Suppose that B1 ,. Then apply the multiplication rule 2. Identity 2. Venn diagram representation of decomposition 2. Example 2. There are three types of coins in circulation.
The type of a coin cannot be determined from its appearance. What is the probability I get tails? If we knew the type of the coin the answer would be one of the probabilities given above. But the type of the coin is now random and consequently we use 2. Now apply 2. As in Example 2. An urn is picked randomly and a ball is drawn from it. Given that the chosen ball is red, what is the probability that the ball came from urn I?
Fact 2. Assume that 0. If a random person tests positive for the disease, what is the probability that they actually carry the disease? The key point is that a doctor would seldom order a test on a random person.
Tests are typically ordered in response to symptoms 50 Conditional probability and independence or a judgment by the doctor. The outcome of the calculation is very sensitive to the inputs. In this example the disease is so rare that even if we see a positive test on a randomly selected person, an error in the test is still a likelier explanation than the disease.
Now the disease is the very likely explanation of a positive test result. Return to Example 2. We hold a coin of unknown type.
These are the prior probabilities. One cannot ask about the independence of events from different sample spaces. We choose two balls with replacement. Let Example 2. Is it true that A and B are independent? What if we sample without replacement? When we sample with replacement, the contents of the urn are the same for each draw.
Consequently the conditional probability of the second draw should differ from the unconditional probability. Let us check this intuition by doing the math. Sampling with replacement.
In the last step we counted the red-green and green-green picks to get the number of ways to get green in the second draw. This gives Sampling without replacement. There is a symmetry going on here which we will study in Chapter 7.
Suppose that A and B are independent. Then Proof. B are independent. Knowing whether A happened is exactly the same as knowing whether Ac happened. Simply because A happens if and only if Ac does not happen, and vice versa. Thus if knowledge about A does not alter the probability of B, neither should knowledge about Ac. Find the probability that exactly one of the two events is true. Using the disjointness of ABc and A c B and the fact that both intersections involve independent events we get Example 2.
Events A1 ,. Suppose events A 1 ,. Then for every collection Ai1 ,. Let us illustrate again with three independent events identities that Fact 2. Exercise 2. This model was introduced in Example 1. Consider these events: Example 2.
The next example shows that there are events which are pairwise independent but not mutually independent. Independence Consider Example 2. The same argument shows that B and C are independent and this is also true for A and C. Thus A , B and C are pairwise independent. However these events are not mutually independent. So for example Example 2. The picture below represents an electric network. Assume that the switches are open or closed independently of each other, and that switch i is closed with probability pi.
In the last step we used 2. For example, if the random variables are discrete, then it is enough to check the condition for particular values. Conversely, assume that the condition in Fact 2.
Decompose the probability and then do some algebra. In the sums below the indices x1 ,. Continuation of Example 2.
In Example 2. Using this we can verify the product property for X1 , X2 , X3. Thus by Fact 2. Denote the outcomes of the successive draws by X 1 , X2 ,. Then if we sample with replacement the random variables X1 ,.
Let us verify these claims. Let us check that X1 , X2 ,. Note that specifying the values of all X1 , X2 ,. Again we begin with the probability mass functions of the individual draws. To show that independence fails, we only need to come up with one devious choice of outcomes x1 ,. We have shown that for sampling without replacement, the random variables X1 ,.
Independent trials A basic setting with independence is that of repeated trials. The simplest kind of trial has two outcomes, success and failure. Let us encode a success as 1 and failure as 0. Suppose we repeat the trial n times. P and 2. We have thus constructed a probability space that models n independent trials with success probability p.
Recall that each random variable has a probability distribution. The most important probability distributions have names. Repeated independent trials give rise to three important probability distributions: Bernoulli, binomial, and geometric.
We introduce them in this section. Bernoulli distribution The Bernoulli random variable records the result of a single trial with two possible outcomes. A sequence of n independent repetitions of a trial with success probability p gives rise to a sequence of n independent Bernoulli random variables with success probability p.
To illustrate how to write things down with Bernoulli random variables, let Xi denote the outcome of trial i. Then probabilities such as the second probability in 2.
