An approach to Calculus of Probabilities through real situations


 Randolph Harris
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1 MaMaEuSch Management Mathematics for European Schools mamaeusch An approach to Calculus of Probabilities through real situations Paula Lagares Barreiro Federico Perea RojasMarcos Justo Puerto Albandoz MaMaEuSch Management Mathematics for European Schools CP DE  COMENIUS  C21 University of Seville This project has been carried out with the partial support of the European Community in the framework of the Sokrates programme. The content does not necessarily reflect the position of the European Community, nor does it involve any responsibility on the part of the European Community.
2 Contents 1 Random and probability Objectives The mus game Random experiments Random events and sample space Outcomes and random events Consistent events and inconsistent events Certain event Impossible event The complement of an event Operations on random events Union of event: one event OR another Intersection of events: one event AND another Difference of events Properties of the operations with events Probability Introduction Definition of probability from the relative frequencies: empirical probability Laplace s rule: theoretical probability Extractions with replacement and extractions without replacement. Tree diagrams Extractions with replacement Extractions without replacement Axiomatic definition of probability Calculus of probabilities in more complex cases Conditional probability Independence of random events Total probability Bayes s rule Answer of the initial question
3 3 OneDimensional Probability Distributions Objectives The example Introduction. Discrete Random Variable and Probability Distributions Cumulative Probability Function The Mode The expectation The variance Resume of the initial question An example of discrete random variable: the binomial distribution Objectives The example Introduction The expectation The variance Continuous distributions: normal distribution Objectives The example Introduction
4 Chapter 1 Random and probability Let us play mus. The cards are dealt and it is time to decide how much to bet. We have to take into account that we do not play alone, but we compete with other participants. Nervously we are going to watch our cards every time we receive one. Which cards will it be? Will they be better than the ones of the others? Before doing anything, we are going to show the objectives that will be covered in this manual and give some rules for playing mus. 1.1 Objectives Understanding the concept of random experiments and distinguishing it from the deterministic ones. Identifying random events after an experiment and telling the difference between event and outcome. Finding some special events: the impossible event and the sure event. Operating with random events and interpreting the events resultants after carried out unions, intersections and differences. Assigning probabilities to a simple random events in two different ways: from relative frequency and from Laplace rule for the case of equally likely outcomes. Understanding the notion of conditional probability an its uses. Understanding the independence of random events and the uses of the calculus of probabilities. Working with the rule of total probability and the Baye s rule, their differences and its applicability in the calculus of probabilities. 3
5 1.2 The mus game We are going to play a simple game of cards. In this game, two teams of two players each will play. The winning team will not receive any money, they will only win a good time with their friends. In order to play a good game, we need a deck of 40 cards, distributed in the following way: Eight aces. Four fours. Four fives. Four sixes. Four sevens. Four jacks. Four horses. Four kings. Deal four cards to each player randomly from the deck. Then, you can do the following moves: If you have two equal cards, and the other two are different within them and different from the first ones, you have a pair. For example, this hand is a pair of horses: (five, horse, horse, ace). Having three cards equals and the fourth one different is a trio. A hand like this (six, king, king, king) is a trio of kings. Two pairs within your four cards is a duplex. They can be different pairs or equal pairs. (ace, king, king, ace) and (ace, ace, ace, ace) are two different duplex. In this game, a duplex has more value than a trio and the trio is better than a pair. In the case of having two pairs, two trios or two duplex, the winning hand will be the one that has the highest cards in the pair, the trio or one of the pairs of the duplex. The cards are classified from the lowest to the highest in the following order: ace, four, five, six seven, jack, horse and king. For example, a duplex of kings and aces will win over a duplex of jacks and horses, because the highest pair of the first duplex (pair of kings) is higher than the highest pair of the second duplex (pair of horses). A pair of jacks is better than a pair of sixes for the reasons above. In the case of having the same duplex, a trio of the same cards or a pair of the same cards, the winner will be the one who has the highest cards apart of the pair or apart of the trio, in each case. If we have two hands with exactly the same four cards, the winner will be the one who first has received his cards, i.e., the person sitting on the right side of to the dealer. 4
