# Module 4 Project Maths Development Team Draft (Version 2)

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1 5 Week Modular Course in Statistics & Probability Strand 1 Module 4

2 Set Theory and Probability It is often said that the three basic rules of probability are: 1. Draw a picture 2. Draw a picture 3. Draw a picture Venn diagrams are particularly useful when answering questions on conditional probability. 1. Complement Rule U A A c The probability of an event occurring is 1 the probability that it does not occur. c P(A) = 1 P(A ) Module 4.1

3 2. Disjoint Sets (Mutually Exclusive Events) Two events are said to be Mutually Exclusive if they have no outcomes in common. U A B For Mutually Exclusive events A and B, the probability that one or other occurs is the sum of the probabilities of the two events. P(A or B) = P(A) + P(B) P(A B) = P(A) + P(B) provided that A and B are mutually exclusive events. Mutually exclusive events are disjoint sets. Module 4.2

4 Example A card is drawn from a pack of 52 cards. What is the probability that the card is either a Queen or a King? K Q U #4 #4 #44 These events are Mutually exclusive "they have no outcome in common" Probability P(Q or K) = P(Q) + P(K) = + = = Module 4.3

5 Example A card is drawn from a pack of 52 cards. What is the probability that the card is either a Queen or a King? K Q U #4 #4 #44 These events are Mutually exclusive "they have no outcome in common" Probability P(Q or K) = P(Q) + P(K) = + = = Module 4.3

6 3. Overlapping Sets (Non Mutually Exclusive Events) Two events are not Mutually Exclusive if they have outcomes in common. U A B P(A or B) = P(A) + P(B) P(A and B) P(A B) = P(A) + P(B) P(A B) A and B are not mutually exclusive events. Module 4.4

7 Example A card is drawn from a pack of 52 cards. What is the probability that the card is either a King or a Heart? U K H #3 #1 #12 These two events have one outcome in common i.e. the king of hearts. The two events are not "Mutually Exclusive" i.e. they have an outcome in common #36 P(King or Heart) = P(King) + P(Heart) P(King which is a Heart) = = 52 4 = 13 Module 4.5

8 Example A card is drawn from a pack of 52 cards. What is the probability that the card is either a King or a Heart? U K H #3 #1 #12 These two events have one outcome in common i.e. the king of hearts. The two events are not "Mutually Exclusive" i.e. they have an outcome in common #36 P(King or Heart) = P(King) + P(Heart) P(King which is a Heart) = = 52 4 = 13 Module 4.5

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12 4.1 Mutually Exclusive Events Which of the following are mutually exclusive events? (a) A thumb tack falling head down and a thumb tack falling head up. (b) A student studying Maths and Physics. (c) Getting both a head and a tail when tossing a coin. (d) Getting a six and a two when throwing a dice. (e) Getting a king and a club when picking a card from a pack of playing cards.

13 4.1 Mutually Exclusive Events Which of the following are mutually exclusive events? (a) A thumb tack falling head down and a thumb tack falling head up. (b) A student studying Maths and Physics. (c) Getting both a head and a tail when tossing a coin. (d) Getting a six and a two when throwing a dice. (e) Getting a king and a club when picking a card from a pack of playing cards.

14 4. The Multiplication Law for Independent Events Two events are said to be independent if one event does not affect the outcome of the other. P(A and B) = P(A) P(B) if A and B are P(A and B) = P(A) P(B A) P(A B) = P(A) P(B) independent events P(A B) = P(A) P(B A) for all events NOTE : P(B A) is the probability of B given A has already occurred. Example A card is drawn from a pack of 52 cards and then replaced. A second card is then drawn from the pack. What is the probability that the two cards drawn are clubs? Solution These two events are independent as the outcome of drawing the second card is not affected by the outcome of drawing the first card because the first card was replaced. P(Club and Club) = P(Club) P(Club) = = 16 Module 4.6