Let Sn be the number of successes in n independent trials with success probability p. From 2. The fact that binomial probabilities add to 1 is a particular case of the binomial theorem see Fact D. The repeated trial is a roll of the die, and success means rolling a six. We shall not worry about the construction of the sample space, but this will not hinder us from doing calculations. For example, 2. Let 0 You can check that the values of the probability mass function add up to one by using the geometric series formula D.
What is the probability that it takes more than seven rolls of a fair die to roll a six? The possible values of Y are 0, 1, 2,. Roll a pair of fair dice until you get either a sum of 5 or a sum of 7. This is not a coincidence. It is a special case of a general phenomenon derived in Exercise 2.
Sampling without replacement is revisited to add the hypergeometric distribution to our collection of named probability distributions. We introduce a probability classic called the birthday problem. Last we leave pure mathematics to illustrate the pitfalls of careless probabilistic reasoning with two cases from recent history. The subsections of this section can be read independently of each other.
It is the same old Definition 2. To investigate this, we compute the probabilities with the law of total probability, as we did previously in Examples 2. Let F denote the event that the coin is fair and B that it is biased. In other words, A1 and A2 are not independent without the conditioning. It was assumed that 0.
We found that if a random person tests positive for the disease, the probability that he actually carries the disease is approximately 0.
What is the probability of disease after two positive tests? Following Example 2. We assume that A1 and A2 are conditionally independent, given D. The probability of actually having the disease increased dramatically because two false positives is much less likely than the disease. The calculation above rested on the assumption of conditional independence of the successive test results. Whether this is reasonable depends on the case at hand. If the tests are performed by the same laboratory, a systematic error in their equipment might render this assumption invalid.
Here is an example where we discover conditional independence. Roll two fair dice. In other words, conditioning can turn dependence into independence. And also vice versa. Further topics on sampling and independence Independence of events constructed from independent events Suppose we know that events A 1 ,.
Then we construct new events B1 ,. Then it stands to reason that the events B1 ,. A general proof of this fact is beyond this course. Here is a small example that illustrates. Let A, B and C be independent events. To show this, we use inclusion-exclusion and the fact that when events are independent, the product rule of independence continues to hold if some events are replaced with complements: Example 2.
However, this is not inherently so, but only because we do not have the right mathematical tools at our disposal. Hypergeometric distribution We revisit sampling without replacement.
Suppose there are two types of items, type A and type B. This is sampling without replacement with order not mattering.
The probability mass function of X can be found through the types of calculations that we saw in Example 1. This leads to one more probability distribution with a name, the hypergeometric distribution. Formula 2. With this convention 2. X A basket contains a litter of 6 kittens, 2 males and 4 females. A neighbor comes and picks 3 kittens randomly to take home with him.
Let X be the number of male kittens in the group the neighbor chose. The probability mass function of X is as follows. Contrast the experiment above with this one. A moment later she comes around again, picks up a random kitten, pets it, and puts it back in the basket with its siblings. She repeats this altogether three times. Let Y be the number of times she chose a male kitten to pet. You might even Example 2. The modest size required often comes as a surprise.
In fact, the 67 2. Further topics on sampling and independence problem is sometimes called the birthday paradox, even though there is nothing paradoxical about it. To solve the problem we set up a probability model. We ignore leap years and assume that each of the possible birthdays are equally likely for each person. We assume that birthdays of different people are independent. Then we can restate the problem as follows. Let pk be the probability that there is repetition in the sample.
The complementary event that there is no repetition is easier to handle. One way to think about it is the following. Thus, if there are approximately pairs of people, there should be a good probability of success. The following approximation will shed more light on why this guess is reasonable. Of course, the computation we just did is not fully rigorous because of the approximations. There are plenty of anecdotes related to the birthday problem.
We recount two here. Most of them thought it incredible that there was an even chance with only 22 or 23 persons. Noticing that there were exactly 22 at the table, someone proposed a test.
We got all the way around the table without a duplicate birthday. But I am the 23rd person in the room, and my birthday is May 17, just like the general over there. The second story is about Johnny Carson, the former host of the Tonight Show. It is a good example of a common misinterpretation of the problem. However, instead of looking for a match anywhere in the crowd, he asked for the birthday of one particular guest and then checked if anybody in the audience shared the same birthday.