6 Suppose that four friends have passed a lot of time playing this game and they have noticed that a pair of kings, a trio of kings or aces and any duplex, has been gotten in a similar number of times. They are discussing about which of these moves are the most probably of them. What do you think about this? Do not answer this question, you can do it when you finish this chapter. Good luck. 1.3 Random experiments Example Imagine the following situation: the cards are dealt of the mus game. Will we know which cards we have before see them? As you see, we can not have any certainty of the cards that we receive until we see them. We can get three kings and an ace or four jacks. Both possibilities and many more can occur in the distribution of our cards. The fact of not having certainty of the result after the distribution is called randomness. In our case, we have an experiment: take four cards out of the pack. After the realization of the experiment several results can be given, we say that it is a random experiment. If we know the result of the experiment before doing it, we could say that it is a deterministic experiment. For instance, if we drop a stone from our hand, we know that it will fall to the ground. Here there is no possibility of having different results, only one: the stone will fall to the ground So we can say that the biggest difference between a random experiment and a deterministic one, is that the first one can give different results and we can not predict which one will happen, and in the deterministic experiment we only have one possibility, and this one is that will occur. Exercise Describe two random experiments and two deterministic ones. Definition (Random experiment) A random experiment is any procedure or situation that produces a definite outcome that may not be predictable in advance. 1.4 Random events and sample space When we understand the idea of what a random experiment is, naturally we will ask ourselves questions about what will occur. It s possible to see that after the card distribution we can obtain a large number of combinations containing different cards. We can get hands like (ace, king, ace, jack) or (four, king, five, seven). Each of these possible hands will be called random events. That is to say, in the random experiment we earlier described (dealing four cards of the mus deck), the hand (ace, seven, jack, six) is a random event. The set of all random events of a random experiment is called Sample Space, and we denote it by E. In our experiment, the sample space is the set consists in all the possible events. Exercise Let us imagine the following random experiment: taking a card randomly out of the mus deck. Describe the sample space of that experiment by the enumeration of all its random events. 5
7 Definition (Sample space and event) The set of all the outcomes that we can obtain after doing a random experiment is called sample space, and we denote it by E. Any subset of the sample space is an event Outcomes and random events We distinguish the different results we can obtain into two groups: outcomes and random events Imagine that you watch the first card of your hand and it is an ace. Now, imagine the followings results: the card watched is an ace, the card watched is lower than seven. You can notice that between these two possible results there is a big difference. In the first one we specify what kind of card it is, while in the second case we say that the card can be an ace, a four, a five or a six. That is to say, the second event described (the card is lower than seven) includes several random events within it. Then we can say that the first possible result described is an outcome while the second one is a random event. Definition We say that a result of a random experiment is an outcome when it consists in only one element of the sample space. In other cases we will call it a random event. Example Consider the following random experiment: take out of a Spanish deck a card randomly. The elements of the sample space are: ace of gold, two of gold,..., king of basto The result obtaining the king of gold is an outcome. But if we choose the result obtaining a king, we have to say that it is a random event, because it is made up of four different outcomes: obtaining the king of cup, obtaining the king of gold, obtaining the king of sword and obtaining the king of basto Consistent events and inconsistent events Let us get back to the dealing of cards. We are going to get four cards each. Let us imagine two random events: A event = Two of the four cards are kings, B event = Two of the four cards are aces. Is it possible to obtain the A and the B events at the same time? That is to say, is it possible to obtain two kings and two aces in the same hand? The answer is yes. We can get a valuable duplex of kings and aces. As the A event and the B event can be obtained at the same time, we say that they are consistent events. However it isn t always is like this, there are a couple of events that can not occur at the same time. As an example, imagine these two new events: 6
8 Event C = Three out of our four cards are kings, Event D = Two out of our four cards are aces. And we ask ourselves again, is it possible to get the C event and the D event at the same time? That is to say, is it possible to obtain three kings and two aces in the same hand? In this case, it is not, because if we have three kings and two aces in the same hand, this would mean that we have at least five cards, and in this game four cards is the maximum. Therefore, this is impossible. So, because the C event and the D event can not appear in the same random experiment, we say that they are inconsistent events. Exercise Find a pair of consistent events and a pair of inconsistent events, they must be different to the pairs given above, in the same random experiment. Definition If we have two random events of a random experiment, we will say that they are consistent events if they can occur at the same time, and we will say that they are inconsistent events if they can not occur at the same time Certain event Let us imagine that we take our deck and we divide it into two parts. In one of the parts we have the aces (eight cards) and in the other part all the other cards (thirty two cards). We choose the part of aces and we take a card out of this group whit randomness. Can we assure anything? We can. We can assure that the card chosen is an ace. So simple is the certain event, it is the event that always occur. Exercise In the dealt of the mus s cards, find a certain event in your hand, that is to say, within your four cards. Definition A certain event is that random event of a random experiment that always occurs. Is possible to say that certain event is the event made up of every outcome of the sample space Impossible event However, in the same random experiment we can sure not to get a six, because there is no six in the part of the deck that we have chosen. Then we say that the random event the card chosen is a six is an impossible event. Exercise In the dealt of mus, describe an impossible event in your hand, that is, in your four cards. Definition We say that a random event is an impossible event when it never occurs. 7