15 4. The Multiplication Law for Independent Events Two events are said to be independent if one event does not affect the outcome of the other. P(A and B) = P(A) P(B) if A and B are P(A and B) = P(A) P(B A) P(A B) = P(A) P(B) independent events P(A B) = P(A) P(B A) for all events NOTE : P(B A) is the probability of B given A has already occurred. Example A card is drawn from a pack of 52 cards and then replaced. A second card is then drawn from the pack. What is the probability that the two cards drawn are clubs? Solution These two events are independent as the outcome of drawing the second card is not affected by the outcome of drawing the first card because the first card was replaced. P(Club and Club) = P(Club) P(Club) = = 16 Module 4.6

16 5. The Multiplication Law for Non Independent Events Two events are not Independent if one event affects the outcome of the other. NOTE : P(A and B) = P(A) P(B A) for all events P(B A) is the probability of B given A has already occured. Example A card is drawn from a pack of 52 cards. A second card is then drawn from the pack. What is the probability that the two cards drawn are clubs? (no replacement) Solution These two events are not independent as the outcome of drawing the second card is affected by the outcome of drawing the first card because the first card was not replaced. P(Club and Club) = P(Club) P(Second card is a club given that the first card was a club) P(Club and Club) = P(Club) P(Club Club) = = = This is Conditional Probability: The probability of the second event is conditional on the first Module 4.7

17 5. The Multiplication Law for Non Independent Events Two events are not Independent if one event affects the outcome of the other. NOTE : P(A and B) = P(A) P(B A) for all events P(B A) is the probability of B given A has already occured. Example A card is drawn from a pack of 52 cards. A second card is then drawn from the pack. What is the probability that the two cards drawn are clubs? (no replacement) Solution These two events are not independent as the outcome of drawing the second card is affected by the outcome of drawing the first card because the first card was not replaced. P(Club and Club) = P(Club) P(Second card is a club given that the first card was a club) P(Club and Club) = P(Club) P(Club Club) = = = This is Conditional Probability: The probability of the second event is conditional on the first Module 4.7

18 Conditional Probability Conditional Probability is the probability of an event which is affected by another event. P(A and B) = P(A) P(B A) if A and B are not independent events This is often written as: P(B A) = P(A B) P(A) This reads as: "the probability of B given A equals the probability of A and B over the probability of A". Module 4.8

19 Example 1 A bag contains 9 identical discs, numbered from 1 to 9. One disc is drawn from the bag. Let A = the event that an odd number is drawn. Let B = the event a number less than 5 is drawn (i) What is the probability that the number drawn is less than 5 given that it is odd i.e. P(B A)? (ii) What is P(A B)? Explain in words first of all what this questions means and then evaluate it. (iii) Are the events A and B independent, Justify your answer? Solution (i) Students should work with a Venn diagram and then explain what they have done by formula. U A B #(A B) 2 P(B A) = = #A 5 More generally we can say P(A B) P(B A) = = = P(A) Module 4.9

20 Example 1 A bag contains 9 identical discs, numbered from 1 to 9. One disc is drawn from the bag. Let A = the event that an odd number is drawn. Let B = the event a number less than 5 is drawn (i) What is the probability that the number drawn is less than 5 given that it is odd i.e. P(B A)? (ii) What is P(A B)? Explain in words first of all what this questions means and then evaluate it. (iii) Are the events A and B independent, Justify your answer? Solution (i) Students should work with a Venn diagram and then explain what they have done by formula. U A B #(A B) 2 P(B A) = = #A 5 More generally we can say P(A B) P(B A) = = = P(A) Module 4.9

21 (ii) P(A B) = probability that the number drawn is odd given that it is less than 5 U A B #(A B) 2 1 P(A B) = = = #B 4 2 More generally we can say P(A B) P(A B) = = = P(B) (iii) No the events A and B are not independent. Is P(B) = P(B A)? or Is P(A) P(B) = P( A B)? #B P(B) = = P(A) P(B) = = #U P(B A) = P(A B) = Since either of these i.e. or are not equal the two events are not independent i.e. the fact that the event A happened changed the probability of the event B happening. Module 4.10