It turned out that there was no match even though the size of the audience was roughly Where did Johnny Carson go wrong? He performed the following, very different experiment.
October 1, Below we describe two cases of faulty reasoning with serious real-world consequences. The Sally Clark case is a famous wrongful conviction in England.
In she was charged with murder. At the trial an expert witness made the following calculation. Population statistics indicated that there is about a 1 in chance of an unexplained infant death in a family like the Clarks. The jury convicted her. Much went wrong in the process and the reader is referred to [SC13] for the history.
Let us consider these two points in turn. Here is a simple illustration of how that can happen. Suppose a disease appears in 0. Suppose further that this disease comes from a genetic mutation passed from father to son with probability 0. What is the probability that both sons of a particular family have the disease?
If the disease strikes completely at random, the answer is 0. Hence the conditional probability that the second son also falls ill is 0. Thus the correct answer is 0. To put a different view on this point, consider a lottery. Yet there are plenty of lottery winners, and we do not automatically suspect cheating just because of the low odds. In a large enough population even a low probability event is likely to happen to somebody.
The crisis originated with mortgages. We present a brief explanation to highlight an independence assumption that failed when circumstances changed. Mortgages are loans that banks make to people for buying homes.
These securities are sold to investors. As the original borrowers pay interest and principal on their loans, the owner of the security receives income.
The investor might be for example a pension fund that invests contributions from current workers and pays out pensions to retirees. Some homeowners default, that is, fail to pay back their loans. That obviously hurts the owner of the security. But defaults were assumed random unrelated rare events. Under this assumption only a tiny proportion of mortgages default and the mortgage-backed security is a safe investment.
Yet the danger in this assumption is evident: if a single event brings down many mortgages, these securities are much riskier than originally thought. Developments in the early s undermined this assumption of safety.
A housing bubble drove home values high and enticed homeowners to borrow money against their houses. To increase business, banks lowered their standards for granting loans. Investors bought and sold mortgage-backed securities and other derivative securities related to mortgages without fully understanding the risks involved. Eventually home values fell and interest rates rose. Borrowers began to default on their loans. A cycle of contagion ensued.
Banks loaded with bad mortgages lost money and could no longer extend credit to businesses. Businesses suffered and people lost jobs. People out of jobs could not pay back their loans. Hence more mortgages defaulted and more mortgage-backed securities declined in value.
The assumption of independent and rare defaults was no longer valid because events in the overall economy were causing large numbers of mortgages to default. Ultimately the crisis led to the worst global economic downturn since the s.
Many actors made mistakes in this process, both in industry and in government. Complex mathematical models were used to analyze mortgage-backed securities. Decision-makers are often not the people who understand the limitations of these models. As we saw in Section 1. Entry si of the sequence represents the outcome of the ith trial. From this you can already surmise that the sequence space is an example of an uncountable space.
This is exactly as in Example 1. This implies that the X1 , X2 , X3 ,. Section 2. We roll two dice. Find the conditional probability that at least one of the numbers is even, given that the sum is 8. What is the probability that a randomly chosen number between 1 and is divisible by 3, given that the number has at least one digit equal to 5? The second urn contains three balls labeled 3, 4 and 5. We choose one of the urns at random with equal probability and then sample one ball uniformly at random from the chosen urn.
What is the probability that we picked the ball labeled 5? The second urn contains four balls labeled 2, 3, 4 and 5. Then we sample one ball uniformly at random from the chosen urn.
What is the probability that we picked a ball labeled 2? When Alice spends the day with the babysitter, there is a 0. Her little sister Betty cannot turn the TV on by herself. But once the TV is on, Betty watches with probability 0.
Tomorrow the girls spend the day with the babysitter. Find P Ac B. We return to the setup of Exercise 2. Suppose that ball 3 was chosen. What is the probability that it came from the second urn? I have a bag with 3 fair dice. One is 4-sided, one is 6-sided, and one is sided. I reach into the bag, pick one die at random and roll it. The outcome of the roll is 4. What is the probability that I pulled out the 6-sided die? The Acme Insurance company has two types of customers, careful and reckless.
A careful customer has an accident during the year with probability 0. A reckless customer has an accident during the year with probability 0. Suppose a randomly chosen customer has an accident this year. What is the probability that this customer is one of the careful customers?
