Chapter 13 & 14 - Probability PART

Chapter 13 & 14 - Probability PART IV : PROBABILITY Dr. Joseph Brennan Math 148, BU Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 1 / 91

Why Should We Learn Probability Theory? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 2 / 91

Let s Make a Deal! Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 3 / 91

Uncertainty Usually the results of a study, observational or experimental, are uncertain: if we repeat a study, we will not get exactly the same results. Example 1 (A coin). You toss a coin. Will it land heads or tails? Example 2 (A die). A regular die, a D6, is a cube with six faces: What number will a die show when it is rolled once? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 4 / 91

Uncertainty Example 3 (Box with marbles) A box contains 7 marbles: 2 red and 5 green. Each marble has an equal chance to be selected. One marble is to be drawn at random from the box. What color will it be? Example 4 (Box with tickets) A box contains 10 tickets labeled 1 through 10. What will be the number of a randomly selected ticket? Example 5 (Height of students) Seven students will be selected at random from the Math 148 class list and their heights will be measured. What will be the average height? Will it change if we choose different seven students? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 5 / 91

Probability What is the chance or probability that we will make the same conclusions every time when we replicate a study? What is the chance or probability that the histogram will change? Knowledge of probability theory will help us answer these questions. NOTE: In this part, the words chance and probability will have the same meaning. chance = probability Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 6 / 91

Inferential Statistics & Probability Theory In Parts II and III we used random samples to collect evidence and make inferences about population. Sample mean, x, estimates unknown population mean µ. Sample standard deviation, s, estimates unknown population standard deviation σ. Regression equation is a mathematical model which approximates the true relationship between x and y. In this part we will think in the opposite direction; we will reason from a known population to randomly selected samples. Probability Theory Inferential Statistics P opulation Sample Sample P opulation Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 7 / 91

Outcomes and Sample Space Probability theory deals with studies where the outcomes are not known for sure in advance. Usually, there are many possible outcomes for a study, we just do not know which particular outcome we will observe. Sample Space: The set of all possible outcomes of a study. The sample space of a study is denoted by S. Every repetition of a study, or a trial, produces a single outcome. Usually an outcome is computed from the values of the response variables. In Example 5 (Height of students) the outcome is the average height, ȳ, which is computed from the values of the response variable (y a student s height). Conclusions from a study are based on its outcome. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 8 / 91

Sample Spaces Example 1 (A coin) When you toss a coin, there are only two possible outcomes: a head or a tail. Then the sample space is S ={ head, tail} or simply S ={H, T}. Example 2 (A die) The sample space is S = {1, 2, 3, 4, 5, 6}. Example 3 (Box with marbles) We can draw either a red marble or a green one: S={R,G}. Example 4 (Box with tickets) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 9 / 91

Sample Spaces Example 5 (Height of students) Of all our current examples, example 5 is the only continuous quantitative variable. The outcome of this study is ȳ, an average height. For any sample of 7 students the average height could be any number between 40 inches and 100 inches. Then the sample space is the interval: S = [40, 100]. Example 6 (Children) For every randomly chosen family the number of boys and girls is recorded. The outcomes in the sample space are pairs of the form (number of boys, number of girls) Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 10 / 91

Sample Space Example 6 (Two dice) Two dice are thrown. The sample space contains 36 outcomes as shown below: Mathematically, the sample space can be written as S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),, (6, 6)}. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 11 / 91

Probability of Outcomes Probability: The theoretical probability of an outcome is the proportion (or percent) of times an outcome occurs in the sample space. Every outcome has a probability. If an experiment is repeated a large number of times, one would expect the ratio of occurrences of an outcome to number of experiments to be approximately the probability of that outcome. Notation: We will denote the probability of an outcome as where P( ) stands for probability. P(outcome), Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 12 / 91

Example 1 (A coin) A single coin flip has a simple sample space: S = {H, T }. There are only two potential outcomes: {H}, {T }. P(H) = number of H in sample space size of sample space = 1 2 P(T ) = number of H in sample space size of sample space = 1 2 Is it correct to assume that after flipping 10 coins, 5 would have come up heads? While imprisoned by the Germans during World War II, the South African statistician John Kerrich tossed a coin 10, 000times. Heads was the outcome 5,067 times! We have 67 more heads then expected. The difference between the observed percentage and the anticipated result is known as error. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 13 / 91

Example 1 (A coin) The graph shows the percentage of heads minus 50% versus the number of trials in Kerrich s experiment. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 14 / 91

Example 1 (A coin) From the graph we can see that the error approaches 0 as the number of tosses increases. After 10,000 tosses there are 67 extra heads, so the error is: 67 100% = 0.67%. 10, 000 It is already very close to 0! If we continue Kerrich s experiment, we can expect the error to become even smaller. We say the error converges to 0, which implies that the observed percentage of heads approaches 50% which suggests that P(H) = 0.5 has been calculated correctly. Question Why is the proportion of heads in Kerrich s experiment not EXACTLY equal to 0.5? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 15 / 91

