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# Probability All Notes for first exam MATH 3160

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Math 3160 – Textbook Notes A First Course in Probability Chapter 1 – Combinatorial Analysis Introduction o Many problems in probability theory can be solved simply by counting the number of different ways that a certain event can occur o Enumeration is basic laying out of arrangements o Mathematical theory of counting is known as Combinatorial Analysis The Basic Principle of Counting o One experiment can result in m number of outcomes o One experiment can result in n number of outcomes o Together there are mn possible outcomes of the two experiments Permutations o How many different ordered arrangement of the letters a, b, and c are possible o When there are n objects, the number of permutations or combinations is n! or n factorial or n(n-1)(n-2)…(3)(2)(1) o If there are repetitions of objects, or non distinct objects n! / r1!r2! where r1 and r2 are the number of repetitions of objects 1 and 2 Combinations o Combinations are the number of different groups of r objects that could be formed from n objects o ( ) r Self Test Problems: o 1. How many different linear arrangements are there of the letters A, B, C, D, E, F for which a. A and B are next to each other 5!2! = 240 b. A is before B 6! / 2 = 360 c. A is before B and B is before C 6! / 6 = 120 d. A is before B and C is before D 180 o 2. If 4 Americans, 3 French people, and 3 British people are to be seated in a row, how many seating arrangements are possible when people of the same nationality must sit next to each other? 4!3!3!3! = 5,184 o 3. A president, treasurer, and secretary, all different, are to be chosen from a club consisting of 10 people. How many different choices of officers are possible if a. there are no restrictions 10*9*8 = 720 b. A and B will not serve together 672 c. C and D will serve together or not at all 384 d. E must be an officer 216 o 4. A student is to answer 7 out of 10 questions in an examination. How many choices has she? How many if she must answer at least 3 of the first 5 questions? 120, 110 o 5. In how many ways can a man divide 7 gifts among his 3 children if the eldest is to receive 3 gifts and the other two receive 2 gifts. 210 o 6. How many different 7-place license plates are possible when 3 of the entries are letters and 4 are numbers? Assume that repetition of letters and number is allowed and that there is no restriction on where the numbers or letters can be placed. 35 * (26)^3 * (10)^4 o 7. Give a combinatorial explanation of n =( n ) r) n−r n! / r!(n-r)! = n! / r!(n-r)! o 8. Consider n-digit numbers where each digit is one of the 10 integers 0, 1, …9. How many numbers are there for which a. no two consecutive digits are equal 10 * 9^(n-1) b. 0 appears as a digit a total of I times, I = 0, …, n (n choose i) 9 ^ (n-i) o 9. Consider three classes, each consisting of n students. From this group of 3n students, a group of 3 students is to be chosen. a. How many choices are possible 3n choose 3 b. how many choices are there for which all 3 students are in the same class 3 times (n choose 3) o 10. How many 5-digit numbers can be formed from the integers 1, 2, … , 9 if no digit can appear more than once 9*8*7*6*5 o 11. From 10 married couples, we want to select a group of 6 people that is not allowed to contain a married couple. a. how many choices are there b. how many choices are there if it must be 3 women and 3 men o 12. A committee of 6 people is to be chosen from a group containing 7 men and 8 women. If the committee must consist of at least 3 women and 2 men, how many different combinations are possible. o 13. An art collection on auction consists of 4 Delis, 5 Van Gogh, 6 Picasso. At the auction were 5 art collectors. If the report only noted the number of D, VG, and P acquired by each collector, how many reports could there have been. o 16. How many subsets of size 4 of the set S = [1, 2…,20] contain at least one of the elements 1,2,3,4,5. o 18. In a certain community, there are 3 families consisting of a single parents and 1 child, 3 families consisting of a single parents and 2 children, 5 families consisting of 2 parents and a single child, 7 families consisting of 2 parents and 2 children, and 6 families consisting