For each of the following choices decide whether the two events in question are independent or not. Note that 1 is not a prime A C number. Decide whether A and B are independent or not. Let A and B be two disjoint events. Under what condition are they independent? Every morning Ramona misses her bus with probability 1 10 , independently of other mornings. What is the probability that next week she catches her bus on Monday, Tuesday and Thursday, but misses her bus on Wednesday and Friday?
Check whether the events A1 , A 2 , A 3 are independent or not. Show that X and Y are independent random variables. Show that X and Z are not independent. We have an urn with balls labeled 1,. Two balls are drawn. A fair die is rolled repeatedly. Use precise notation of probabilities of events and random variables for the solutions to the questions below.
Jane must get at least three of the four problems on the exam correct to get an A. She also assumes that the results on different problems are independent. Ann and Bill play rock-paper-scissors. Each has a strategy of choosing uniformly at random out of the three possibilities every round independently of the other player and the previous choices.
Remember that the round could end in a tie. Exercises Exercise 2. Find the probability that there is no accident during the next 4 days, but there is at least one by the end of the 10th day. Let X be the number of women on the team.
I have a bag with 3 fair dice: a 4-sided die, a 6-sided die, and a sided die. I reach into the bag, pick one die at random and roll it twice. The rolls of a given die are independent.
Suppose events A , B, C , D are mutually independent. Show that events AB and CD are independent. Urn I has 1 green and 2 red balls. Urn II has 2 green and 1 yellow ball. What is the probability that the ball is green? Then repeat the entire experiment: choose one of the urns uniformly at random and sample one ball from this urn. What is the probability that both picks are green?
We play a card game where we receive 13 cards at the beginning out of the deck of We play 50 games one evening. For each of the following random variables identify the name and the parameters of the distribution. Further exercises Exercise 2. There is a softball game at the company picnic. His probability of hitting a single is 0. Once on base, his probability of scoring after hitting a single is 0.
What is the probability that Uncle Bob will be able to score in this turn? Assume that 1 3 of all twins are identical twins. You learn that Miranda is expecting twins, but you have no other information. What is the probability that the girls are identical twins? Explain any assumptions you make. Suppose a family has two children of different ages. We assume that all combinations of boys and girls are equally likely.
Precisely: we learn that there is at least one girl. What is the probability that the other child is a boy? What is the probability that the older child we have not yet seen is a boy?
Suppose a family has three children of different ages. Assuming we have no other information beyond that at least two of the children are girls, what is the probability that the child we have not yet seen is a boy?
What is the probability that the oldest child we have not yet seen is a boy? We choose an urn randomly and then draw a ball from it. You play the following game against your friend.
You have two urns and three balls. One of the balls is marked. You get to place the balls in the two urns any way you please, including leaving one urn empty. Your friend will choose one urn at random and then draw a ball from that urn. If he chose an empty urn, there is no ball.
His goal is to draw the marked ball. A bag contains three kinds of dice: seven 4-sided dice, three 6sided dice, and two sided dice. A die is drawn from the bag and then rolled, producing a number. For example, the sided die could be chosen and rolled, producing the number Assume that each die is equally likely to be drawn from the bag. An urn contains one 6-sided die, two 8-sided dice, three 10sided dice, and four sided dice. One die is chosen at random and then rolled. Determine the probability mass function of X.
Incoming students at a certain school take a mathematics place- ment exam. The possible scores are 1, 2, 3, and 4. What is the probability that he or she scored a 4 on the placement exam? Suppose the phone is not defective. What is the probability that it came from factory II?
Urn A contains 2 red and 4 white balls, and urn B contains 1 red and 1 white ball. A ball is randomly chosen from urn A and put into urn B, and a ball is then chosen from urn B. What is the conditional probability that the transferred ball was white given that a white ball is selected from urn B? We have an urn with 3 green balls and 2 yellow balls. We pick a sample of two without replacement and put these two balls in a second urn that was previously empty.
Next we sample two balls from the second urn with replacement. We have two bins. The second bin has 3 blue marbles and 4 yellow marbles. We choose a bin at random and then draw a marble from that bin. This marble is also yellow. These are professional football teams.