22 (ii) P(A B) = probability that the number drawn is odd given that it is less than 5 U A B #(A B) 2 1 P(A B) = = = #B 4 2 More generally we can say P(A B) P(A B) = = = P(B) (iii) No the events A and B are not independent. Is P(B) = P(B A)? or Is P(A) P(B) = P( A B)? #B P(B) = = P(A) P(B) = = #U P(B A) = P(A B) = Since either of these i.e. or are not equal the two events are not independent i.e. the fact that the event A happened changed the probability of the event B happening. Module 4.10

23 Example 2 A game is played with 12 cards, 5 of the cards are red {1, 7, 8, 11, 12} and 7 are yellow. The cards are numbered from 1 to (i) Given that a card is red what is the probability that the number on it is even? (ii) Given that a card is red what is the probability that the number on it is odd? Work out this answer in 2 ways. (iii) Given that the number on the card is even what is the probability that it is red? (iv) Is P(E R) = P(R E) where E means card has an even number on it and R means that the card is red? (v) Is P(E R) = P(E)? What does your answer tell you about the events of being red and being even? Solution (i) P(E R) = probability that the number drawn is even given that it is red. E R U P(E R) = 5 More generally we can say 2 P(E R) 12 2 P(E R) = = = P(R) Module 4.11

24 (ii) P(O R) = probability that the number drawn is odd given that it is red. O R 6. U P(O R) = 5 More generally we can say 3 P(O R) 12 3 P(O R) = = = P(R) We could also say that it is 1 P(E R) = 1 = 5 5 (iii) P(R E) = probability that the number drawn is red given that it is even. U E R P(R E) = 6 More generally we can say 2 P(R E) 12 1 P(R E) = = = P(E) Module 4.12

25 (ii) P(O R) = probability that the number drawn is odd given that it is red. O R 6. U P(O R) = 5 More generally we can say 3 P(O R) 12 3 P(O R) = = = P(R) We could also say that it is 1 P(E R) = 1 = 5 5 (iii) P(R E) = probability that the number drawn is red given that it is even. U E R P(R E) = 6 More generally we can say 2 P(R E) 12 1 P(R E) = = = P(E) Module 4.12

26 (iv) Is P( E R ) = P(R E)? 2 P(E R) = from part (i) 5 1 P(R E) = from part (iii) 3 they are not equal (v) Is P(E R) = P(E)? What does your answer tell you about the events of being red and being even? 2 P(E R) = from part (i) 5 1 P(E) = 2 they are not equal Hence the events E and R are not independent. Module 4.13

27 (iv) Is P( E R ) = P(R E)? 2 P(E R) = from part (i) 5 1 P(R E) = from part (iii) 3 they are not equal (v) Is P(E R) = P(E)? What does your answer tell you about the events of being red and being even? 2 P(E R) = from part (i) 5 1 P(E) = 2 they are not equal Hence the events E and R are not independent. Module 4.13

28 Landlines versus Mobiles According to estimates from the federal government s 2003 National Health Interview Survey, based on face-to-face interviews in 16,677 households, approximately 58.2% of U.S. adults have both a land line and a mobile phone, 2.8% have only mobile phone service, but no landline, and 1.6% have no telephone service at all. (a) What proportion of U.S. households can be reached by a landline call? (b) Are having a mobile phone and a having a landline independent events? Explain. Solution: (a) Since 2.8% of U.S. adults have only a mobile phone, and 1.6% have no phone at all, polling organisations can reach = 95.6% of U.S. adults. (b) Using the Venn diagram, about 95.6% of U.S. adults have a landline. The probability of a U.S. adult having a land line given that they have a mobile phone is 58.2/( ) or about 95.4%. It appears that having a mobile phone and having a land line are independent, since the probabilities are roughly the same. P(L) = 95.6% 58.2 P(L M) = 95.4% Land line 37.4% 58.2% 2.8% Mobile 1.6% Taken from : Stats, Data and Models (2 nd Ed.), pg. 387 Module 4.14