of 2 parents and 3 children. If a parent and child from the same family are to be chosen, how many possible choices are there. o 19. If there are no restrictions on where the digits and letters are placed, how many 8-place license plates consisting of 5 letters and 3 digits are possible if no repetition of letters or digits is allowed? What if the three digits must be consecutive? Chapter 2 – Axioms of Probability Introduction o Introduce probability as an event Sample Space and Events o Suppose that you know the set of all possible outcomes o This is referred to as the sample space o Denoted by S o S = {option 1, option 2, …} o Any subset E of the sample space is known as an event An event is a set consisting of possible outcomes of the experiment If the outcome of the experiment is contained in E, then E has occurred o U either o N intersection o C contained o Laws Commutative E U F = F U C EF = FE Associative (E U F) U G = E U (F U G) (EF)G = E(FG) Distributive (E U F)G = EG U FG EF U G = (E U G)(F U G) Axioms of Probability o Relative frequency – a way of defining probability of an event n(E) = the number of times in the first n repetitions of the experiment that E occurs o P(S) = 1 o P(E) = #E / #S o When the sample space is an uncountable infinite set, P(E) is defined only for a class of events called measurable Propositions o P(Ec) = 1 – P(E) o If E C F, then P(E) < P(F) o P(E U F) = P(E) + P(F) – P(EF) o Matching problem Sample Spaces having Equally Likely Outcomes o If equally likely then P({i}) = 1 /N where i=1,2,3,…N o Otherwise: P(E) = #E / #S Chapter 3 – Conditional Probability and Independence Introduction o Often interested in calculating probabilities when some partial information concerning the result of the experiment is available o Desired probabilities are conditional o Conditional probabilities can normally be used to compute desired probabilities easier Conditional Probabilities o P ( E | F) o Conditional probability that F occurs given that E occurs o P (E | F) = P(EF) / P(F) Baye’s Them o See class notes Indepdence o Two events are independent if o P(EF) = P(E)P(F) o Three events must be tested for all duplicates and the total multiplication of the three MATH 3160 - Video Lecture Notes 1a – Counting and Permutations o Why flipping classroom? Lots of material in a little amount of time Can pause, rewind, and more individualized to your pace Videos allows us to offload more passive uses of class time More time on problem solving o What is probability Mathematics of uncertainty Relatively young compared with other areas of math Allows you to quantify uncertainty and risk Laplace: “Common sense reduced to calculation…The most important questions of life are, for the most part, really only problems of probability” Applications include: Weather modeling Financial and actuarial analysis And games of chance o Counting: Rules of Sum and Product In order to determine probabilities, we need first to be able to count how many ways certain events can occur. Basic principles: Principle – Rule of Sum o If there are a ways that A can occur and b ways that B can occur, the the number of ways that A or B can occur is a + b o These have to be disjoint properties, different things with no overlap Principle – Rule of Product o If there are a ways that A can occur and b ways that B can occur, then the number of ways that A and B can occur is ab o Permutations: How many orderings are possible for the three letters A,B,C: There are six n! (n factorial) ways n(n-1)(n-2)(n-3)(n-4)…(3)(2)(1) = n! o Permutations with some elements indistinguishable Some elements that are being arranged are not different Ex. Innkeeper (n!)/(a!)(b!) where a and b are the number of elements that are the same ex. There are 2 n’s and 3 e’s so it would be 9!/(2! 