Example 1 (A coin) Answer The number of heads is approximately half the number of tosses (5000), but it is off by 67 because of the chance error. The error 67 seems to be large in absolute units, but in relative units it is just 0.67%, which is a very small error. CONCLUSION: Kerrich s experiment supports the theoretical P(H) = 0.5. The observed number of heads may be interpreted as number of heads = half the number of tosses + chance error, where the error may appear to be large in absolute terms, but small relative to the number of tosses. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 16 / 91

Example (Blood Distribution) Data lifted from Wikipedia. The following Table shows the distribution of Blood Types of people in Australia. O+ A+ B+ AB+ O- A- B- AB- Total 40% 31% 8% 2% 9% 7% 2% 1% 100% The above distribution was obtained by computing the proportions of blood types for a huge representative group of Australian people. Therefore, we can say that the numbers in the table are the probabilities of different blood types for Australians. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 17 / 91

Example (Blood Distribution) 1 What is the probability of AB+ blood for Australians? Answer: P(AB+) = 0.02 If we randomly sample 1000 Australians, approximately 20 of them are expected to have AB+ blood type. However, due to chance variability, we will not observe exactly 20 people with an AB+ blood type. 2 In July 2009 the population of Australia was 28,395,716 people. Approximately, how many people in Australia have an A+ blood type? Answer: As the probability of having type A+ blood is 31%, the population, disregarding chance variability, of is 0.31 28, 395, 716 = 8, 802, 671.96 8, 802, 672 people Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 18 / 91

Example 2 (A die) An ace is the face of a die with one spot. Question What is the probability of getting an ace in a single roll? swer: number of faces with 1 P(ace) = = 1 number of faces 6 Indeed, a die has 6 faces and every face has the same chance to be observed in a single roll. Then P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1 6. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 19 / 91

Equally Likely Outcomes Assigning correct probabilities to individual outcomes often requires long observation of the random phenomenon. In some circumstances, however, we are willing to assume that individual outcomes are equally likely because of some balance in the phenomenon. For the equally likely outcomes: Probability of a single outcome = 1 Number of outcomes in S Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 20 / 91

Equally Likely Outcomes Ordinary coins have a physical balance that should make heads and tails equally likely. So the outcomes in Example 1 are equally likely: P(H) = P(T ) = 1 2 A fair die in Example 2 is equally likely to show any of its 6 faces in a single roll. P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1 6 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 21 / 91

Equally Likely Outcomes In Example 4, every one of the 10 tickets has the same chance to be drawn: P(1) = P(2) = P(3) = = P(8) = P(9) = P(10) = 1 10 In Example 6 (two dice), there are 36 equally likely outcomes. The probability of each outcome is 1 36. In Example 3, the outcomes of drawing a red marble and drawing a green marble are not equally likely. There are more green marbles in the box, so the probability of selecting a green marble is greater. What is this probability? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 22 / 91

Computing Probabilities of Outcomes RULE: In problems involving drawing at random, the probability of getting an object of a particular type in a single draw is equal to proportion of objects of this type in the population. Example 3 (Box with marbles) The population is the box with marbles. The proportion of red marbles is 2/7, the proportion of green marbles is 5/7. This implies that in a single draw P(R) = 2 P(G) = 5 7 7 Notice that P(R) + P(G) = 1. We will explore this phenomenon later in this chapter. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 23 / 91

Events We will give the definition of an event in words and mathematically. Definition of an event in words: An event is some statement about the random phenomenon. Events are usually denoted by capital letters as A, B, etc. Example of Events: (a) A = At least one head in two tosses of a coin, (b) B = Sum of the numbers on the dice is five, (c) C = A die rolled once shows an even number. Mathematical definition of an event: An event is a set of outcomes from the sample space S. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 24 / 91

Examples of Events (a) A = At least one head in two tosses of a coin. The sample space for this experiment is S = {HH, HT, TH, TT }. The event A = {HH, HT, TH} (b) B =Sum of the numbers on the dice is five. The sample space for this experiment contains 36 equally likely outcomes which are discussed in Example 2. The event B is B = {,,, } (c) C = A die rolled once shows an even number. The sample space is S = {1, 2, 3, 4, 5, 6}. The event C is C = {,, } Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 25 / 91

Occurrence of an Event We say that an event has occurred if ANY of the outcomes that constitute it occur. For instance, in Example (c) if we roll a die and it shows 4, we say that the event C has occurred. If the die shows 3, the event A does not occur because 3 is an odd number, so the event C does not contain this outcome. Let A be the event of landing at least one heads when tossing a coin 2 times, as in Example (a). If we toss and get {H, T } or {T, H} or {H, H} then we ll say that A has occurred. Only when we have two tails appearing does A not occur. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 26 / 91