29 Landlines versus Mobiles According to estimates from the federal government s 2003 National Health Interview Survey, based on face-to-face interviews in 16,677 households, approximately 58.2% of U.S. adults have both a land line and a mobile phone, 2.8% have only mobile phone service, but no landline, and 1.6% have no telephone service at all. (a) What proportion of U.S. households can be reached by a landline call? (b) Are having a mobile phone and a having a landline independent events? Explain. Solution: (a) Since 2.8% of U.S. adults have only a mobile phone, and 1.6% have no phone at all, polling organisations can reach = 95.6% of U.S. adults. (b) Using the Venn diagram, about 95.6% of U.S. adults have a landline. The probability of a U.S. adult having a land line given that they have a mobile phone is 58.2/( ) or about 95.4%. It appears that having a mobile phone and having a land line are independent, since the probabilities are roughly the same. P(L) = 95.6% 58.2 P(L M) = 95.4% Land line 37.4% 58.2% 2.8% Mobile 1.6% Taken from : Stats, Data and Models (2 nd Ed.), pg. 387 Module 4.14

31 4.2 The table below shows the percentages of boys and girls in a class and the percentages of each gender who wear glasses. Glasses No Glasses Totals Boys Girls Totals The teacher chooses a student at random from the class. (a) What is the probability she chooses a boy? (b) What is the probability that she chooses a boy who is not wearing glasses? (c) What is the probability that she chooses a boy given that the person she chooses is not wearing glasses? (d) What is the probability that the person chosen is not wearing glasses given that it is a boy. (e) Is P(NG B) = P(B NG)? (f) What is the probability that she chooses a girl given that the person she chooses is wearing glasses? (g) Is P( Girl Glasses)= P(Glasses Girl)? Answer the above questions using the contingency table above and then using the formula for conditional probability.

32 4.3 Games Console Internet (a) P(Games console and no internet) = 0.31 (b) P(Games console or internet but not both) = = 0.48 (c) P(neither a games console nor internet access) = 0.31

33 4.3 Games Console Internet (a) P(Games console and no internet) = 0.31 (b) P(Games console or internet but not both) = = 0.48 (c) P(neither a games console nor internet access) = 0.31

34 4.4 Garage Pool (a) P(Pool or a Garage) = = 0.68 (b) P(neither) = 0.32 (c) P(Pool but no garage) = 0.04 (d) P(Pool and Garage) 0.17 P(Pool Garage) = = P(Garage) 0.64 (e) Having a pool and a garage are not independent events. 26.6% of homes with garages have pools. Overall, 21% of homes have pools. If having a garage and a pool were independent these would be the same. (f) No, having a garage and a pool are not disjoint events. 17% of homes have both.

35 4.4 Garage Pool (a) P(Pool or a Garage) = = 0.68 (b) P(neither) = 0.32 (c) P(Pool but no garage) = 0.04 (d) P(Pool and Garage) 0.17 P(Pool Garage) = = P(Garage) 0.64 (e) Having a pool and a garage are not independent events. 26.6% of homes with garages have pools. Overall, 21% of homes have pools. If having a garage and a pool were independent these would be the same. (f) No, having a garage and a pool are not disjoint events. 17% of homes have both.