9 1.4.5 The complement of an event Suppose now that the deck has been divided into two parts or groups: in the first part we have all the kings and all the aces (sixteen cards) and in the second part we have the rest of the cards (twenty four cards). Then, we choose a card from the first group. Let us look at the following events: A = The card chosen is an ace, B = The card chosen is a king. As a special characteristic of these events, we can say that if the event A does not occur then the event B will occur, and if the event B does not occur then the event A will occur. That is, one of them always occurs. You can also notice that they are inconsistent events, that is to say, they can not occur at the same time. These two qualities have to be complied with a pair of random events in order to say that they are complement events or they are the complement of the other. Exercise Getting a pair of complement events in the random experiment of taking a card out of the complete mus deck with randomness. Definition Two events are said to be complement events if they are inconsistent (they can not occur at the same time) and when one of them always occurs. That is to say, the complement of any event made up of some outcomes consists in all other outcomes. If we denote by A a random event then its complement is denoted by A or A c. 1.5 Operations on random events In the same form as we operate with numbers (sums, differences, multiplications,...), we can operate with random events. But now the operations are different than the number operations, in this way we will talk about unions of events, intersections of events or differences between events Union of event: one event OR another Imagine that we deal the cards again in the mus game, that is, four cards to each player. So that, we have the random experiment of taking four cards out of the deck and watching them. Let us describe two possible events of this random experiment: A = Taking two kings out, B = Taking an ace out. Suppose that after the dealt, our cards are: (ace, jack, seven, seven). After having knowledge of our cards, has the event A occurred? It has not, because we do not have two kings in our cards. And, has the event B occurred? It has, because we have obtained an ace. In this case, we will say that the event A or B has occurred, and we will denote it A B. So, if we have two random events and one of the events or maybe both occurs, then we will say that the event A B has occurred. 8
10 Exercise We have again the previous random experiment. We deal four cards. Imagine the following events: A = We obtain three kings, B = We obtain three aces. From this, describe two distributions in which the event A B has occurred. Definition Given the event A and the event B, we define the event A or B, and we denote it as A B, as the event consisting in occurring at least one of them. Note that if both events are given, the event A B is given too Intersection of events: one event AND another Let us describe now two new random events after the realization of the random experiment consisting in dealing four cards of the mus deck. A = Taking two aces out, B = Taking a seven out. Imagine that after the distribution of the cards, we have this hand: (ace, king, seven, ace). After observing it, has the event A occurred? Yes, because we have two aces within our cards. And, has the event B occurred? It has too, because our third card is a seven. Since both events have occurred, we say that in this distribution the event A and B has occurred, and we denote this event as A B. Exercise Think about the following random events after taking the four cards out of the deck A = Taking two kings out, B = Taking two aces out. Describe an event in which the event A B has occurred and another in which the event A B has not occurred. Definition Given the event A and the event B, the event A AND B is defined, and we denote it by A B, as the random event consisting in both events, event A and event B, happens at once. Notice that if the intersection between two random events is the impossible event, ( ), these events are inconsistent, (remember the definition of inconsistent events given in the previous section). If the intersection between two random events is not the impossible event, then the events are consistent. 9