Probabilities of Events The probability of an event A is denoted by P(A). In all our examples, except for Example 6, the sample space consisted of finitely many outcomes which we could list. In this case the sample space is called finite. An event is a collection of outcomes which support it. Theorem 1 (Probability of an event in a finite sample space) The probability of any event is the sum of probabilities of the outcomes making up the event. In the special case that the outcomes in the sample space are equally likely, the probability of an event E is computed as P(E) = # of outcomes supporting E Totat # of outcomes in S. (1) Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 27 / 91

Probabilities of Events (a) A = At least one head in two tosses of a coin. There are 4 equally likely outcomes in this experiment. The probability of every outcome is 1 4. Then P(A)= P(HH) + P(HT ) + P(TH)= 1 4 + 1 4 + 1 4 = 3 4. (b) B = Sum of the numbers on the dice is five. There are 36 equally likely outcomes. Out of them, 4 outcomes support the event B. P(B) = 4 36 = 1 9. (c) C = A die rolled once shows an even number. There are 6 equally likely outcomes, 3 of them support C. P(C) = 3 6 = 1 2. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 28 / 91

Example 3 (Box with Marbles) A box contains 7 marbles, 2 red and 5 green. A marble is drawn at random, its color is recorded, and the marble is put back in the box. Then, the next marble is drawn at random. This example is of sampling with replacement, which we will define later. Find the probability that 2 red marbles are drawn. Solution: The sample space for this experiment is: S = {RR, RG, GR, GG} The outcomes are not equally likely since the red and green marbles have unequal chances to be selected. Intuitively, the RR is the least likely outcome. This outcome can be described: A = First two randomly chosen marbles are red Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 29 / 91

Example 3 (Box with Marbles) We can use the following two events to describe event A: A 1 = The first selected marble is red A 2 = The second selected marble is red Then it follows that A = A 1 and A 2. Note that if we replace the marble back to the box, a red marble has the same chance to be selected at every draw: P(A 1 ) = P(A 2 ) = 2 7 We have calculated P(A 1 ) and P(A 2 ), but how can we use this to calculate P(A 1 and A 2 )? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 30 / 91

Special Events We have the ability to calculate the probability of simple events; those resulting from one repetition of an experiment with equally likely outcomes. To tackle more complex problems, we must define some special events. Some special events: Case 1: Sure-to-happen or certain event. Case 2: Impossible event. Case 3: Opposite event. Case 4: Mutually exclusive or disjoint events. Case 5: Independent events. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 31 / 91

Certain Event Certain Event: An event which is guaranteed to happen at every repetition of the experiment. The event A = A head or a tail in a single toss of a coin. In every toss a coin lands up either heads or tails. There are no other possibilities. So A is a certain event. Notice that A = {H, T } which coincides with the sample space S for this experiment. The event B = rolling a die once will show a number less than 7, is a certain event. Notice that B = {1, 2, 3, 4, 5, 6} is the whole sample space. A certain event is equal to the sample space. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 32 / 91

Impossible Event Impossible Event: An event which never can occur. The event A = a coin lands up neither heads nor tails is an impossible event. In all the coin-related experiment we usually assume that the coin may not lay on the edge. Eliminating this possibility makes event A impossible. The event B = the number that shows up when a die is rolled once is 2 is an impossible event. Mathematically an impossible event is written as, the empty set. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 33 / 91

Opposite Event Opposite Event: An event B is opposite to A if it happens whenever A does not happen. The event opposite to event A contains all the outcomes from the sample space S which do not belong to A. The opposite event for event A is called not A event. An opposite event for A is denoted as Ā. A and Ā split the sample space into two parts. (Also called a partition) Quite often, events are plotted on a graph as certain areas in the sample space. Such graphs are called the Venn diagrams. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 34 / 91

Opposite Event on a Venn diagram The Venn diagram shows the opposite event to the event A as the shaded region. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 35 / 91

Opposite Events: Examples (a) Event A = At least one head in two tosses of a coin. The opposite event Ā = It is NOT true that at least one H in two coin tosses = No heads in two tosses of a coin = {TT } The probability is given by P(Ā) = P(TT ) = 1 4 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 36 / 91

Opposite Events: Examples (b) Event B = Sum of five when rolling two dice. The opposite event: B = Sum of the numbers on the dice is NOT equal to five. B = {(1,1), (1,2), (1,3), (1,5),(1,6), (2,1), (2,2), (2,4), (2,5), (2,6), (3,1), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. There are 32 equally likely outcomes in B. The sample space for this experiment contains 36 equally likely outcomes. This implies P( B) = 32 36 = 8 9 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 37 / 91

