Introduction to Probability and Counting

Transcription

1 Introduction to Probability and Counting SJSU August 15, 2016

2 Probability- Chapter 3 Population and Sample Statistics: a branch of mathematics dealing with the collection, analysis, interpretation, and presentation of masses of numerical data Probability: the relative frequency with which an event occurs or is likely to occur Probability allows us to assess the uncertainty in experimental results or random events Population - overall group of objects about which conclusions are to be drawn Sample - a subset of the population that is actually obtained Example: population - every student at SJSU, sample - a subset of 100 students 2/1

3 Probability- Chapter 3 Probability- 3.2 Probabilities are numbers between 0 and 1 which is the chance of a physical event occurring Probability is denoted as P(A) where A is an event Percentages are probabilities multiplied by 100 Example: The probability that a fair coin will yield heads is 0.50 or there is a 50% chance of heads. 3/1

4 Sample Spaces and Events Sample Spaces and Points sample space: a sample space for an experiment is a set S with the property that each physical outcome of the experiment corresponds to exactly one element of S. sample point: an element of S event: any subset A of a sample space is called an event. The empty set is called the impossible event (null or empty set); the subset S is called the certain event. Example: During a space flight there are three computers (two for backup) that operate independently. We want to discuss the operability of the computer at launch time. Computer 1: yes or no, computer 2: yes or no, computer 3: yes or no. Sample space: S = {yyy,yyn,yny,yyn,nny,nyn,ynn,nnn}, yny is a sample point. Event A: computer 1 s yes A = {yyy,yyn,yny,ynn}, Event B: computer 2 s yes B = {yyy,yyn,nyn,nyy}, Event C: computer 3 s yes C = {yyy,nyy,nny,yny}. 4/1

5 Sample Spaces and Events Union and Intersection and Complement- 3.2 and 3.4 union: A B (A or B) intersection: A B (A and B) compliment: C (not) 5/1

6 Sample Spaces and Events Union and Intersection and Complement Example: Event A: computer 1 s yes A = {yyy,yyn,yny,ynn} Event B: computer 2 s yes B = {yyy,yyn,nyn,nyy} Event C: computer 3 s yes C = {yyy,nyy,nny,yny} A B computer 1 or 2 are operable A B computer 1 and 2 are operable (A B) C computer 1 or 2 are operable but computer 3 is inoperable 6/1

7 Sample Spaces and Events Mutually Exclusive mutually exclusive (disjoint): two events A 1 and A 2 are mutually exclusive if and only if A 1 A 2 = or P(A 1 A 2 ) = 0. Events A 1,A 2,A 3,... are mutually exclusive if and only if A i A j = for i j. Example: Mutually exclusive is defined for events (which have probabilities associated with them) not for components. The computers cannot be defined as mutually exclusive. Event A (computer 1 is operable) can be mutually exclusive of the event A (computer 1 is inoperable). Both A and A cannot happen at once. A={yyy,yyn,yny,ynn} and A ={nnn,nyn,nny,nyy} are mutually exclusive, they do not share sample points. A A =. Event A: computer 1 s yes A = {yyy,yyn,yny,ynn} Event B: computer 2 s yes B = {yyy,yyn,nyn,nyy} Question: Are events A and B mutually exclusive? 7/1

8 Sample Spaces and Events Trees Example: Computer 1 is inoperable 51% of the time, computer 2 is inoperable 30% and computer 3 is inoperable 10%. What is the probability that all three computers are operable? What is the probability that computer 1 is operable but computers 2 and 3 are inoperable? 8/1

9 Permutations and Combinations Multiplication Principle Multiplication Principle: Suppose a procedure can be broken into m successive (ordered stages, with r 1 outcomes in the first stage, r 2 in the second stage,..., and r m in the m th stage. If the number of outcomes at each stage is independent of the choices in previous stages and if the composite outcomes are all distinct, then the total procedure has r 1 r 2... r m different composite outcomes. 1. DNA is formed from four nucleotides (A,T,G,C). How many different chains of 6 nucleotides is possible? A-T-T-G-C-A, A-T-T-G-C-C How many three letter words can be created (26 letters in this alphabet)? 9/1

10 Permutations and Combinations Factorial! Factorial n! n is a positive integer n! = n(n 1)(n 2) (n factorial) By convention the empty product: 0! = 1 Examples: 7! = = ! 5! = = 7 6 = 42 10/1