Suppose that the probability of a 7-year-old fan of the Chicago Bears going to a game in a given year is 0. A 7 year old is selected randomly from Madison. Consider three boxes with numbered balls in them. Box A contains six balls numbered 1,. Box B contains twelve balls numbered 1,.
Finally, box C contains four balls numbered 1,. One ball is selected from each box uniformly at random. A medical trial of 80 patients is testing a new drug to treat a certain condition. This drug is expected to be effective for each patient with probability p, independently of the other patients.
You personally have two friends in this trial. Given that the trial is a success for 55 patients, what is the probability that it was successful for both of your two friends? A crime has been committed in a town of , inhabitants. The police are looking for a single perpetrator, believed to live in town. DNA evidence is found on the crime scene. Before the DNA evidence, Kevin was no more likely to be the guilty person than any other person in town.
What is the probability that Kevin is guilty after the DNA evidence appeared? Reason as in Example 2. The king has chosen one of the three uniformly at random to be pardoned tomorrow, while the two unlucky ones head for the gallows. The guard already knows who is to be pardoned. Prisoner A begs the guard to name someone other than A himself who will be executed. Prisoner C learned conditional probability before turning to a life of crime and is now hopeful.
What is his new probability of being pardoned? B Exercise 2. Decide whether A, B, and C are mutually independent. Peter and Mary take turns throwing one dart at the dartboard. Peter hits the bullseye with probability p and Mary hits the bullseye with probability r.
Find the possible values and the probability mass function of X. Check that the function you give sums to 1. Is this a familiar named distribution? Show that if independent of A. We have a system that has two independent components. Both components must function in order for the system to function. The second component has 4 independent elements that each work with probability 0.
If at least 3 of the elements are working then the second component will function. Given that information, what is the probability that the second component is not functioning? Alex, Betty and Conlin are comparing birthdays. A a Are events A, B and C pairwise independent?
Two towns are connected by a daily bus and by a daily train that go through a valley. What is the probability that travel is possible between the two towns tomorrow?
Suppose an urn has 3 green balls and 4 red balls. Let X be the number of times a green ball appeared. Identify by name the probability distribution of X. Let N be the number of draws that were needed. Identify by name the probability distribution of N.
Is there a reason these should be the same? We have a bin with 9 blue marbles and 4 gold marbles. Draw 3 marbles from the bin without replacement and record their colors.
Put the 83 Exercises marbles back in the bin. Perform this procedure 20 times. Let X be the number of times that the three draws resulted in exactly 3 blue marbles. Find the probability mass function of X and identify it by name. You win a prize if you get exactly 2 heads. How should you choose n to maximize your chances of winning?
What is the probability of winning with an optimal choice of n? Find both. Let fn n n n Derive precise conditions for n Exercise 2. On a test there are 20 true-or-false questions.
If the student does not know the answer he chooses true or false with equal probability. What is the probability that he will get the correct answer for at least 19 questions out of 20?
Three dice are rolled. What is the conditional probability that at least one lands on a 4 given that all three land on different numbers? You are given a fair die. You must decide ahead of time how many times to roll.
If you get exactly 2 sixes, you get a prize. How many rolls should you take to maximize your chances and what are the chances of winning? There are two equally good choices for the best number of rolls. In other words, the earlier n failures are forgotten. On an intuitive level this is a self-evident consequence of the independence of trials. In the context of Examples 2. Use conditional independence. Flip a coin three times. For which values of p are events A and B independent?
This generalizes Example 2. The population of a small town is Assume that the 14 customers are chosen uniformly at random from the population of the town. The results of the drug test are known to be independent from test to test for a given person. What is the probability he is a drug user?
Now what is the probability he is a drug user? You know that the store where you buy your phones is supplied by one of the factories, but you do not know which one. You buy two phones, and both are defective. What is the probability that the store is supplied by factory II? Make an appropriate assumption of conditional independence. Consider again Examples 2. This exercise gives an alternative solution to Example 2. Let D be the event that a random person has the disease. Using this new probability as the prior in equation 2.
Compare your solution with that of Example 2. Let and B be two disjoint events for a random experiment. Perform independent repetitions of the experiment until either A or B happens. P A Hint. Decompose as in Example 2. In a randomly chosen group of 22 people there are no matching birthdays, but a randomly chosen 23rd person has a birthday that matches the birthday of one in the original group of Find the probability that among 7 randomly chosen guests at a party no two share a birth month.