Opposite Events: Examples (c) Event C = A die rolled once shows an even number. Event C can be described via its outcome space as C = {2, 4, 6}. The sample space for this experiment is S = {1, 2, 3, 4, 5, 6}. Then the event C will contain all the outcomes from S that C doesn t contain. In particular, C = {1, 3, 5} In words, C = A die rolled once shows an odd number. The probability P( C) = 1 2. Rule: For an event A, P(Ā) = 1 P(A). Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 38 / 91

Example 3 (Box with marbles) A box contains 7 marbles: 2 red and 5 green. Two marbles are chosen with replacement. We found the sample space S = {RR, RG, GR, GG}. We reconsider the event A =two red marbles in two draws with replacement = {RR}. What is the opposite event? It follows that Ā = {RG, GR, GG} can be expressed as: Ā = NOT exactly two red marbles in two draws = At most one red marble in two draws = Not all the marbles in two draws were red = At least one green marble in two draws P(Ā) = 3 4 = 1 1 4 = 1 P(A) Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 39 / 91

Mutually Exclusive Events Mutually Exclusive Events: Two events A and B are mutually exclusive if they both cannot happen at the same time. Examples of mutually exclusive events: Events A = An odd numbered face showing and B = The 2 face showing are mutual events in a single die role. Events A = The coin lands up heads and B = The coin lands up tails are mutually exclusive events in a single coin flip experiment. Events A and B are mutually exclusive if they do not have common outcomes. We say A and B do not overlap or do not intersect. For this reason, mutually exclusive events are also called disjoint events. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 40 / 91

Mutually Exclusive Events A Venn diagram depicting two mutually exclusive events A and B in the sample space S. Single outcomes of an experiment which make up the sample space are always mutually exclusive. An event and its opposite are mutually exclusive: A and Ā are mutually exclusive. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 41 / 91

Example (Two dice) A pair of dice is rolled once. Consider the following events: A = Sum of numbers on the dice is 11. B = Both dice show an even number. C = Both dice show the same number. Are the events A and B mutually exclusive? Are the events B and C mutually exclusive? Let s express the events as sets of outcomes and see if they have common outcomes. A = { ; } B = { ; ; ; ; ; ; ; ; } C = { ; ; ; ; ; }. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 42 / 91

Example (Two dice) A = { ; } B = { ; ; ; ; ; ; ; ; } C = { ; ; ; ; ; }. Events A and B are mutually exclusive since there are no common outcomes. Events B and C are not mutually exclusive since they have three common outcomes. What are the probabilities of events A, B, and C? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 43 / 91

Intersection of Events If A and B are not mutually exclusive events, they have some common outcomes. The set of common outcomes is an event which is called the intersection of events A and B. We will denote the intersection of events A and B by A and B Example: B = { ; ; ; ; ; ; ; ; } C = { ; ; ; ; ; }. Denote by D the intersection of events B and C. Then B and C = D = { ; ; }. D = The face showing on each die are the same and even. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 44 / 91

Intersection of Events When we connect two events with AND, we are interested in their intersection. The intersection of two mutually exclusive events is the impossible event. If A and B are mutually exclusive, then A and B = Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 45 / 91

Example Suppose you toss a fair coin twice. You are counting heads, so two events of interest are A = First toss is a head, B = Second toss is a head. What is the probability of the intersection P(A and B)? The events A and B are both disjoint, they occur together when both tosses give heads. Indeed, the sample space in this experiment is S = {HH, HT, TH, TT } All the outcomes are equally likely. The events are A = {HH, HT } The intersection is A and B = {HH}. B = {HH, TH} P(A and B) = 1 4 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 46 / 91

Independent Events Independent Events: Two events A and B are independent of each other if knowing that one event occurs does not change the probability that the other event occurs. Tosses of a fair coin are independent events. Further, it is equally likely that each side show after a toss. Whether there are two heads in a row or twenty, the chance of getting a head next time is still 0.5. It is the memoryless property of the coin. A fair die also has the memoryless property, i.e., the rolls are independent. For instance, the probability of an ace at every roll is 1 6 no matter how many aces has appeared in the previous rolls. We will assume that child births are independent events. Child births are independent events. The actual probability of a boy is slightly higher than 0.5, but we will assume it to be 0.5. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 47 / 91

Independent Events Consider two events arising from rolling two dice: A = Getting a 6 on the first roll. B = The sum of the numbers seen on the first and second rolls is 11 Are A and B independent events? Solution 1: The event B consists of the following outcomes: B = { ; }. The probability of event B is P(B) = 2 36 = 1 18. Suppose event A has happened. Then the possible outcomes for the second trial are 1, 2, 3, 4, 5, 6. Event B will happen if the second outcome was 5. The probability of this is 1 6 Knowing that event A has happened changes the probability that event B will happen. Hence, A and B are not independent events. which is larger than 1 18. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 48 / 91

