1. PROBABILITY

2. Probability Simulations - Simulations are used when the actual event cannot be reproduced or it is difficult to calculate the theoretical probability. -A simulation will often either involve the calculation of: a) The long run relative frequency of an event happening b) The average number of ‘visits’ taken to collect a full set Probability Tools Tools that can be used include: DiceCardsCoinsSpinners Random Number TableRandom Generator on a Calculator The probability tool chosen must always match the situation appropriately

3. Using the Random Number Generator - The Ran# button on the calculator yields a 3 digit number between 0.000 and 0.999 e.g. To simulate LOTTO balls in NZ use the calculator as follows 40Ran# + 1 (truncating the number to 0 d.p. every time) Number of lotto balls Rounding the numbers up to between 1 and 40 Reading only the whole number (not digits after decimal point) e.g. To simulate a situation where there is a 14% chance of success 100Ran# + 1 (truncating the number to 0 d.p. every time) For success we use the digits For failure we use the digits 1 – 14 15 - 100

4. Describing Simulations - Use the TTRC method to describe the four most important aspects T T R C ool - Definition of the probability tool - Statement of how the tool models the situation rial - Definition of a trial - Definition of a successful outcome of the trial esults - The results will be placed on a table - Statement of how many trials should be carried out alculation - Statement of how the calculation needed for the conclusion will be done Either: Long-run relative frequency = Number of ‘successful’ results Number of trials Or: Mean = Sum of trial results Number of trials

5. Simulation Example 1 What is the probability that a four child family will contain exactly 2 boys and 2 girls? Design a simulation to solve the problem. Tool - A calculatorusing 10Ran# + 1where 1 – 5 = boy 6 – 10 = girl Trial - One trial will consist of4 randomly generated numbers to simulate a 4 child family Results TrialOutcome of trialResult of trial 1 5, 7, 3, 9  2 2, 2, 7, 1 X - Results will be recorded on a table as follows: Calculation - A successful outcome would be2 numbers between 1 – 5 (boys) and 2 numbers between 6 – 10 (girls) - 30 trials would be sufficient = number of ‘Successful’ results number of trials P(2 boys & 2 girls) - To calculate the estimated probability use:

6. Simulation Example 2 e.g. A cereal manufacturer includes a gift coupon in each box of a certain brand. These coupons can be exchanged for a gift when a complete set of six coupons has been collected. What is the expected number of boxes of cereal you will have to buy before you obtain a complete set of six coupons? Design a simulation. Tool - A calculatorusing 6Ran# + 1where 1 – 6 = each of the coupons Trial - One trial will consist ofrandomly generated numbers until all 6 numbers are obtained (all coupons) Results - Results will be recorded on a table as follows: Calculation - 30 trials would be sufficient = Sum of Number of Boxes number of trials Average number - To calculate the estimated number of boxes use: TrialOutcome of trialResult of trial 1 3121144156 10 2 352334653245321 15

7. Experimental Probability - Also known as long run relative frequency - When experimental results are used to determine the probability - Is calculated using: P(A) = Number of Times Event Occurs Total Number of Trials e.g. A drawing pin is tossed 300 times and lands pin down 204 times. Find the probability a tossed pin will land pin down P(Landing Pin Down) = 204 300 Number of Times Event Occurs = 204 Total Number of Trials = 300 17, 0.68, or 68% also acceptable 25 e.g. A basketball player has 26 shots at the hoop and converts 22 of them. Find the probability the player will miss a shot. P(Misses shot) = 4 26 Number of Times Event Occurs = 4 (26 – 22) Total Number of Trials = 26 2, 0.15 (2 d.p.), or 15% also acceptable 13

8. Two Way Tables - When tables are used to calculate experimental probabilities. - Always make sure to check which group the person/item is being chosen from. e.g. A class of 35 students recorded the following sports they each played. FootballRugbyBasketballTotal Boy1062 Girl827 Total Calculate the probability that: a) A student chosen at random is a boy 359818 17 18 b) If a girl is chosen at random, she plays football c) A student chosen at random plays rugby or basketball P(Boy) = 18 35 P(Football Girl) = 8 17 P(Rugby/BB player) = 17 35

9. Theoretical Probability - Is calculated using: P(A) = Number of Successful Outcomes Number in Sample Space - Can only be calculated if events are equally likely to occur e.g. Find the probability of getting an even number on a dice. P(even on dice) = 3 6 Number of Successful Outcomes = 3 (2, 4, or 6) Number in Sample Space = 6 (1, 2, 3, 4, 5, or 6) 1, 0.5, or 50% are also acceptable 2 e.g. Find the probability of a nine being drawn from a full deck. Number of Successful Outcomes = 4 (4 suits) Number in Sample Space = 52 (4 suits of Ace to King) P(Nine) = 4 52 1, 0.08 (2 d.p.), or 8% are also acceptable 13 - When prior knowledge is used to determine the probability