11 1.5.3 Difference of events Let us describe now two new events after the distributions of cards in the mus game: A = Taking three jacks out, B = Taking a king out. After the dealt we have obtained the following cards: (jack, ace, jack, jack). Has the event A occurred? Of course it has, because we have exactly three jacks within our four cards. And, has the event B occurred? Now we have to say no, because in our cards there are no kings. Then we say that the event A minus B has occurred, and we will denote it by A \ B. So every time that an event has occurred and another has not, we will say that has happened the event difference between the first one and the second one. Exercise We are in facing the previous experiment. Think about the following events: A = We take two aces out, B = We take two jacks out. From this, propose a possible distribution in which the event A \ B occurs, and propose another in which the event B \ A occurs. Definition Given the event A and the event B, the event A \ B is defined as the event consisting in the occurrence of the event A and the nonoccurrence of the event B Properties of the operations with events The properties most utilized are the following. We have to take the event E in account which is the certain event, the event is the impossible event. The event A, the event B and the event C are any three events, subsets of the sample space and A c is the complement of the event A. Union: A B = B A, A E = E, A = A, A A c = E. Intersection: Difference: De Morgan s laws: A B = B A, A E = A, A =, A A c =. A \ B = A B c. (A B) c = A c B c, (A B) c = A c B c. Other properties: A (B C) = (A B) (A C), A (B C) = (A B) (A C). 10
12 Chapter 2 Probability 2.1 Introduction In the previous chapter we said that when we realize a random experiment, we have no certainty about the outcomes that we will obtain. In other words: there is uncertainty and we will measure this uncertainty into a number that we will associate with each random event and it will be called probability. For instance, if we take a card of the deck with randomness, we will not know how to predict certainly which card it will be. But we know that there are more aces than sevens, it would be more logical to think that there is a bigger chance of taking out an ace than a seven. So we will say that, after the experiment, the random event the card is an ace has more chances to occur than the other event which is the card is a seven. In this chapter we will show several techniques to measure the frequency of occurring that the different events have, that is to say, we will introduce several methods for assigning probabilities. Exercise Write down two events after the distribution of the mus cards that you think they very different probabilities or possibilities of occurring and explain why Definition of probability from the relative frequencies: empirical probability In the previous sections we said that probability is a number that we assign to each event after a random experiment, and with this number we want to show the frequency that the event has of occurring. A straightforward way to obtain the probability of a random event is from the table of relative frequencies of these experiment. This probability is called empirical probability because it is obtained after realizing the experiment. So, if we have done the experiment n times and after observing the results we see that in k of these times the event that we are studying has happened, we call it A, 11
13 we say that the probability for the event A to occur is k n, and we will denote it P (A), that is: P (A) = k n. Example Suppose that we are taking cards out of the deck one by one and after seeing one we replace it in the deck before taking the next one out. So we have obtained the following frequency table: Card Relative frequency ace four five six seven jack horse king From this table we could say that if we take a card out of the mus deck, we will obtain each card with the following probabilities: P( the card is an ace )= = 0 19, P( the card is a four )= = 0 085, P( the card is a five )= = 0 105, P( the card is a six )= = 0 12, P( the card is a seven )= = 0 105, P( the card is a jack )= = 0 115, P( the card is a horse )= = 0 09, P( the card is a king )= = Imagine that we do 1000 extractions, then we obtain the following relative frequencies: Card Relative frequency ace four five six seven jack horse king
14 From this table we could say that if we take a card out of the deck with randomness, we will obtain the different cards with the new following probabilities: P( the card is an ace )= = 0 192, P( the card is a four )= = 0 111, P( the card is a five )= = 0 109, P( the card is a six )= = 0 085, P( the card is a seven )= = 0 087, P( the card is a jack )= = 0 116, P( the card is a horse )= = 0 091, P( the card is a king )= = After seeing these probabilities, and once that we understand that the probability is a number that we assign to each random event to measure how likely it is, we can say that the event the card drawn is a king is more likely, that is to say, it has more chances to occur, than the event the card drawn is a seven, because of P(the card is a king) > P(the card is a seven). Using the same reasoning, we will say the event the card is a jack is more likely than the event the card is a horse,... Mark This method to assign probabilities is based on the Law of Large Numbers for Relative Frequencies, which comes to say that every event has a special number (called its probability) so that if the random experiment is repeated a large number of times, then the relative frequency of the event will be close to the probability of the event. The more times the random experiment is repeated, the closer the relative frequency will tend to be to this special number. That is to say, if we take 100 cards out in the previous example, the reliability of the probabilities obtained would be less than the ones obtained taking 200 cards out and vice versa, if in the previous example we take cards out instead of the 1000, the probabilities should be closer to the real probabilities of these events than the ones obtained taking 1000 cards out. In any case, the table gotten after the performance of the 1000 extractions is more reliable than the obtained after 200 extractions. Exercise Perform the following experiment: carry out twenty distributions of the mus game, that is to say, four cards of the deck without replacement, and write if in each distribution has been obtained: a pair, a trio, a duplex or none of them. Make the table of relative frequencies of these experiment and after assign probabilities to these random events (pair, trio, duplex and none). Do you think that these probabilities are reliable? Why? In the following section, we will assign probabilities of events in other way. We will not need perform the experiment for knowing these probabilities so in some cases, this method will be easier than the method explained before (relative frequencies). 13