Independent and Disjoint Events If events A and B are independent, they can not be mutually exclusive. Independent events always have common outcomes (intersect). The opposite statement is also true: Mutually exclusive events are not independent (dependent). Explanation: If we know that A and B are mutually exclusive events with nonzero probabilities P(A) > 0 and P(B) > 0, and we know that event A has happened, this implies that the probability that event B also happened is 0. Example: Toss a coin twice. Two events A = First outcome is a head and B = Second outcome is a head are independent, but not mutually exclusive. A = {HH, HT }, B = {TH, HH} Events A and B are not mutually exclusive since they have a common outcome {HH}. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 49 / 91

Union of Events Union: The union of events A and B is the event which happens when either event A or event B or both happen. The union of two events is expressed as A or B In mathematics, the word or is not inclusive, rather it is exclusive. Inclusive means either or, but not both; as in a restaurant. Exclusive means either or both. In other words, the union of events A and B is the event which happens when at least one of events A or B happens. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 50 / 91

Union of Events: Venn diagram Mathematically, the union of two events A and B is found by combining the outcomes from A and B into a single set A or B. The following Venn diagram illustrates the union of events A and B. The shaded region corresponds to A or B. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 51 / 91

Example Find the probability of event E = at least one 3 in two successive rolls of a die. Solution: The event E consists of the following outcomes: ; ; ; ; ; ; ; ; ; ; The sample size for the two dice problem has 36 equally likely outcomes. As a consequence: P(E) = 11 36. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 52 / 91

Example Observe that event E can be represented as the union of two events: A = First outcome is 3 = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)} B = Second outcome is 3 = {(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)} Therefore: E = At least one 3 in two successive rolls of a die = First outcome is 3 or second outcome is 3 = A or B When combining the outcomes from two events into a single event, do not repeat the same outcomes twice. Observe that each of the events A and B has 6 outcomes, but their union (event E) has 11 outcomes. This is because A and B have one common outcome (3, 3) which is written just once in E. Notice A and B= {(3, 3)}. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 53 / 91

Rules of Probability The following rules simplify many probability computations: Rule 1: The probability P(A) of any event A satisfies 0 P(A) 1. In other words, the probability of any event is between 0 and 1. Rule 2: If S is the sample space for an experiment, then P(S) = 1. The probability of a certain event is 1. Rule 3: The probability of an impossible event is 0. Hence, P( ) = 0. Rule 4: If events A and B are independent, then the probability that they both happen is the product of their probabilities: P(A and B) = P(A) P(B) Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 54 / 91

Rules of Probability Rule 5: For any events A and B, the probability of their union is equal to the sum of their individual probabilities minus the probability of their intersection: P(A or B) = P(A) + P(B) P(A and B) Subtracting the probability of the intersection is needed to avoid double counting. A special case: if A and B are disjoint events, then the probability of their union is the sum of individual probabilities: Rule 6: For any event A P(A or B) = P(A) + P(B) P(Ā) = 1 P(A) Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 55 / 91

A Few Remarks Some important notes about the rules of probability: 1. The closer the probability of an event is to 1, the more likely this event is to happen. 2. Unlikely events have probabilities close to 0. 3. Outcomes partition the sample space: Every cell in the above picture is an outcome. 4. Probabilities of all the outcomes add up to 1 because the set of all outcomes is S and P(S) = 1. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 56 / 91

Example We reconsider the following events arising from rolling a die twice: E = At least one 3 in two rolls of a die A = The first outcome is 3 B = The second outcome is 3. It should be clear that E = A or B, or that E is the union of both A and B. The experiment which consists in throwing 2 fair dice has 36 equally likely outcomes. P(A) = 6 36 = 1 6 = P(B) The intersection of events A and B is { P(A and B) = 1 36. } with probability P(E) = P(A or B)= P(A) + P(B) P(A and B)= 1 6 + 1 6 1 36 = 11 36 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 57 / 91

Example Problem 1. A total of 30% of American males smoke cigarettes, 7% smoke cigars, and 5% smoke both cigars and cigarettes. What percentage of males smoke neither cigars nor cigarettes? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 58 / 91

Example (Blood Type - 2) From Moore and McCabe All human blood can be ABO-typed as one of A, B, O, or AB, but the distribution of the types varies among groups of people. Here is the distribution of blood types for a randomly chosen person in the US: Blood type O A B AB U.S. probability 0.45 0.40 0.11? (a) What is the probability of type AB blood in the US? P(AB) = 1 0.45 0.40 0.11 = 0.04 (b) Maria has type B blood. She can safely receive blood transfusions from people with blood types O and B. What is the probability that a randomly chosen American can donate blood to Maria? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 59 / 91