3!) o Problem Solving Probability is all word solving Show answer as expression and numerical value Can combine rule of product and rule of sum 1b – Combinations and Binomial Coefficients o Combinations – unordered collections of objects ( ) = # r-element subsets of an n-element set r n! (n−r)r! n n ()=( ) same thing as r sub committee and r non sub committee r n−r n= n−1 +(n−1 ) (r (r−1 r this leads to the famous triangle of Khayyam- YangHui-Pascal o Pascal’s triangle each row n is sum 2^(n-1) o Binomial Theorem Proofs: Induction Combinatorial reasoning o › o Combinatorial Reasoning Summing across a row of Pascal’s triangle 1c – Multinomials & Compositions o Multinomial Coefficients Q: At insurance company, ten incoming actuaries are to be assigned to divisions as follows: 3 to auto, 1 to pensions, 2 to life, 4 to residential. How many ways can this be done? 10 7 6 4 = 10! = 12,600 (3)(1 2 4) 3!1!2!4! o auto, pension, life, residential Same as number of permutations of AAAPLLRRR Definition (Multinomial coefficient): n ( )≔ # of ways n distinct object can be divided up into r1,r2,r3,rl…. l (distinct) classes of size r1, r2, … , rl, where r1 + r2 + r3 +… + rl = n When l = 2, binomial coefficient Pascal’s triangle o Compositions weak allows any to be 0 strict means no zero a composition can be weak and strict When creating al the different compositions of a number the order DOES matter To find the number of ordered pairs it is all the weak L-compostion of n use the ball and bar example o draw n balls o and L-1 bars o if you put two bars together then there is a zero o could be to the left of the first ball, or to the right of the last ball o # of L-composition of n = # of ways to arrange n balls and L-1 bars o for strict: no bars at the ends or bars together 2a – The Probability Set up o Examples of : A finite sample space: A single coin flip S={H,T} #S = 2 A roll of a die: S={1,2,3,4,5,6} #S = 6 A countable infinite sample space Flipping a coin indefinitely many time S = {H,T}*{H,T}*… #S = infinite Random walk on the integer lattice An uncountable sample space Throwing a dart at a dartboard (while blindfolded) can be a infinite number of laces on the dartboard S = disc of some radius Time to next accident on I-84: S=[0, inifinity) o For countable (finite or infinite) sample spaces, it seems natural to assign probability to single points (outcomes) Example: roll of a die. Put P(1) = P(2) = … = P(6) = 1/6 One could assign unequal probabilities by rigging the die, say replacing the faces 5 and 6 with 1 Then P(1) = 1/2 , P(2)=P(3)=P(4) = 1/6 o But for uncountable sample spaces Throw a dart while blindfolded: natural to assume that the dart is equally likely to hit anywhere on the board. The probability of hitting some region of the board is proportional to the region’s area Probability of hitting single points on the board? Zero: points have zero area Add up all the probability of all points in S? Zero o A better notion: Events Instead of single points, consider subsets of S, also called events Example: roll of a die. The event {1} means “roll a 1” while {2,4,6} means roll an even number An event consisting of a single event is called a singleton What types of events should be included? Intersections (AND) & Unions (OR) events o Venn diagrams o Sigma Field: o The Kosher way to assign a probability These axioms were introduced by Kolmogorov Why isn’t the probability of the empty sets =0 part of the axioms? o Frequentist approach vs. axiomatic approach to probability You may be used to thinking about probability from the frequentist point of view Example why is the probability of heads ½? You flip a coin many many times and find out that approximately half the time it lands on heads While there is nothing wrong with this approach, it is HARD to use. The axiomatic approach is easier to use. In fact, one can recover the frequentist interpretation from the axioms Above example falls out as a consequence of the Law of Large Numbers o Consequences of the axioms 2b – Uniform discrete probability problems o When all outcomes are equally likely You have identified a sample space S with a finite number of #S = N of outcomes Here an outcome may be an “ordered arrangement” (permutation) or an “unordered arrangement” (combination. You need to recognize which one to use depending on the word problem. In any case, suppose that all outcomes are equally likely. Then is easy to see that P({W}) = 1/N for each W If any event E contains #E outcomes, then P(E) = #E/N = #E/#S o Example 1: Dice A pair of fair dice is rolled POV that you can distinguish die 1 and die 2 #S = 6*6 = 36 E1 = Probability of a 3 on the first die P(E1) = #E1 / #S = (1*6)/36 = 6/36 = 1/6 E2 = Probability of a sum of 7 E2 = {(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)} = #E2 = 6 P(E2) = #E2 / #S = 6/36 = 1/6 E3 = Probability of an even