15 2.1.2 Laplace s rule: theoretical probability As you could see, assigning probabilities from relative frequencies is too tedious, because it is necessary to repeat the experiment many times to get a good approximation of the real probability of an event, and even so, we will never be sure of getting the real probability using this method. For this reason it is necessary to introduce an alternative method for the calculus of probabilities which has to be more handy. Let us imagine the previous example: we have a mus deck and we are going to take a card out of the deck. We want to know the different probabilities of all the possible events. Well, is logical to think that the deck is well made and we will take a card out with the same probability as the others. That is to say, there are no cards bigger than another, there are no folded cards,... in other words, the deck has not any faults and so we can take any of the forty cards out of the deck with equal probability. In this case we say that they are equally likely. Another equally likely outcome can be the number that we have after throwing a dice (with equal probability it will be a one, two,..., six) or the fact of obtaining a head or a tail after throwing a coin (with equal probability it will be head or tail), always supposing that the dice and the coin have no faults. Return to our example with the mus deck. We have forty cards, all of them with equal weight, equal form,... There are eight aces within them, so is logical to think that after forty extractions, in eight of them we will receive an ace. This is something theoretical. You can see that you will not always obtain eight aces after forty extractions (you could obtain three aces or twelve aces, depending of the randomness). But the fact of having eight aces within the deck gives us an idea of how likely is to obtain an ace after an extraction. So we will say that the probability of an ace was taken out is It is a theoretical probability, we have to repeat it, in practice we do not always obtain eight aces after forty extractions. In this experiment, as there are forty cards in total, we will say that in this experiment there are forty possible outcomes (we can take forty different cards out of the deck) and eight outcomes in the event that we analyze (taking an ace out), because it has eight chances of occurring. Now, after these concepts have been introduced, we can enunciate the Laplace rule for the calculus of probabilities: Definition If all outcomes of a random experiment are equally likely, and we are studying a random event of this random experiment called A, it holds: P (A) = Number of outcomes in A Total number of outcomes, A. where the Number of outcomes in A are the real possibilities for the occurrence of the event This definition was the first formal definition given in the history, and it was given by Pierre Simon de Laplace in the beginning of the nineteenth century. After this, you can correctly do the following exercise: Exercise Calculate, using the Laplace rule, the probabilities of receiving each card in the experiment consisting in taking a card out of the mus desk with randomness. 14
16 2.2 Extractions with replacement and extractions without replacement. Tree diagrams In this section we are going to propose some new and more complex random experiments, instead of only taking one card out, we are going to take several cards out. After the study of this section we will be able to analyze better than before all the different moves in the mus game Extractions with replacement Let us start with a simple situation, we take two cards out of the mus deck, one after the other and replacing the card in the deck after looking at it. This process will be called extraction with replacement. If we call A 1 the event The first card is a king and we call A 2 the event The second card is a jack, we could ask ourselves what the probability of these two events occurring at the same time is, that is to say, the first card taken is a king and the second card taken is a jack. To calculate the probability of this event, we sketch the following diagram, it is called a tree diagram: The probability of the first card being a king and the second card being a jack is the multiplication of the probabilities of the path which leads to the result (the product rule), that is to say, = However, if the only thing we want is for both cards to be a king and a jack, and the order of occurrence is without importance, we have to propose that we can receive (king, jack) in this order, or the same cards in the inverse order (jack, king). If we call B 1 the event The first card is a jack and B 2 the event The second card is a king, for obtaining the combination (jack, king) it is necessary the event B 1 B 2 occurs. For the same reasoning as above, using a similar tree diagram as before, we get that the probability of receiving (jack, king) is = So, for obtaining the probability of receiving a jack and a king and without taking the order in account we add up the probabilities from before (king, jack) + (jack, king) (adding up rule), and 15