10. Tree Diagrams - Display all outcomes of a probability event - Always multiply along the branches and add the resulting probabilities that match the question e.g. A student walks to school 40% of the time and bikes 60%. If he walks he has an 80% chance of being late, if he bikes it is 10%. Calculate a) P(Student is late) b) P(Student is on time and biking) 1. Set up the tree diagram 2. Add probabilities (decimals/fractions) 3. Determine appropriate branch/branches Walk Bike Late On Time Late On Time 0.6 0.4 0.8 0.2 0.1 0.9 a) P(Student is late) = 0.4 × 0.8 = 0.38 b)P(Student is on time and biking) = 0.6 × 0.9 = 0.54 + 0.6 × 0.1

11. Tree Diagrams Continued - Sometimes harder questions can involve conditional probability e.g. From the previous example, it is found that if a student is late, the probability that they will be given a detention is 0.9. Walk Bike Late On Time Late On Time 0.6 0.4 0.8 0.2 0.1 0.9 Detention No Detention Detention No Detention 0.9 0.1 a)A student is chosen at random. Calculate the probability that the student biked to school and received a detention for being late P(Student biked and received a detention)= 0.6 × 0.1 × 0.9 = 0.054 b) A student that walks to school is randomly chosen. Calculate the probability that they received a detention for being late. P(Student received a detention given they walked)= 0.8 × 0.9 = 0.72

12. Expected Number - If the probability of an event is known it can be used to predict outcomes or explain events. Expected number = Probability × Number of trials e.g. A coin is tossed 60 times. How many heads would you expect? Expected number = 0.5 × 60 = 30 e.g. If 12 cards are randomly selected from a full deck, how many diamonds would you expect? Expected number = 0.25 × 12 = 3 Note: Expected numbers are only estimations, if you actually carry out the event the results can be quite different and can be used to help decide if the probability tool is biased or not. = P(Head) × Number of trials = P(Diamond) × Number of trials

13. Sampling Without Replacement - When the probability of an event changes. e.g. Two marbles are drawn in succession from a bag that contains 3 white marbles and 6 red marbles. Calculate the probability that one of each colour is drawn. 1. Set up the tree diagram 2. Add probabilities (decimals/fractions) 3. Determine appropriate branch/branches White Red White Red White Red 6/9 3/9 2/8 6/8 3/8 5/8 P(One marble of each colour) = 3/9 × 6/8 = 0.5 + 6/9 × 3/8

14. Venn Diagrams - Used when events concerned are not independent (probabilities cannot be multiplied) AB 1. The Total Probability Is Always Equal To 1 A A’ A’ is the complimentary event to A. A’ means the event ‘not in A’. e.g. The probability it will hail tomorrow (A) is 0.15. The probability that it won’t hail tomorrow (A’ ) is: P(A) + P(A’ ) = 1 0.15 + P(A’ ) = 1 P(A’ ) = 1 – 0.15 P(It won’t hail tomorrow) = 0.85 Probability of event A occurring P(A) Probability of both event A and event B occurring P(A  B) Probability of event B occurring P(B) P(A) + P(A’ ) = 1 HAIL = 0.15 NO HAIL

15. 2. Mutually Exclusive Events Have Nothing In Common AB - Where sets do not overlap and are called disjoint. P(A  B) = P(A) + P(B) union (or) The intersection of these sets is called the empty set or P(A  B) = 0 intersection (and) e.g. There are 20 marbles in a bag, six of which are white (A) and five of which are red (B). Calculate the probability of choosing a white or red marble when one is chosen at random. These two events are mutually exclusive as you cannot pick a white marble and red marble at the same time when choosing one. P(A  B) = P(A) + P(B) union (or) = 6/20 + 5/20 P(Red or White Marble) = 11/20 (or 0.55) WHITE = 6/20 RED = 5/20

16. 3. Intersecting Events Overlap AB P(A  B) = P(A) + P(B) – P(A  B) P(A  B) We need to subtract P(A  B) to avoid counting it twice e.g. A survey showed that in NZ homes 30% have a dog (A), 55% have a cat (B) and 20% have both P(A  B). Calculate the probability that a randomly chosen house has a) At least one of these pets b) None of these pets e.g. If P(A) = 0.5, P(B) = 0.2 and P(A  B) = 0.1, calculate P(A  B) P(A  B) = P(A) + P(B) – P(A  B) P(A  B) = 0.5 + 0.2 – 0.1 P(A  B) = 0.6 BOTH = 0.2 CAT = 0.55 DOG = 0.30 a) P(At least one of these pets) = P(A) + P(B) – P(A  B) = 0.3 + 0.55 – 0.2 P(At least one of these pets) = 0.65 b) P(None of these pets) = 1 – P(A  B) = 1 – 0.65 P(None of these pets) = 0.35