A Martian year is Martian days long. Solve the birthday problem for Mars. Challenging problems Exercise 2. There are N applicants for an open job. Assume that the applicants can be ranked from best to worst. The interviewer sees the N applicants sequentially, one after the other, in a random order. She must accept or reject each applicant on the spot without a chance to go back and reverse her decisions. If there is no such applicant then she has to hire the last applicant.
She chooses the parameter k ahead of time. Initially we have 1 red and 1 green ball in a box. At each step we choose a ball randomly from the box, look at its color and then return it in the box with an additional ball of the same color. Let X denote the number of red balls in the box after the th step. Find the probability mass function of X. Recall the rules of the game from Exercise 1. Suppose again that your initial choice is door 1.
We need now one extra bit of information. If the prize is behind door 1, Monty has to open either door 2 or door 3. Do the same conditioned on Monty opening door 3. Exercises b Assume that you will switch once Monty has opened a door. Compute the probability of winning, conditioned on Monty opening door 2, and then conditioned on Monty opening door 3.
Check that, regardless of the value of p, and regardless of which door you see Monty open, deciding to switch is always at least as good as deciding not to switch. If it comes up heads, you win a dollar. If it comes up tails, you lose a dollar. Suppose you start with N dollars in your pocket. In other words, the coin in the game is biased. Use it to produce a fair coin. We have a coin with an unknown probability 87 3 Random variables We have already encountered several particular random variables.
Now we begin a more general treatment of probability distributions, expectation, and variance. The last section of the chapter introduces the centrally important normal, or Gaussian, random variable. In this section we discuss three concrete ways of achieving this description: the probability mass function that we have already encountered, the probability density function, and the cumulative distribution function. Probability mass function We review some main points from Section 1.
The probability distribution of a discrete random variable is entirely determined by its probability mass function p. The probability mass function is a function from the set of possible values of X into [0, 1]. If we wish to label the probability mass function with the random variable we write pX k. Recall from 1. It is sometimes useful to extend the domain of the probability mass function to values k that are not possible for X.
This does not affect any calculations. The probability mass function can be represented as a bar chart see Figure 3. Probability density function We have seen many examples of discrete random variables, including some of the most important ones: Bernoulli, binomial and geometric. This class is studied with the tools of calculus. Let X be a random variable. We prefer 3. On the left is a graphical representation of the probability mass function p X of a discrete random variable X.
The column at position k has height pX k. On the right is the graph of the probability density function fY of a continuous random variable Y. Figure 3. Probability distributions of random variables The set B in 3.
Note especially the following case. It also follows that probabilities of intervals are not changed by including or excluding endpoints.
For example, in 3. Example 3. Which ones are probability density functions? To evaluate the integral, note that it is half the area of a disk of radius a. It is important in the subject called random matrix theory. If c is nonzero, then f3 takes both positive and negative values. Hence no choice of c can turn f3 into a probability density function. The answer is essentially no. Section 3. Random variables that have a density function are sometimes called continuous random variables.
We will use it nevertheless as the terminology is commonly accepted. In Example 1. Let [a, b] be a bounded interval on the real line. Recall again that by 3. Probability distributions of random variables Example 3. Find the probability that its absolute value is at least 1.
A density function f gives probabilities of sets by integration, as in 3. However, multiplied by the length of a tiny interval, it approximates the probability of the interval. Suppose that random variable X has density function continuous at the point a. See Figure 3. A limit like 3. The dotted curve is the probability density function 3. The approximation is P 0. You can also search for others PDF ebooks from publisher Cambridge University Press , as well as from your favorite authors.
We have thousands of online textbooks and course materials mostly in PDF that you can download immediately after purchase. This classroom-tested textbook is an introduction to probability theory, with the right balance between mathematical precision, probabilistic intuition, and concrete applications. Introduction to Probability covers the material precisely, while avoiding excessive technical details. After introducing the basic vocabulary of randomness, including events, probabilities, and random variables, the text offers the reader a first glimpse of the major theorems of the subject: the law of large numbers and the central limit theorem.
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