number on at least one of the dice Note: E3^c = event of rolling both odds P(E3^c) = #E3^c / #S = (3*3)/36 = 9/36 = ¼ P(E3) = 1 – P(E3^c) = 1 – ¼ = ¾ o Example 2: 5-card poker cards S = space of all possible 5-card poker hands (unordered) #S = (52 choose 5) Pick a combination of 5 cards from a well-shuffled standard deck of 52 cards. E1 = Probability of one pair (Two of the same rank, the rest are of different ranks) #E1 = 13 * (4 choose 2) * (12 choose 3) * 4 * 4 * 4 then divide the above by (52 choose 5) E2 = Probability of a full house (three of the same rank and two of another rank) #E2 = 13 * (4 choose 3) * 12 * (4 choose 2) the divide the above by (52 choose 5) 3a - Independence o Definition Two events E and F are said to be independent if and only if P (E N F) = P(E)P(F) Probability of the intersection = product of the probabilities of the two events If don’t meet this requirement then dependent o (In)dependence vs. disjointness A common confusion: Does independence imply disjointness or vice versa? Can be disjoint and independent Can be disjoint and dependent Can be not disjoint and independent Can be not disjoint and dependent NO RELATIONSHIP o Another fact about independence If E and F are independent, then E^c and F are independent So are E and F^c So are E^c and F^c o Independence of more than 2 events Just checking pairwise is not enough Ei and Ej, Ej and Ek, Ek and Ei, all three If pairwise is not independent then the total is not independent o Bernoulli trials Demand independence amongst certain individuals Include a combinatorial phrase Refers to a sequence of independent experiments with binary outcomes 3b - Conditional Probability o Definition of conditional probability Setting: Given that event F happens, what is the probability that another event E happens Idea: Introduce a different probability, tentatively called Q, such that Q(F) =1. Q should be proportional to P, in the sense that there is a constant C such that for all events E: Q(E) = C*P(E N F) Q(F) = 1 = C*P(F) C = 1 / P(F) The gist behind conditional probability is taking proportions Definition: The conditional Probability of an event E given an event F is dash (|) means given after bar is given o Remarks on the definition Can rewrite the definition into the form: P(E N F) = P(E | F)P(F) This form is handy for solving many world problems By symmetry: P(E N F) = P(F | E)P(E) By the axioms: P(E) = P( E N F) + P(E N F^c) Combine to get P(E) = P(E | F)P(F) + P(E | F^c)P(F^c) o A notational comment: Do not write E | F only P( E | F) o The Multiplication Rule Proposition: Let E1, E2, …, En be a sequence of events. Then o Revisiting independence Assuming that E and F are independent: If one of P(E) AND P(F) is 0, then P(E N F) = 0 If neither P(E) nor P(F) is 0, we may rewrite this definition using conditional probability as follows: o The second case has a nice interpretation: o 3c – Bayes Theorem o Test Positive False positives overwhelm the number of true positives since there are a very few amount of people with the disease Look at it in a table o Bayes’ Formula We can think of a “given event” as providing new information (test result) to update our estimate of the likelihood of another event (have disease). Although it’s often called “Bayes’ Theorem” it is a simple consequence of the basic probability axioms and definitions Recall by conditional probability: P( E N F) = P(E | F)P(F) By symmetry: P(E | F)P(F) = P(E N F) = P(F N E)=P(F | E)P(E) P(E | F) = Chapter 3 – Supplemental Material – Numberphile (Monty Hall Problem) o Choose Door 1, 2 , or 3 o There was a dream car behind one o There were “donks” behind two other doors o He would open one of the doors you didn’t pick Always the one that had a “donk” o Could stick or switch o Switch every time o 1/3 chance that it is behind the door you chose o 2/3 behind one of the other two but not behind the one shown therefore the one you would switch to would be a 2/3 probability o Imagine having 100 doors You pick one He shows you what’s behind 98 Then you have to choose to stay or switch b/c 99/100 concentrated behind the one you would switch too 4a – Random Variables o random variable some statistic of interest associated to the possible outcomes of an experiment for example X = sum when we roll two cubical dice Y = number of Heads when we flip