17. NORMAL DISTRIBUTION

18. Normal distributions are described by the mean µ and standard deviation σ These sets of data have most measurements near the middle and a few extreme values above and below the mean. When these sets of data are graphed it forms a: Normal Distribution Curve µ1 σ 3 σ2 σ3 σ2 σ 34% Of the data: - 68% of the data lies within 1 standard deviation of the mean. It is likely or probable that data will fall in this region - 95% of the data lies within 2 standard deviations of the mean. It is very likely or very probable that data will fall in this region - 99% of the data lies within 3 standard deviations of the mean. It is almost certain that data will fall in this region 47.5% 49.5% 68% 95% 99%

19. e.g. The weights of school bags are normally distributed with a mean of 4.2 kg with a standard deviation of 0.9 kg. 4.25.13.31.52.46.96.0 a) What percentage of bags weigh between 4.2 kg and 6.0 kg? 47.5% b) What percentage of bags weigh more than 3.3 kg? 50%+ 34% = 84% c) 400 bags are weighed. Estimate how many will weigh less than 2.4 kg? 50%- 47.5% = 2.5% 2.5% of 400= 0.025 x 400 = 10 bags

20. The Standard Normal Distribution To convert an ordinary normal measurement (x) to a standard normal measurement (z) use: mean standard deviation measurement The z-value tells us how many standard deviations x is away from the mean.. Standard Normal Tables Once we know the z-score, we can use these tables to calculate the probability of a certain event. e.g. Calculate the following probability: a) P(0 < Z < 1.32) = 0.4066 Note: Normal Distribution tables give the probabilities between the mean (0) and the z-score. 01.32 1. Draw curve 2. Add in boundaries 3. Shade in correct area 4. Find probability on table and use it to calculate final answer

21. e.g. b) P(-1.32 < Z < 0) 0-1.32 = 0.4066c) P(Z < -1.32) -1.320 = 0.5 - 0.4066 = 0.0934 d) P(Z > -1.32) 0-1.32 = 0.5 + 0.4066 = 0.9066 e) P(Z < 1.32) 01.32 = 0.5 + 0.4066 = 0.9066 - Using the differences columns e.g. Find P(0 < Z < 0.473)= 0.1808+ 0.0011 = 0.1819 Symmetrical properties of the curve can then be used to calculate other probabilities.

22. Application Questions To convert an ordinary normal measurement (x) to a standard normal measurement (z) use: e.g. A distribution X has a mean of 50 and standard deviation of 12. Calculate the probability that X has a value less than 70. 1. Write probability statementP(X < 70) 2. Change to z-score using above formula = P(z < 70 - 50) 12 = P(z < 1.667) 3. Draw curve etc. 01.667 4. Calculate appropriate probability using table. = 0.5 + 0.4522 = 0.9522

23. Application Questions Continued e.g. A flock of two-year old sheep have a mean weight of 32.5 kg and a standard deviation of 5.5 kg. Find the probability of a randomly selected sheep being: a) between 32.5 kg and 40 kg b) over 40 kg. c) In the flock there are 400 sheep. How many would you expect to weight under 37 kg? a) P(32.5 < X < 40)= P(32.5 – 32.5 < z < 40 – 32.5) 5.5 5.5 = P(0 < z < 1.364) 01.364 = 0.4137 b) P(X > 40)= P(z > 40 – 32.5) 5.5 = P(z > 1.364) 01.364 = 0.5 - 0.4137 = 0.0863 c) P(X < 37)= P(z < 37 – 32.5) 5.5 = P(z < 0.818) 00.818 = 0.5 + 0.2932 = 0.7932 Number of sheep =0.7932 × 400= 317.28 = 317 sheep

24. Inverse Normal - Using a known probability to calculate a z-score or X-score e.g. Calculate the value of k for which P(0 < Z < k) = 0.3485 0k 0.3485 Look up 0.3485 in main table to find z-score k =1.03 - You may need to use the differences column e.g. Calculate the value of k for which P(0 < Z < k) = 0.165 0k 0.165 If the exact probability is not listed in the main table, look for the nearest probability below. Then check the differences column to find the remainder needed to get the exact total k = (0.1628) (22) 0.426

25. Inverse Normal Applications e.g. The weights of school bags are normally distributed with a mean of 4.2 kg and a standard deviation of 0.9 kg. A bag is considered to be heavy if it is in the top 10% of weights. What is the lowest weight a bag can be to still be considered heavy? 0k 0.1 1. As the table gives probability between 0 and k we first need to find the correct probability to look up. = 0.5 – 0.1 2. Find nearest probability in the main table (0.3997) 3. Check differences column to find remainder needed, without going over desired probability (2) P(0 < z < k) = 0.4 k =1.281 4. Use the z-score and equation below to find the necessary measurement (X) 1.281 = X – 4.2 0.9 × 0.9 1.1529 = X – 4.2 + 4.2 5.3529 = X Lowest weight = 5.3529 kg If k is below 0, make the z-score negative before substituting into equation