Example (Blood Type - 2) Consider the events: Blood type O A B AB U.S. probability 0.45 0.40 0.11 0.04 O = Randomly selected person has O blood type B = Randomly selected person has B blood type. Events O and B are disjoint since a person can not have type O and type B of blood at the same time. We are interested in the probability that a randomly selected person has a blood type of either O or B type. Then P(O or B)= P(O) + P(B) = 0.45 + 0.11 = 0.56. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 60 / 91

Example (Blood Type - 3) The probability of randomly choosing an individual with blood type A is 0.43 in the US and 0.22 in China. What is the probability of randomly and independently choosing two people, one from the US and the other from China, that both have blood type A? Solution: Define the following events: X = An American has A blood type Y = A Chinese has A blood type Notice that events X and Y are independent. We are interested in the probability of the event X and Y. By the multiplication rule for independent events P(E and F ) = P(E) P(F ) = 0.43 0.22 = 0.0946 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 61 / 91

Extending the Rules of Probability 1 The multiplication rule for a collection of m mutually independent events. For any collection of events A 1, A 2,..., A m that are mutually independent, the probability of the intersection of all the events is equal to the product of their individual probabilities: P(A 1 and A 2 and... and A m ) = P(A 1 ) P(A 2 )... P(A m ). 2 The addition rule for m disjoint events. If events A 1, A 2,..., A m are disjoint in the sense that they do not have common outcomes, then P(A 1 or A 2 or... or A m ) = P(A 1 ) + P(A 2 ) + + P(A m ). Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 62 / 91

Example (Universal Donors) From Moore and McCabe People with O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only 7% of the American population have O-negative blood. If 3 people appear at random to give blood, what is the probability that at least one of them is a universal donor? Solution: Let s code the outcomes of the experiment in the following way: Y means a randomly selected person has O-negative blood N means a person does NOT have O-negative blood. Then the sample space for this experiment is as follows: S = {NNN, NNY, NYN, YNN, NYY, YNY, YYN, YYY } Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 63 / 91

Example (Universal Donors) We are given that P(Y ) = 0.07. Notice that Y and N are opposite events! Therefore P(N) = 1 P(Y ) = 1 0.07 = 0.93. We are interested in the probability of the following event: A = at least one of the three people is a universal donor = at least one Y in the outcome = { NNY, NYN, YNN, NYY, YNY, YYN, YYY} A has all the outcomes from S except {NNN}, which should be the opposite event. Ā = None of the three people is a universal donor = No Y in the outcome = {NNN} Clearly it ll be easier to find the probability of Ā rather than A! Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 64 / 91

Example (Universal Donors) S = {NNN, NNY, NYN, YNN, NYY, YNY, YYN, YYY } A = {NNY, NYN, YNN, NYY, YNY, YYN, YYY } Ā = {NNN} Recall that each outcome is the result of three independent events. Therefore, we may use the multiplication rule of mutually independent events: P(Ā) = P(N and N and N) = P(N) P(N) P(N) = (0.93) 3 = 0.804357. As Ā is opposite A we have P(A) = 1 P(Ā) = 1 0.804357 = 0.195643. Notice that P(A) can be calculated without finding P(Ā) by using multiplication and addition formulas: P(A) = P(NNY or NYN or YNN or NYY or YNY or YYN or YYY ) = P(N)P(N)P(Y ) + P(N)P(Y )P(N) + P(Y )P(N)P(N) +... Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 65 / 91

Two Sampling Schemes We will consider two important sampling schemes: Sampling with Replacement Sampling without Replacement We explore these sampling techniques through the following problem: Problem. A box contains r red and g green marbles. Define the following events: R i = The marble selected on the ith draw is red. For example, R 1 is the event for which the first drawn marble is red. We are interested in the probabilities of events R 1, R 2, R 3,.... Since every marble has the same chance to be picked, we see that P(R 1 ) = r r + g Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 66 / 91

Two Sampling Schemes What are the probabilities of events R 2, R 3,...? That depends on the sampling scheme: Sampling with replacement: Suppose we return the marble to the box after each draw. In this case the conditions of experiment do not change since each subsequent draw is made from the same population. 1 Subsequent draws are independent. 2 The probability to select a red marble is the same at every draw: P(R i ) = r, i = 1, 2,... r + g Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 67 / 91

Two Sampling Schemes Sampling without replacement: After each draw the drawn marble is not returned to the box. Subsequently, conditions of the experiment change from draw to draw because the population of marbles in the box changes after each draw. 1 The probability of selecting a red marble changes with each draw! 2 Subsequent draws are NOT independent. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 68 / 91

Example (Box with Marbles) A box contains 7 marbles - 2 red and 5 green. We have considered the event A: A = The first two chosen marbles are red. We used the following two events to describe event A : We found that A = A 1 and A 2 A 1 = The first selected marble is red, A 2 = The second selected marble is red. We know A 1 but we are not sure how to calculate A 2 ; is the sampling done with or without replacement? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 69 / 91