17 we obtain = Exercise In the random experiment before calculate the probability of both cards being aces, sketching the respective diagram tree. In the following section we will see another kind of extraction in which the cards are not replaced after they have been looked at Extractions without replacement In the random experiments performed above, we put the card back after looking at it but, what would happen if we do not put the card back into the deck? Well, the things changes but the reasoning is similar, the only things that change are the second s probabilities, that is to say, the probabilities relating to the second extraction. It is logical, because if we before the first extraction have forty cards, just before the second extraction we have thirty nine, because we do not put the first card that we took back. Then, in the experiment above, if we want that the first card to be a king and the second one to be a jack the diagram tree is the same, but the probabilities of the second extraction change. Let us calculate again the probabilities of taking a king and a jack out in this order, but this time with different extractions, without putting back the cards after looking at them. These extractions are called extractions without replacement. The diagram tree of this experiment is quite similar to the diagram tree before, let us see it: In this case we have that the probability of taking (king, jack) out is Notice that in this case the second factor is 4 39 because when we do not put back the first card that we have taken we only have thirty nine cards, and within them there are four jacks. In a similar way we can calculate the probability of receiving the combination (jack, king) after two extractions without replacement, and it would be And again, if we want to calculate the probability of receiving a jack and a king without taking the order in account, after two extractions, we will only have to apply the add up rule, and we will get that the probability of this is =
18 Exercise Do the same as in the exercise of the section before but supposing that we do extractions without replacement. 2.3 Axiomatic definition of probability Now, in this section, we are going to introduce a definition more abstract of probability. We will do it according to some principles that we will accept as evident (we will call them axioms). These axioms are: 1. For each event A, its probability is a number between 0 and 1, that is to say, 2. P (E) = 1, where E is the certain event. 0 P (A) If A and B are two inconsistent events, it holds that P (A B) = P (A) + P (B). From these axioms, we can deduce a great number of properties that the probability has to comply with: 1. If we denote by A c the complement of the event A, it holds that P (A c ) = 1 P (A). Exercise Prove this property from the axioms above. 2. If we have a set of events A 1, A 2,..., A n, they are inconsistent in pairs (A i A j =, i j), it holds that n n P ( A i ) = P (A i ). i=1 As a particular case we can study the case when the set of events A 1, A 2,..., A n complies that n i=1 A i = E too, where E is the certain event. In this case we say that the set of event A 1, A 2,..., A n is a complete set of events, and it holds that n i=1 P (A i) = If the sample space can be broken down into n outcomes or single events, E = {x 1,..., x n }, then it holds that i=1 P (x 1 ) + P (x 2 ) P (x n ) = n P (x i ) = 1. As a particular case, if the probability in each single event or outcome is the same, that is, P (x i ) = 1/n, and A is an event consisting of k outcomes, it holds that P (A) = k/n, which is the Laplace rule again. i=1 17
19 4. If A and B are any random events, it holds that P (A B) = P (A) + P (B) P (A B). This property can be used in a case of having three events, and we will get that P (A B C) = P (A) + P (B) + P (C) P (A B) P (A C) P (B C) + P (A B C). Example Suppose that we carry out the following experiment: we take four cards out of the mus deck with replacement. We take a note of whether the card taken is a face card (jack, horse or king) or if it is not after every draw and in the end we take a note of the number of face cards taken out. Consider as the sample space the number of face cards that we have, (0,1,2,3 ó 4). a) Describe the outcomes and calculate theirs probabilities. The outcomes are A i = we have i face cards, i = 0, 1, 2, 3, 4. As there are sixteen face cards in total we know that, applying the Laplace rule, that the probability of taking a card out of the mus deck with randomness is = 2 5 and the probability of the card not being a face card is = 3 5. From this and making calculations as the calculations that we made before in this section, we can see that the probabilities of these events are, respectively: P (A 0 ) = ( ) 4 ( 3 3, P (A 1 ) = ) 3, P (A 2 ) = 6 ( 2 5 ) 2 ( 3 5 ) 2, P (A 3 ) = 4 ( 2 5 ) 3 3 5, P (A 4) = b) If B = We have taken a face card out of the deck, calculate P (B). We have that B = A c 0, so, P (B) = P (A c 0) = 1 P (A 0 ) = 1 ( 3 4 5) = c) If C = We have taken three or more face cards out of the deck, calculate P (C). C = A 3 A 4, and also A 3 A 4 =. Then we have that P (C) = P (A 3 A 4 ) = P (A 3 )+P (A 4 ) = 4 ( ) ( 2 4 5) = Calculus of probabilities in more complex cases Conditional probability ( ) We deal the cards again. They are dealt one by one and we will receive our card in the forth turn, that is to say, the last one. The first player has received a king, the second player has received and ace and the third player has received a jack. What is the probability that we receive an ace? If we apply the Laplace rule we can say that the probability of receiving an ace is 7 37, because three cards has been dealt already and so thirty seven cards remain in the deck, and one of the aces was given to the second player, so there are seven aces in the deck. Imagine now that no player has received an ace, what would the probability be of receiving an ace in this case? In this case, there would be eight aces in the deck so the probability would be But, if two of the players have an ace? What is the probability of receiving an ace? In this case the probability would be As you can see the value of the probabilities of receiving an ace are changing depending of the cards that our rivals have. That is to say, the probability of an event can depend of the information that we 18