a coin six times Z = rim it takes for a light bulb to burn out A random variable (r.v.) is a real-valued function on S, the set of possible outcomes A r.v. is discrete if it takes on only countably many values o Probability Mass Function For a discrete r.v. the probability mass (aka density) function (pmf/pdf) p(x) is given by p(x) = P(X-x):=P({w E S : X(w) = x}) o Cumulative distribution function: (cdf) F(x) of a r.v. X is given by F(x): = P{X < x}, for x E (-infinity, infinity) On distribution add the previous block to the current block 4b – Expectation o Expectation Recall that a random variable can be though of as a kind of probability histogram, that indicates how probable a certain result is X = sum when we roll two cubical dice Y = number of heads when we flip a coin six times A natural question to ask, is what should you expect on average your result to be. This is the intuition behind the notion of expectation. Definition ( Expectation): The expectation of a discrete random variable X is the average value of X, weighted by the probability of each outcome: o Linearity of Expectation A very useful property of expectation is that (unlike many other statistics) it is linear Proposition (Linearity of Expectation) o Proposition (Expectation of g(x)) Let X be random variables, g a real-valued function. Then 4c – Random Variables arising from Coin Toss o Tips for mastering random variables Remember the story Learn how to easily compute the expectation of each o Bernoulli X is a Bernoulli random variable with parameter p if it counts the number of heads shown up in a single coin toss with probability of heads being p P[X=1] = p, P[X=0] = 1-p o Binomial Y is a binomial random variable with parameters n and p if it counts the number of heads shown up in n coin tosses with probability of heads being p Y takes integer values k between 0 and n. From the independence section we already learned that Binomial theorem o Geometric T is a geometric variable with parameter p if it counts the number of trials needed to obtain he first heads, with probability of heads being p. It represents the “waiting time” for something to happen in a discrete-time setting Thus T can be any integer k between 1 and infinity P[T = k] = p(1-p)^(k-1) 4d – Hypergeomatric and Poisson Random Variables o Hypergeometric Example: Draw 4 lottery game 80 balls in an urn of which 10 are winning draw 4 w/o replacement let X denote the number of winning balls drawn What is P[X = k] MATH 3160 – Class 1 Notes January 20 , 2016 th Combinatorics: o The study of counting o Make counting efficient Permutations: o Ordered arrangements o Factorial (n!): for any positive number n, n! = n*(n-1)*(n-2)*…*3*2*1 o The number of ordered arrangements of n distinct objects o 0! = 1 number of ways you can order 0 objects, 1 way aka nothing Examples: o Chipotle – How many custom orders are available with one base and one protein Base: taco, burrito, salad, bowl Protein: beef, chicken, pork 1, pork 2, tofu 4 base 5 protein multiply the base by protein 4*5=20 basic principle of counting o License Plate numbers – How many combinations are available First three have letters Second three have numbers 26 * 26 * 26 * 10 * 10 * 10 o Variation on License Plate numbers – How many combinations are available First three have letters Second three have numbers No repeats in numbers or letters 26 * 25 * 24 * 10 * 9 * 8 o Seating Arrangements – How many seating arrangements are possible 4 people 4 chairs ordered permutation 4 * 3 * 2 * 1 = 4! = 4 factorial MATH 3160– Class 2 Notes nd January 22 , 2016 3 white balls, 2 green balls -> that are indistinguishable 5! o 3!2! 5 people in 2 groups -> group 1 (3 people), group 2 (2 people), Find number of group combinations 5! o 3!2! o 5 choose 3 -> number of ways to choose 3 out of 5 When choosing use binomial coefficient When arranging use factorials Don’t simplify on hw MATH 3160 – Class 3 Notes January 25 , 2016 th (n m) o Combinations o Read “n choose m” o Can also be written as: n C m o Order doesn’t matter o m is not distinguishable o Is equivalent to (n n-m) o Is equal to n! / (n-m)!m! o 0 < n < m Combinations from a committee point of view o Can pick who’s on committee and who’s not on committee o Therefore, probability of picking who’s on the committee is the same probability as picking who