Example (Box with Marbles) Sampling with replacement: Subsequent draws are independent. P(A 1 ) = P(A 2 ) = 2 7 P(A) = P(A 1 ) P(A 2 ) = 2 7 2 7 = 4 49 Sampling without replacement: The probability of A 1 is still 2 7. The probability of A 2 depends on the outcome of the first draw: The probability to select a red marble on the second draw, given that the first selected marble was red is 1 6. The probability to select a red marble given that the first selected marble was green is 2 6. Sampling multiple times without replacement is not independent! The multiplication rule for independent events cannot be used! Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 70 / 91

Conditional Probability Conditional Probability: Let A and B be events. If the events are not independent, then the occurrence of B alters the probability that A will occur. The conditional probability of event A given that event B has happened is denoted P(A B). Example: (Single roll of a fair die) Consider two events: A = The die shows 3 B = The die shows odd. The ordinary, unconditional, probabilities of events A and B are P(A) = 1 6 P(B) = 1 2 Suppose we know that event B happened. Then the conditional probability of A given B is P(A B) = 1 3 since there are only 3 odd outcomes possible: 1, 3, and 5, all equally likely. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 71 / 91

Conditional Probability The conditional probability of an event A given an event B is P(A B) = P(A and B) P(B) A B S When we say given B ; we are only considering outcomes covered by event B. Ignore everything outside of B! Then find the probability that an A outcome lies in B. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 72 / 91

Conditional Probability: Examples (a) Two machines, I and II, produce bolts. Five percent of those from I and ten percent of those from II are defective. This can be written as P(defective machine I) = 0.05 P(defective machine II) = 0.10. Notice that we do not know P(defective), nor can we calculate it with the information supplied. (b) Suppose a mortality table shows that the probability of dying within one year for a 25-year-old male is 0.00193. This can be stated as P(male dying within one year age 25) = 0.00193. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 73 / 91

Conditional Probability Note 1: Knowledge that B has occurred effectively reduces the sample space from S to B. This is sometimes called the reduced sample space. Therefore, when interpreting the area of an event on the Venn diagram as its probability, P(A B) is the proportion of the area of B occupied by A. Note 2: If events A and B are mutually exclusive, then P(A B) = 0 as P(A and B) = 0. Note 3. Recall the formula: P(A B) = P(A and B) P(B) With algebraic manipulation we find: P(A and B) = P(A B) P(B) This formula is the general multiplication rule. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 74 / 91

The General Multiplication Rule As and is commutative, P(A B) P(B) = P(A and B) = P(A and B) = P(B A) P(A) The order of conditioning may be changed if needed. Extension of the General Multiplication Rule: The events A 1, A 2,..., A n are not necessarily independent. P(A 1 and A 2 and A 3 and A 4 and...) = = P(A 1 ) P(A 2 A 1 ) P(A 3 A 1 and A 2 ) P(A 4 A 1 and A 2 and A 3 )... Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 75 / 91

Intersection of 3 Events on a Venn Diagram The intersection of three events A, B, and C has probability P(A and B and C) = P(A) P(B A) P(C A and B) A B C S Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 76 / 91

Example (Athletes) (From Moore and McCabe) Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years. Define these events A = Competes in college, B = Competes professionally, C = Professional career longer than 3 years. What is the probability that a high school male athlete competes in college, reaches a professional level and then goes on to have a pro career of more than 3 years? Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 77 / 91

Example (Athletes) We are given that The probability we want is, therefore, P(A) = 0.05 P(B A) = 0.017 P(C A and B) = 0.4. P(A and B and C) = P(A) P(B A) P(C A and B) = 0.05 0.017 0.4 = 0.00034 Interpretation: Only about 3 of every 10,000 high school male athletes can expect to compete in college, reach a professional level and have a professional career for more than 3 years. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 78 / 91

Conditional Probability and Independence When events A and B are independent, knowing that event B has occurred gives no additional information about the occurrence of event A. This can be expressed as P(A B) = P(A) To check whether two events A and B are independent, you should check either equality: P(A B) = P(A) or P(A and B) = P(A) P(B). Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 79 / 91

Example (Degrees) (From Moore and McCabe) The counts (in thousands) of earned degrees in the US in the 2005-2006 academic year, classified by level and by the sex of the degree recipient: Bachelor s Master s Professional Doctorate Total Female 784 276 39 20 1119 Male 559 197 44 25 825 Total 1343 473 83 45 1944 Tables of this type are called cross-tabulation or contingency tables. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 80 / 91