36 4.5 Quality Control A consumer organisation estimates that 29% of new cars have a cosmetic defect, such as a scratch or a dent, when they are delivered to dealers. This same organisation believes that 7% have a functional defect something that doesn t work properly and that 2% of new cars have both kinds of problems. (a) If you buy a new car, what s the probability that it has some kind of defect? (b) What s the probability it has a cosmetic effect but no functional defect? (c) If you notice a dent on a new car, what s the probability it has a functional defect? (d) Are the two events mutually exclusive? Explain. (e) Do you think the two kinds of defects are independent events? Explain. Taken from : Stats, Data and Models (2 nd Ed.), pg. 428

37 4.5 Quality Control A consumer organisation estimates that 29% of new cars have a cosmetic defect, such as a scratch or a dent, when they are delivered to dealers. This same organisation believes that 7% have a functional defect something that doesn t work properly and that 2% of new cars have both kinds of problems. (a) If you buy a new car, what s the probability that it has some kind of defect? (b) What s the probability it has a cosmetic effect but no functional defect? (c) If you notice a dent on a new car, what s the probability it has a functional defect? (d) Are the two events mutually exclusive? Explain. (e) Do you think the two kinds of defects are independent events? Explain. Solution: (a) = 0.34 (b) 0.27 P(F C) 0.02 (c) P(F C) = = P(C) (d) The two kinds of events are not disjoint events, since 2% of cars have both kinds. (e) Approximately 6.9% of cars with a cosmetic defects also have functional defects. Overall, the probability that a car has a functional defect is 7%. The probabilities are estimates, so these are probably close enough to say that the two types of defects are independent. Cosmetic Functional Taken from : Stats, Data and Models (2 nd Ed.), pg. 428

38 4.6 A B (a) P(A B) = P(A) + P(B) P(A B) P(A B) = or directly from venn diagram P(A B) = = 0.9 (b) P(A B) 0.3 P(A B) = = = 0.6 P(B) 0.5 (c) Not independent or Not independent since P(A B) P(A) since P(A B) P(A).P(B) i.e i.e

39 Tree Diagrams Getting it Right! Event A Event B Outcomes A P(B) B AB P(A) C P(B ) B C AB C C P(A ) P(B) B A C B A C Independent C P(B ) B C A C B C

40 Tree Diagrams Getting it Right! Event A Event B Outcomes P(B A) B AB A P(A) C P(B A) B C AB C C P(A ) C P(B A ) B A C B A C C C P(B A ) B C A C B C Not Independent

42 A bag contains 4 red sweets and 5 blue sweets. Two sweets are taken out of the bag at random Draw a probability tree diagram, when: (i) the sweets are taken with replacement (ii) the sweets are taken without replacement 1 st Sweet 2 nd Sweet Outcomes 1 st Sweet 2 nd Sweet Outcomes P(R) = 4 9 R RR P(R R) = 3 8 R RR P(R) = 4 9 R P(B) = 5 9 B RB It s important to recognise what the outcomes mean. P(R B) The first is red and the second is blue. P(R) = 4 9 R P(B R) = 5 8 B RB P(R) = 4 9 R BR P(R B) = 4 8 R BR P(B) = 5 9 B P(B) = 5 9 B P(B) = 5 9 B BB P(B B) = 4 8 B BB P(R and B) = P(R).P(B) Choosing red followed by a blue are independent events because the first sweet is replaced. P(B R) = P(B) 4.7 multiply along branch P(R and B) = P(R).P(B R) Choosing red followed by a blue are not independent events as the first sweet is not replaced. P(R and B) P(R) = P( B R) In General: P(A B) = P(A B) P(B) multiply along branch dividing both sides by P(R)

43 Susan goes to work by one of two routes A or B. The probability of going by route A is 30%. If she goes by route A the probability of being late for school is 5% and if she goes by route B, the probability of being late is 10%. Draw a probability tree diagram, and then (i) Find the probability that Susan is late for school (ii) Given that Susan is late for school, find the probability that she went via route A. Solution P(A) = 0.30 P(B) = 0.70 Route Lateness Outcomes A B C L' = L = complement of L i.e. not late P(L A) = 0.05 P(L' A) = 0.95 P(L B) = 0.10 P(L' B) = 0.90 L L L L AL AL BL BL It s important to recognise what the outcomes mean. P(A L) (i) P(L) = P(A L) or P(B L) P(L) = P(L) = = P(A L) (ii) P(A L) = P(L) P(A L) = P(A L) = = (3 d.p.) 7 This shows clearly that P(A L) P(L A) 4.8 Taken from:

44 4.9 Cancer Test Result Outcomes P(P C) = 0.98 P CP (a) P(P) = P(C and P) + P(C' P) P(P) = = P(P) = = P(C) = P(C') = C C P(N C) = 0.02 P(P C') = 0.02 N P CN C P P(C P) (b) P(C P) = P(P) P(C P) = = When a disease occurs in a very small percentage of the population P(N C') = 0.98 N C N (in this case, 0.5%), a test that is only 98% accurate will give a lot more false postives than true positives. In this case 199 false positives for every 49 true positives.

45 Expected Value Random Variable The outcome of an experiment need not be a number, for example, the outcome when a coin is tossed can be 'heads' or 'tails'. However, we often want to represent outcomes as numbers. A random variable is a function that associates a unique numerical value with every outcome of an experiment. The value of the random variable will vary from trial to trial as the experiment is repeated. There are two types of random variable: 1. Discrete 2. Continuous Examples 1. A coin is tossed ten times. The random variable X is the number of tails that are noted. X can only take the values 0, 1,..., 10, so X is a discrete random variable. 2. A light bulb is burned until it burns out. The random variable Y is its lifetime. Y can take any positive real value, so Y is a continuous random variable. A random variable has a probability distribution i.e. an assignment of probabilities to the specific values of the random variable or to a range of its values. A discrete random variable gives rise to a discrete probability distribution and a continuous random variable gives rise to a continuous probability distribution. Module 4.15

46 Example 1 If I toss 3 coins. x = the number of heads I get. Start H T H T H T H T H T H T H T H H H 3 heads H T H 2 heads H T H 2 heads H T T 1 head T H H 2 heads T H T 1 head T T H 1 head T T T 0 heads x P(x) (Total of 8 outcomes) This is a probability distribution. It is similar to a frequency distribution. We calculate mean and standard deviation for a probability distribution in the same way we calculated the mean and standard deviation of a frequency distribution. The mean of the probability distribution µ= As P(x) = 1 µ= xp(x). This is called the EXPECTED VALUE. xp(x) P(x) µ= = = Module 4.16

47 Example 1 If I toss 3 coins. x = the number of heads I get. Start H T H T H T H T H T H T H T H H H 3 heads H T H 2 heads H T H 2 heads H T T 1 head T H H 2 heads T H T 1 head T T H 1 head T T T 0 heads x P(x) (Total of 8 outcomes) This is a probability distribution. It is similar to a frequency distribution. We calculate mean and standard deviation for a probability distribution in the same way we calculated the mean and standard deviation of a frequency distribution. The mean of the probability distribution µ= As P(x) = 1 µ= xp(x). This is called the EXPECTED VALUE. xp(x) P(x) µ= = = Module 4.16

48 Fair games and expected values Example 2 If the above example represented a game at a carnival where you tossed 3 coins together and you win 2 euro for each head which shows you could get 0, 1, 2 or 3 heads and hence win 0,2, 4, or 6 euro. We consider 3 headings: (i) the outcome (ii) the probability associated with each outcome (iii) the value associated with each outcome (the random variable) Outcome Probability of each Value associated with xp(x) (no. of heads) outcome P(x) each outcome (in euro) (x) Expected Value µ= xp(x) = = = The Expected Value is 3 i.e. the carnival owner must charge more than 3 to make a profit. If he charges 3 euro the average profit per player is 0, and the game is fair. However he must make a profit so he will charge more than the Expected Value. If he charges 3.50 per game the average profit per game is 50 cent. While individual players may make a profit, or lose or win back some of their costs, in the long run the carnival owner will win. Module 4.17