is not on it Binomial Theorem o (x+y)^n = (x+y) (x+y) (x+y) (x+y) ….. repeats n times o Before combining like terms, how many factors will there be o Each grouping has two choices (x and y) and each of these have n cycles therefore there are 2^n factors o Each term is of the form x^j * y^n-j How many ways can we get a term in the form above (n j) OR (n n-j), first is picking x’s, second is picking y’s n n n n n j n−j Therefore ∑ (j=2 and ∑ (j x y j=0 j=0 Probability of an event o Take the probability of an event over the probability of the other occurrence o Add up to 1 o Uses big P MATH 3160 – Class 4 Notes th January 27 , 2016 Multinomial o More than 2 groups o The number of ways to sort n things into k groups o N choose r1, r2, r3, r4, … , rk o Where n = r1 + r2 +r3 + r4 +…+rk Therefore, no thing is left out of a group o N! / R1! R2! R3! Rk! Composition o EX. How many terms can you generate with (x+y+z)^4 o Use box and balls situation – how many ways can you separate the balls into boxes o N items getting split into k groups k-composition of n o No constraint means that you can have some groups with 0 items Ball and bar argument K-1 bars and n balls n + k – 1 choose k-1 o Constraints where every group must have at least one Ball and bar argument 2 bars can’t be next to each other and can’t be on outside have to a bar in between balls n-1 choose k-1 MATH 3160 – Class 5 Notes th January 29 , 2016 Axioms o Are built on sets o Ground rules for defining a lot of things Axioms of probability o S = sample space = space of outcomes o Sigma field = Collection of subsets Contains all possible sets that could happen Includes empty sets, sets that everything happens, complements, unions, etc. Probability -> P: sigma[0,1] o Satisfies the two rules: 1. P(S) = 1 2. For any pairwise disjoint subsets, all disjoint with no subsets, countable addivity Set Theory o The number of points in a set E is denoted #E o The complement of a set E, denoted by E^c, is the set of points which do not lie in E o The union of two sets E and F, denoted by E U F, is the set of points which lie in either E or F (can be both) o The intersection of two sets E and F, denoted by E n F, is the set of points which lie in BOTH E AND F DeMorgan’s Laws o MATH 3160 – Class 6 Notes st February 1 , 2016 Always keep in mind when items are using combination or permutations Be consistent To use combinations, things must be similar Remember the card trick MATH 3160 – Class 7 Notes rd February 3 , 2016 The Matching Problem: o Sample space is S = matching people with hats o #S = 5! o Same as the taylor expansion of e^x where x = -1 -1/2! +1/3! -1/4! +1/5! MATH 3160 – Class 8 Notes February 5 , 2016h No notes – snow day MATH 3160 – Class 9 Notes th February 8 , 2016 No notes – snow day MATH 3160 – Class 11 Notes February 12 , 2016th Bayes’s Theorem o Tree diagram o P(E | F) = P(E N F) / P(F) o P(E N F) = P(F | E)P(E) o P(E | F) = P(F | E)P(E) / P(F) o P(F) = P(E N F) +P(E^c N F) o P(E | F) = P(F | E)P(E) / P(E N F) + P(E^c N F) o P(E N F) = P(F | E)P(E) o P(E^c N F) = P(F| E^c)P(E^c) o P(E | F) = P(F | E)P(E) / P(F | E)P(E) + P(F| E^c)P(E^c) o Final Formula: MATH 3160 – Class 12 Notes th February 15 , 2016 Exam review information o Monday, February 22, 5PM – 7PM o Tuesday, February 23, office hours by email appointment Exam Rules o No calculator o Covers material on chapters 1 – 3 o One-page double sided of formulas Monty Problem o Use conditional probability to solve o Use tree diagrams to solve MATH 3160 – Class 13 Notes th February 17 , 2016 Random Variables o Not random or a variable o Is a function o X: S -> R o Function from the sample space with an output of what you want to know and only real numbers Example: roll of a die -> identity function o 1 – 1 o 2 - 2 o 3 - 3 o 4 - 4 o 5 - 5 o 6 – 6 Example: roll of a die assigning 0 to even and 1 to odd o 1 - 1 o 2 - 0 o 3 - 1 o 4 - 0 o 5 - 1 o 6 – 0 {X=k} : event P(X=k) o =set of outcomes o in a sample space o ={w E S: X(w)=k} o w = omega Discrete random variables: the values of the function are countable, countable many values, both finite and countable infinite Probability Mass Function (pmf) Cumulative Density Function (CDF) o Mass is in between each jump o Resembles a step function o Right continuous MATH 3160 – Class 15 Notes February 22 , 2016 nd Exam Stuff o Covers