20 have before performing the experiment. In this case, the information before are the cards that our rivals have received, that is, we know which cards will not be in the deck when our turn comes. In these cases it is very simple to calculate the probabilities, but there are other cases in which we will need a formula for calculating these probabilities. Let us go back to the example explained before. If we call the event My card is an ace A and we call the event The three first players have received: a king, an ace and a jack respectively B, we want to calculate P (A), but knowing the cards that the other player have, that is, knowing that the event B has occurred. We want to calculate the probability of the event A given the event B, and we will denote it as A/B. For calculating this we can apply the formula of the conditional probability which says: P (A/B) = P (A B), P (B) Therefore, we would have to calculate P (A B) and P (B). For calculating P (B) we apply the concepts that we learned before about the extractions without replacement, and with a similar reasoning we obtain that P (B) = , while for A B to occur it is necessary for the four players to receive a king, an ace a jack and an ace, that is to say, the probability of A B is P (A B) = Therefore, we have to apply the formula of the conditional probability that P (A/B) = P (A B) P (B) = = 7 37, as we knew before. In this case, we could have solved the question without the conditional probability formula, but in other cases its use is necessary. Definition We will denote the probability of the event A to occur given the event B as A/B, which gives: P (A B) P (A/B) =. P (B) Independence of random events Let us go back to our explanations and let us think about the example that we saw in the section about the tree diagrams. If you remember it, we had to do a lot of operations for calculating the probabilities asked for, in spite of this, it was one of the simplest cases. Imagine that if instead of having two possible outcomes after each extraction (king or not king in the first extraction and jack or not jack in the second extraction) we have three outcomes, there would be nine possibilities in total, if we have four possible outcomes after each extraction we would have sixteen possibilities after 19
21 the second extraction. In general, if we have n outcomes in each extraction, after two extraction we will have to analyse n 2 cases, that is a lot. And it is only if we talk about two extractions, if the number of extractions is three, we would have n 3 possible outcomes, if we have four there will be n 4,... The tree diagram technique is only usable in very simple cases, when the numbers are getting bigger, the tree is almost impossible to sketch. Is there another simpler way for calculating the probability of this kind of events? There is, but we have to study a new concept before: the independence of random events. Definition Given a random experiment and any two events of this experiment, let us call them A and B, we say that these two events are independent if for one of them to occur it does not matter if the other one has occurred or not. In other words, two events, A and B, are independent if the probability of A is equal to the conditional probability of A given B and the opposite, that is P (A/B) = P (A) and P (B/A) = P (B). In another words two events, A and B, are said to be independent if the probability of A and B is equal to the product of the probability of A times the probability of B. P (A B) = P (A) P (B). This result is of great use for the calculus of probabilities in the repetition of random experiments. So, if we repeat n times a random experiment and we know that the result after any time is independent of the previous results, and we want to calculate the probability of the occurrence of the event A i in each repetition i = 1,..., n, we will get the probability for all of these events to occur, this will be the event intersection between all of them A 1 A 2... A n, is P (A 1 A 2... A n ) = P (A 1 ) P (A 2 )... P (A n ). Example If we take two cards of the mus deck randomly and with replacement, what will be the probability of both cards being face cards? What is the probability of no card being face card? What is the probability of one of them being a face card and the other card not? If we call the event The first card is a face card A and the event The second card is a face card B, we get that the event Both cards are face cards is the event A B. Clearly they are independent events, because we have done extractions with replacement so the conditions before each extraction are the same. So, it holds that P (A B) = P (A) P (B) = The event None of the two cards are face cards is represented according to the events A and B in the following way: A c B c. They are also random events. Therefore, the probability of this event can be calculated like this: P (A c B c ) = P (A c ) P (B c ) = (1 P (A)) (1 P (B)) = The event One card is a face card and the other card is not is represented according to the events A and B in the following way: A c B and A B c, because it is possible that the first card 20