11 Permutations and Combinations Permutation Permutation - is an arrangement of objects in a definite order: np r = n! (n r)! ex. twenty amino acids commonly form peptides. Five amino acids A-V-G-C-T ordering gives a different pentapeptide than A-G-V-C-T. Permutation - is an arrangement of objects in a definite order Given 5 amino acids how many pentapeptides can be made? There is no repetition of amino acids in this pentapeptide ! = 120 Extra: Given 20 amino acids how many pentapeptides can be made? 20*19*18*17*16= 1,860,480 11/1

12 Permutations and Combinations Combination Combination - is a selection of objects without regard to order: n C r ex. There are 20 components, 3 are selected at random to be tested for quality. We only care about which three are selected and not their order. If components (1,6,10) are selected, it is the same as components (6,1,10) being selected. 20C 3 = ( ) /1

13 Permutations and Combinations Multiplication Principle 1. Two dice are rolled, one green and one red (a) How many different outcomes of this procedure are there? (b) What is the probability that there are different values on the two dice (no doubles)? 13/1

14 Permutations and Combinations Permutation How many ways can we arrange the seven letters in the word SYSTEMS? 14/1

15 Permutations and Combinations Combination A committee of k people is to be chosen from a set of 7 women and 4 men (a) The committee has 5 people: 3 women and 2 men (b) The committee can be any positive size but must have equal men and women (c) The committee has 4 people and one of them must be Mr. Baggins (d) The committee has 4 people and at least 2 are women 15/1

16 Permutations and Combinations Overview sample space: a sample space for an experiment is a set S with the property that each physical outcome of the experiment corresponds to exactly one element of S. sample point: an element of S event: any subset A of a sample space is called an event. The empty set is called the impossible event (null or empty set); the subset S is called the certain event. mutually exclusive (disjoint): two events A 1 and A 2 are mutually exclusive if and only if A 1 A 2 = or P(A 1 A 2 ) = 0. Events A 1,A 2,A 3,... are mutually exclusive if and only if A i A j = for i j. union: A B (A or B) intersection: A B (A and B) compliment: C (not) 16/1

17 Axioms of Probability Axioms of Probability Let S denote a sample space for an experiment: P(S)=1 Let P(A) 0 for every event A Let A 1,A 2,A 3,... be a finite or an infinite collection of mutually exclusive events. Then P(A 1 A 2 A 3...) = P(A 1 )+P(A 2 )+P(A 3 )+... Example: There are 4 blood types (A,B,AB,O) with percentages (41%,9%,4%,46%). What is the probability that a person will have type A, B, or AB blood? 17/1

18 Axioms of Probability Theorems P( ) = 0 the probability assigned to the empty set is zero. P(A ) = 1 P(A) the probability an event will not occur A is equal to one minus that it will occur Addition rule: P(A B) = P(A)+P(B) P(A B) if events A and B are mutually exclusive then: P(A B) = 0 Example: Lead and mercury can be found in water; near industrial plants 32% of streams have toxic levels of lead and 16% have toxic levels of mercury. 38% of streams have toxic levels of lead or mercury. What is the probability that a randomly selected sample will contain toxic levels of lead only? 18/1

19 Conditional Probability Conditional Probability Let A and B be events such that P(A) 0. The conditional probability of B given A, denoted by P(B A), is defined by: P(B A) = P(A B) P(A) Example: 49% of all infections involve anaerobic bacteria. 70% of all anaerobic infections are polymicrobic. What is the probability that a given infection involves anaerobic bacteria and is polymicrobic? Multiplication rule: P(A B) = P(B A)P(A) Example: Suppose we toss a die once. Find the probability of a 1, given that an odd number was obtained. 19/1

20 Independence Independence Independent - informally - events/objects acting irrespective of each other. Independent events: events A and B are independent if and only if P(A B) = P(A)P(B) Independent events: events A,B,C... are independent if and only if P(A B C...) = P(A)P(B)P(C)... Let A and B be events such that at least one of P(A) or P(B) is non-zero. A and B are independent if and only if P(B A) = P(B) if P(A) 0 and P(A B) = P(A) if P(B) 0 Example: Draw a card from a deck. Event A: spade is drawn and Event B: a 10,J,Q,K,A is drawn. (If I knew the card was a spade (event A) would that give me any information about event B?) and vice-versa Example: Two children have 3/4 chance of having brown eyes. Are these events independent? Yes 20/1

21 Bayes Rule Bayes Theorem Let A 1,A 2,A 3,...,A n be a collection of mutually exclusive events whole union is S. Let B be an event such that P(B) 0. Then for any of the events A j, j = 1,2,3,...,n, P(A j B) = P(B A j )P(A j ) n i=1 P(B A i)p(a i ) We usually use the simpler case when there are only two events A and A : P(A B) = P(B A) P(B) = P(B A)P(A) P(B A)P(A)+P(B A )P(A ) 21/1