chapters 1 – 3 o Go over homework problems and quizzes o Topics: conditional probability, Bayes’s Formula, independence, counting, permutations, combinations, Monty hall problem o Cheat Sheet can have anything on it Expectation: E[X] is essentially the mean of variable X Definition 1: o E[X] = k∗P[X−k] ∑ Note 1: o X and Y are random variables o E[X + Y] = E[X] + E[Y] o ****The expectation of the sum is the sum of the expectations*** Definition 2: o E[X] = ∑ X[w]P[{w}] Definition 2 = Note 1 o E[X + Y] = ∑ [X[ ]+Y[w]]P(w) Note 2: o E[a*X]=a*E[X] Expectation is a linear operation o Proven by Notes 1 & 2 o Integrals are also linear operations Standard Deviation is the spread Variance is the square of standard deviation E[X^p] =pth moment of X o When p=2: second moment of X o When p=3: third moment of X o When p=1: mean o When p=0: 1 VarX = E[X^2] – [E[X]]^2 o Variance is quadratic o Same under adding by constant Var(a*X) = (a^2)*Var(X) MATH 3160 – Class 17 Notes th February 29 , 2016 Geometric o Discrete waiting time o p models the number of trials needed to get the first heads o Ex. Coin toss: p=probability of heads o So if X~Geom(p) P[X=k] = p(1-p)^(k-1) TTTTT….TH o E[X] = 1/p Negative Binomial o Y~NB(r,p) if Y models the number of trials needed to get the rth heads o By the time the k-1 trial happens there must be r-1 heads o P[Y=k] = (k-1 choose r-1)(p^r)(1-p)^(k-r) o E[Y] = r/p Hypergeometric (urn games) o N balls o N-m not winning o M winning o Choose n balls o X is the number of winning balls drawn out of n balls o P[X=k] = (m choose k)(N-m choose n-k) / (N choose n) MATH 3160 – Class 18 Notes nd March 2 , 2016 MATH 3160 – Exam 1 Review Sheet February 24 , 2016 th Review Items: Quiz 1 (Chapter 1) o Number of ways to seat 9 people in a row is 9! o Number of ways to seat 9 people with James and Susan next to each other is 8! * 2! o So where they don’t sit next to each other -> 9! – 8! * 2! o 9 choose 4 is number of ways to choose a committee of 4 people out of 9 o if James and Susan must serve together then 7 choose 2 o so where they don’t serve together -> 9C4 – 7C2 Quiz 2 (Chapter 2) o #Ways to get 4 of a kind in 6 card poker hand 13 * 1 * 12C2 * 4 * 4 13 for choosing the rank of the quad 1 for assigning all 4 suits to the quad 12C2 for assigning the ranks to the two loners 4*4 for assigning suits to the two loners o Divide #E / #S o #(A U B U C) = #A + #B + #C - #(A N B) - #(B N C) - #(C N A) + #(A N B N C) Quiz 3 (Chapter 2 and 3) o #Ways to generate one pair from 8 pairs 8 * 7C4 * 2^4 8 for choosing one pair to keep whole 7C4 to choose 4 other pairs 2^4 to choose either left or right o Divide #E / #S o Probability that at least 2 heads appears 1 – probability of no heads – probability of 1 heads no heads -> (2/3)^7 one heads -> 7 * (1/3) * (2/3)^6 7 to choose which position the tails appears Quiz 4 (Chapter 3) o Use Bayes’s Formula Chapter Materials: Chapter 1 – Combinatorial Analysis o Permutations: ordered n! where r1, r2, r3, …n are the number of each similar r1!r2!r3!…n! item in n n! = (n)(n-1)(n-2)…(3)(2)(1) o Combinations: unordered n (r) where n is total and r is the number of n you want to choose Think of forming committees o Binomial Theorem: (x+y)^n = use the coefficients from the nth row of pascals triangle and y^n going down and y^0 going up o Compositions Ball and bar Weak – allows any to be 0 Strict – allows none to be 0 Chapter 2 – Axioms of Probability o S = sample space or the set of all possible events o P(E) = ¿E ¿S o P(S) = 1 o P(E ) = 1 – P(E) o P(E U F) = P(E) + P(F) – P(EF) Inclusion / Exclusion test o U = lie in either of the sets or both o N = lie in both of the sets Chapter 3 – Independence o Independent: P(E N F) = P(E)P(F) Disjoint and independent are not the same o Conditional Probability P(E | F) Probability of E given F P(E N F) = P( ) o Use tree diagrams o Bayes’s Theorem P(E | F) = o Monty Hall Problem Problems: Chapter 1 o 1P o 21P o 5ST o 5P o 25P o 9ST o 9P o 29P o 13ST o 13P o 33P o 17ST o 17P o 1ST o 20ST Chapter 2: o 1P o 25P o 49P o 5P o 29P o 53P o 9P o 33P o 1ST o 13P o 37P o 5ST o 17P o 41P o 9ST o 21P o 45P o 13ST o 17ST Chapter 3 o 5P o 15P o 25P o 7P o 17P o 27P o 9P o 19P o 29P o 11P o 21P o 31P o 13P o 23P Other: o Take Fall ’14 Exam o Take Spring ’15 Exam

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