22 is a face card and the second one not and the opposite. So, the event that we want study is the union of both events, (A c B) (A B c ).As they are inconsistent events, (disjoints because of (A c B) (A B c ) A A c = (A c B) (A B c ) = ), the probability of this event can be calculated in this way P ((A c B) (A B c )) = P (A c B)+P (A B c ) = P (A c ) P (B)+P (A) P (B c ) = You can see which of these three events is more likely to occur. Exercise Take the following experiment into account: take two cards out of our deck randomly and with replacement, that is to say, putting back the card to the deck after looking at it and before taking the next card out the deck. Then add up the numeric values that they have in the mus game. Answer the following questions: a) Describe the sample space, the certain event and an impossible event of the experiment. b) Calculate the probability of the sum of the cards value being twenty. c) Calculate the probability of the sum of the cards value being six or less than six. Exercise Do the same as in the previous exercise but supposing that we take the cards randomly but without replacement, that is to say, without putting the card back to the deck Total probability Imagine that we draw any two cards out of the deck without replacement. We look at the first card and then at the second card. What is the probability that the second card was a king? With the knowledge that we already have, we can easily say that if the first card was a king then the probability that the second card was a king too is However, if the first card is not a king, then we have the probability of the second card being a king is As you can see, depending of which kind of card the first card taken was, we can assure something about the second card. In the section about conditional probability, we started with an advantage, we knew which card the first one was. But now we do not have this advantage. How can we solve this problem? We can do it taking both possibilities into account, the first card can be a king or other than a king. Let us see how we solve this problem: Consider the following random events: 1. A 1 = The first card is a king. 2. A 2 = The second card is a king. We want to calculate P (A 2 ). Well, we will take if the event A 1 happens or not into account. How can we do it? We divide P (A 2 ) into several probabilities which makes it easier to calculate. For doing it we have to resort to the properties of the operations with random events. If we consider A 1 as the complement of the event A 1, clearly we get that So, it holds that A 1 A 1 = E, A 1 A 1 =. 21
23 Besides, as we hold that A 2 = A 2 E = A 2 = (A 2 A 1 ) (A 2 A 1 ). (A 2 A 1 ) (A 2 A 1 ) =, and P (A 2 ) = P (A 2 A 1 ) + P (A 2 A 1 ). And applying the formula of the conditional probability, we obtain P (A 2 A 1 ) = P (A 2 /A 1 ) P (A 1 ), P (A 2 A 1 ) = P (A 2 /A 1 ) P (A 1 ). And we can calculate these two probabilities easily, applying the Laplace rule and the techniques seen for the extractions without replacement. So, we hold that P (A 2 /A 1 ) P (A 1 ) = , and P (A 2 /A 1 ) P (A 1 ) = =... = 0 2. As you see we have divided the probability into two different addends: in the first one we suppose that the first card is a king, A 1, and in the other one we suppose that the first card is other than a king, A 1. The events A 1 and A 1 have two special characteristics A 1 A 1 = y A 1 A 1 = E. This technique can be used as a general rule: If we have a set of inconsistent events A 1, A 2,..., A n in pairs (A i A j =, i j), and holding that A 1 A 2... A n = E, (if they comply with these two conditions, we say that this set is a complete set of events), then the probability of an event S E is equal to P (S) = P (A 1 ) P (S/A 1 ) + P (A 2 ) P (S/A 2 ) P (A n ) P (S/A n ), and this formula is called formula of total probability. The most difficult part when you apply the formula of total probability is to choose an appropriate complete set of events, because an inappropriate complete set of events only creates more difficulties solving the problem. It will be necessary to study what the convenient events are, because a bad choice of events of the complete set of events will not help us solving the problem. Exercise Find the probability, in the same experiment proposed at the beginning of this section, that the second card was not a face card. 22
24 Exercise Let us propose the following exercise: we deal three cards of the mus deck. Calculate, a previously chosen appropriate complete set of events and applying the formula of total probability, the probability of the third card being an ace. What is the probability that the third card was not an ace? Suggestion: choose as a complete set of events the number of aces taken out as the two first cards Bayes s rule Let us go back to the previous situation proposed as example. We took two cards out without replacement. A new question may be asked: what was the probability of the first card being a king knowing that the second card was a king? This question can look like the question that we asked in the previous section, but this one has a great difference: in this case we have performed the experiment already (we have seen the second card) and we ask ourselves which card was the first one. That is to say, if we call the events A 1 and A 2 just like before, we will calculate: P (A 1 /A 2 ). Applying the conditional probability formula, we obtain that P (A 1 /A 2 ) = P (A 1 A 2 ). P (A 2 ) And developing the denominator according to the total probability formula and applying the conditional probability formula in the numerator, we see that P (A 1 /A 2 ) = P (A 2 /A 1 ) P (A 1 ) P (A 2 /A 1 ) P (A 1 ) + P (A 2 /A 1 ) P (A 1 ). The calculus well be easy from the calculus that we did in the previous section, so it holds that P (A 1 /A 2 ) = = In general, Bayes s formula is obtained in the following way: Given a Complete Set of Events A 1, A 2,..., A n and any event S, we want to calculate the probability of the event A i to occur knowing that after doing the experiment that the event S occurred, that is to say, we will calculate P (A i /S). For the reasoning above it holds that: P (A i /S) = P (A i S) P (S) = P (S/A i ) P (A i ) n i=i P (S/A i) P (A i ), where P (A i ) is the probability a priori of the event A i (it is know before performing the experiment) and P (A i /S) is its probability a posteriori, because it is calculated once the experiment is performed. In the same way that we said in the section of the conditional probability, for using correctly the Bayes s rule is needed to choose an appropriate Complete Set of Events (CSE) which will be the most difficult step for solving a problem. 23
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