Example (Degrees) (a) If you choose a degree recipient at random, what is the probability that the person that you choose is a woman? P(W ) = 1119 thousands 1944 thousands 0.5756. (b) What is the probability of choosing a woman, given that the person chosen received a professional degree? P(W P) = 39 83 0.4699. (c) Are the events choose a woman and choose a professional degree recipient independent? Why or why not? We found in (a) that P(W ) = 0.5756. In (b) we found P(W P) = 0.4699. Since P(W ) P(W P), the events are not independent. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 81 / 91

Example (Degrees) (d) A randomly chosen person is a man. What is the probability that he received a bachelor s degree? P(B M) = 559 825 0.6776. (e) Use the multiplication rule to find the probability of choosing a male bachelor s recipient. Check your result by finding this probability directly from the table of counts. P(M and B) = P(B M) P(M) = 559 825 825 1944 = 559 1944, which is the probability obtained directly from the table of counts. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 82 / 91

Example (Box with Marbles) Recall that there are 2 red and 5 green marbles in a box. We defined the following events: We found A = A 1 and A 2, and A = Two randomly chosen marbles are red A 1 = First selected marble is red A 2 = Second selected marble is red. P(A 1 ) = P(A 2 ) = 2 7. We want to compute the probability of A under sampling without replacement: P(A) = P(A 1 and A 2 ) = P(A 2 A 1 )P(A 1 ) = 1 6 2 7 = 1 21. We previously calculated that the probability of event A under sampling with replacement is 4 49. Since 4 49 > 1 21, event A is more likely to occur under sampling with replacement. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 83 / 91

Examples Example: Two birds are selected at random without replacement from a cage containing five male and two female finches. What is the probability that both are males? P(M and M) = P(M) P(M M) = 5 7 4 6 = 10 21 Example: A large basket of fruit contains 3 oranges, 2 apples and 5 bananas. If two fruit are chosen at random without replacement, what is the probability that one of the selected fruits is an apple and the other one is an orange? P(A and O) = P(A) P(O A) = 2 10 3 9 = 1 15 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 84 / 91

Examples Example: A box of 10 items has 2 defective items. Three items are selected by the researcher without replacement. (a) What is the probability that the researcher will obtain no defective items? P( D and D and D) = P( D) P( D D) P( D D and D) = 8 10 7 9 6 8 = 7 15 (b) Given that the researcher finds the first item selected as defective, what is the probability that the researcher will also have the other defective item? P( ( D and D) D) = P( D D) P(D D and D) = 8 9 1 8 = 1 9 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 85 / 91

Example (Roulette) Example 20 (Roulette), from Moore and McCabe A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and at the same time rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) What is the probability that the ball will land at any one slot? P(any slot) = 1 38 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 86 / 91

Example (Roulette) (b) If you bet on red, you will win if the ball lands on a red slot. What is the probability of winning? P(red) = # of red slots 38 = 18 38 = 9 19 < 1 2 Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 87 / 91

Example (Roulette) (c) The slot numbers are laid on a board on which gamblers place their bets. One column of numbers on the board contains all multiples of 3, i.e., 3, 6, 9,..., 36. You place a column bet that wins if any of these numbers comes up. What is you probability of winning? P(Slot # is a multiple of 3) = 12 38 = 6 19 < 13. In fact, every game in a casino offers options to the gambler that has less than a 50% chance of winning. In later chapters we will discuss probability and payout. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 88 / 91

Example (Lottery) (From Moore and McCabe.) A state lottery s Pick 3 game asks players to choose a three-digit number, 000 to 999. The state chooses the winning three-digit number at random. You win if the winning number contains the digits in you number, in any order. (a) Your number is 123. What is your probability of winning? (b) Your number is 112. What is your probability of winning? Solution: First of all, note that the sample space S for the experiment, which consists in choosing a 3-digit number at random, contains 1000 equally likely outcomes: S = {000, 001,..., 999}. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 89 / 91

Example (Lottery) (a) Consider the event A: a chosen number has digits 1, 2, and 3. A = {123, 132, 213, 231, 312, 321} Then P(A) = 6 1000 = 0.006, quite a small chance! (b) Define the event B: a chosen number has two 1 s and one 2. B = {112, 121, 211} Then P(B) = 3 1000 = 0.003. You do not want the digits to repeat in your lottery ticket! Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 90 / 91

Concluding Remarks Observation 1: and and means that we are looking at the intersection of events. We need to compute the probability that ALL events in the statement occur simultaneously. Probability computations in this case usually involve using one of the multiplication rules. Observation 2: or or means that we are looking at the union of events. We need to compute the probability that at least one event in the union occurs. Probability computations in this case usually involve either one of the addition rules or the rule for opposite events. Observation 3: at least, at most, not exactly, not equal Most of the problems of this type reduce to computing probabilities of the unions of events. It is often easier to compute the probability of the opposite event first. Dr. Joseph Brennan (Math 148, BU) Chapter 13 & 14 - Probability 91 / 91