49 Fair games and expected values Example 2 If the above example represented a game at a carnival where you tossed 3 coins together and you win 2 euro for each head which shows you could get 0, 1, 2 or 3 heads and hence win 0,2, 4, or 6 euro. We consider 3 headings: (i) the outcome (ii) the probability associated with each outcome (iii) the value associated with each outcome (the random variable) Outcome Probability of each Value associated with xp(x) (no. of heads) outcome P(x) each outcome (in euro) (x) Expected Value µ= xp(x) = = = The Expected Value is 3 i.e. the carnival owner must charge more than 3 to make a profit. If he charges 3 euro the average profit per player is 0, and the game is fair. However he must make a profit so he will charge more than the Expected Value. If he charges 3.50 per game the average profit per game is 50 cent. While individual players may make a profit, or lose or win back some of their costs, in the long run the carnival owner will win. Module 4.17

50 Example 3 You and a friend are playing the following game: Two dice are rolled. If the total showing is a prime number, you pay your friend 6, otherwise, your friend pays you 2. (i) What is the expected value of the game to you? (ii) If you played the game 40 times, what are your expected winnings? After playing the game for a while, you begin to think the rules are not fair and you decide to change the game. (iii) How much (instead of 6) should you pay your friend when you lose so that your expected winnings are exactly 0? Solution Outcome Probability of each Value associated with xp(x) outcome, P(x) each outcome( ), x Prime = 36 2 Non Prime = 36 6 Module 4.18

51 Example 3 You and a friend are playing the following game: Two dice are rolled. If the total showing is a prime number, you pay your friend 6, otherwise, your friend pays you 2. (i) What is the expected value of the game to you? (ii) If you played the game 40 times, what are your expected winnings? After playing the game for a while, you begin to think the rules are not fair and you decide to change the game. (iii) How much (instead of 6) should you pay your friend when you lose so that your expected winnings are exactly 0? Solution Outcome Probability of each Value associated with xp(x) outcome, P(x) each outcome( ), x Prime = 36 2 Non Prime = 36 6 Module 4.18

52 (i) Expected Value µ= xp(x) = + = (ii) (iii) 4 Expected winnings after 40 games = 40 = Loss of (x) + (2) = x + 42 = 0 x = 2.8 You should pay your friend 2.80 Module 4.19

53 Example 4 For a particular age group, statistics show that the probability of dying in any one year is 1 in 1000 people and the probability of suffering some sort of disability is 3 in 1000 people. The Hope Life Insurance Company offers to pay out 20, 000 if you die and 10, 000 if you are disabled. What profit is the insurance company making per customer based on the expected value if it charges a premium of 100 to its customers for the above policy? Solution Outcome Value of each Probability of each xp(x) outcome, x outcome, P(x) Dying 20, Disability 10, Neither of the above Expected Value 50 Charging 100 per customer, the company is expecting a profit of 50 per customer. Module 4.20

54 Example 4 For a particular age group, statistics show that the probability of dying in any one year is 1 in 1000 people and the probability of suffering some sort of disability is 3 in 1000 people. The Hope Life Insurance Company offers to pay out 20, 000 if you die and 10, 000 if you are disabled. What profit is the insurance company making per customer based on the expected value if it charges a premium of 100 to its customers for the above policy? Solution Outcome Value of each Probability of each xp(x) outcome, x outcome, P(x) Dying 20, Disability 10, Neither of the above Expected Value 50 Charging 100 per customer, the company is expecting a profit of 50 per customer. Module 4.20

56 4.10 (a) The net change in your finances is 1 when you lose and 35 when you win. Outcome Probability of each Value associated with xp(x) outcome, P(x) each outcome( ), x Get number = Do not get number 38 = Expected Value µ= xp(x) = = This is not a fair game as the expected value is not zero (b) Outcome Probability of each Value associated with outcome, P(x) each outcome( ), x xp(x) Black = Other colour 38 = Expected Value µ= xp(x) = = This is not a fair game as the expected value is not zero.

### Independent and Mutually Exclusive Events

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