26 a. c, b. d, c. 48, d. 3272 Try It 3.26 Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? 3.5 42 Venn Diagrams A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Venn diagrams also help us to convert common English words into mathematical terms that help add precision. Venn diagrams are named for their inventor, John Venn, a mathematics professor at Cambridge and an Anglican minister. His main work was conducted during the late 1870's and gave rise to a whole branch of mathematics and a new way to approach issues of logic. We will develop the probability rules just covered using this powerful way to demonstrate the probability postulates including the Addition Rule, Multiplication Rule, Complement Rule, Independence, and Conditional Probability. Example 3.27 Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A intersect B = A∩B={6} and A union B = A∪B={1, 2, 3, 4, 5, 6, 7, 8, 9}.. The Venn diagram is as follows: 45 Figure 3.7 Try It 3.28 Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A∩B={3, 5}. A∪B={1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation. Example 3.29 A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood. 46 Figure 3.8 The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor. We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor. 47 a. P(O) = b. P(R) = c. P(O∩R)= d. P(O∪R)= e. In the Venn Diagram, describe the overlapping area using a complete sentence. f. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence. Solution 3.29 a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor. Example 3.30 Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time. Figure 3.9 If a student is selected at random, find • the probability that the student belongs to a club. P(C) = 0.40 • the probability that the student works part time. P(PT) = 0.50 • the probability that the student belongs to a club AND works part time. P(C∩PT)=0.05 50 Figure 3.10 Draw a second Venn diagram illustrating that 10 of the male dogs have brown coloring. Solution 3.31 The Venn diagram below shows the overlap between male and brown where the number 10 is placed in it. This represents Male∩Brown: both male and brown. This is the intersection of these two characteristics. To get the union of Male and Brown, then it is simply the two circled areas minus the overlap. In proper terms, Male∪Brown=Male+Brown−Male∩Brown will give us the number of dogs in the union of these two sets. If we did not subtract the intersection, we would have double counted some of the dogs. 51 Figure 3.11 Now draw a situation depicting a scenario in which the non-shaded region represents "No white fur and female," or White fur′ ∩ Female. the prime above "fur" indicates "not white fur." The prime above a set means not in that set, e.g. A' means not A. Sometimes, the notation used is a line above the letter. For example, A¯¯¯ = A'. Solution 3.31 Figure 3.12 The Addition Rule of Probability We met the addition rule earlier but without the help of Venn diagrams. Venn diagrams help visualize the counting process that is inherent in the calculation of probability. To restate the Addition Rule of Probability: 52 P(A∪B)=P(A)+P(B)−P(A∩B) Remember that probability is simply the proportion of the objects we are interested in relative to the total number of objects. This is why we can see the usefulness of the Venn diagrams. Example 3.31 shows how we can use Venn diagrams to count the number of dogs in the union of brown and male by reminding us to subtract the intersection of brown and male. We can see the effect of this directly on probabilities in the addition rule. Example 3.32 55 Figure 3.15 The multiplication rule must also be altered if the two events are independent. Independent events are defined as a situation where the conditional probability is simply the probability of the event of interest. Formally, independence of events is defined as P(A|B)=P(A) or P(B|A)=P(B). When flipping coins, the outcome of the second flip is independent of the outcome of the first flip; coins do not have memory. The Multiplication Rule of Probability for independent events thus becomes: 56 P(A∩B)=P(A)⋅P(B) One easy way to remember this is to consider what we mean by the word "and." We see that the Multiplication Rule has translated the word "and" to the Venn notation for intersection. Therefore, the outcome must meet the two conditions of freshmen and grade of "B" in the above example. It is harder, less probable, to meet two conditions than just one or some other one. We can attempt to see the logic of the Multiplication Rule of probability due to the fact that fractions multiplied times each other become smaller. The development of the Rules of Probability with the use of Venn diagrams can be shown to help as we wish to calculate probabilities from data arranged in a contingency table. Example 3.33 Table 3.11 is from a sample of 200 people who were asked how much education they completed. The columns represent the highest education they completed, and the rows separate the individuals by male and female. Less than high school grad High school grad Some college College grad Total Male 5 15 40 60 120 Female 8 12 30 30 80 Total 13 27 70 90 200 Table 3.11 Now, we can use this table to answer probability questions. The following examples are designed to help understand the format above while connecting the knowledge to both Venn diagrams and the probability rules. What is the probability that a selected person both finished college and is female? Solution 3.33 This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes. P(College Grad ∩ Female) = 30200=0.15 57 What is the probability of selecting either a female or someone who finished college? Solution 3.33 This task involves the use of the addition rule to solve for this probability. P(College Grad ∪ Female) = P(F) + P(CG)− P(F ∩ CG) P(College Grad ∪ Female) = 80200+90200−30200=140200=0.70 What is the probability of selecting a high school graduate if we only select from the group of males? Solution 3.33 Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability. P(HS Grad | Male = P(HS Grad∩Male)P(Male)=(15200)(120200)=15120=0.125 Can we conclude that the level of education attained by these 200 people is independent of the gender of the person?