22 Bayes Rule Bayes Theorem Example: 40% of highway accidents involve speeding and 30% involve alcohol. If alcohol is involved there is a 60% chance that there is also speeding, otherwise this probability is only 10%. An accident involves speeding, what is the probability that alcohol is involved? 22/1

23 Bayes Rule Bayes Theorem Example: The original blood test for HIV was called ELISA; according to a study (Weiss et al. 1985) the conditional probability that a person would test positive given they have HIV was and the conditional probability that a person would test positive given they did not have HIV was The World Almanac at the time cited that the probability of having HIV in North America was Suppose a random person is tested and they test positive; what is the conditional probability they have HIV? 23/1

24 Bayes Rule Bayes Theorem Do the results surprise you? 3.3% chance of having HIV given a positive test result. Let s go a bit further: Out of people we would expect 26 to have HIV. Of the 26 people with HIV, 26*.977 =25.4 would get a positive test Of those 9974 without HIV, 9974*.074= 738 would get a positive test Note that this phenomenon of large unexpected changes in conditional probabilities is not unusual when dealing with rare events/diseases. What happens is the number of false positives is much larger than the number of true positives. Public policy: how does this effect mandatory testing of the population? How about just testing at risk groups? 24/1

25 Extra Problems Multiplication Principle How many ways are there to form a three letter sequence using the letters a, b,c,d,e,f: (a) with repetition of letters allowed? Answer: we have six choices for each letter in the sequence, there are = 216 three-letter sequences (b) without repetition of any letter? Answer: six choices for the first letter, five choices for the second letter, four choices for the third letter: = 120 three-letter sequences. (c) without repetition and containing the letter e? Answer: e... e... e In each case, there are 5 choices for which of the other 5 letters (excluding e) goes in the first remaining position and 4 choices for which of the remaining 4 letters goes in the other positions. Thus there are = 60 three-letter sequences containing an e. (d) with repetition and contain the e? Answer: All possible ways (6 3 = 216) minus and sequences with no e s (5 3 = 125), so =91 25/1

26 Extra Problems Multiplication rule Assume there is a 50% chance of hard drive damage if a powerline is hit in an electric storm. There is a 5% chance that an electrical storm will occur. If there is a.1% chance that the line will be hit during a storm what is the probability that the line will be hit and there will be hard drive damage during the next storm? Answer: H: power line hit D: hard drive damaged P(H D) = p(d H)p(H) = = /1

27 Extra Problems Combinations (a) How many different 8-digit binary sequences are there with six 1s and two 0s? Answer: the position of the 1s and 0s matters. There are 8 places the 1s can go. The zeros fill in the remaining 2 spots: 8 C 6 = 28 (b) What is the probability of getting from a random generator an 8-digit binary sequence with six 1s and two 0s? Answer: 8C = 28/256 = /1

28 Extra Problems Combinations Questions: (a) How many 5-card hands (subsets can be formed from a standard 52-card deck? Answer: 52 C 5 = 52! 5!47! = 2,598,960 different 5 card hands (b) If a 5-card hand is chosen at random, what is the probability of obtaining a flush (all 5 cards in the hand are in the same suit- 13 cards per suit) Answer: There are 4 suits, and a subset of 5 cards from the 13 cards in a suit can be chosen b 13 C 5 = 13! 5!8! = 1287 ways. So there are = 5148 flushes or in probability terms = (c) What is the probability of obtaining exactly 3 aces? Answer: First how many ways are there to choose 3 of the 4 aces? 4C 3 = 4. Then fill out the hand with the other 2 cards from the 48 that are not aces ( 48 C = 1128) ways. There are = 4152 hands with 4152 exactly 3 aces. In probability terms that is = /1

29 Extra Problems Bayes Theorem It is reported that 50% of all computer chips are defective. Inspection ensures that only 5% of chips legally marketed are defective. Unfortunately, some chips are stolen before inspection. If 1% of all chips are stolen, find the probability that a given chip is stolen given that it is defective. Answer: D: chip is defective T: chip is stolen P(D) =.50 and P(T) =.01 P(T D) = P(D/T)P(T) P(D T)P(T)+P(D T )P(T ) = = /1

30 Extra Problems Bayes Theorem Type A: 41%, Type B 9%, Type AB 4%, Type O 46%. During WWII 4% of inductees with type O blood were typed as having type A, 88% of those with type A were correctly typed, 4% with type B were typed as A and 10% with type AB were typed as A. A soldier was tested and typed as A, what is the probability that his true blood type is A? Answer: P(A1 E)? A 1 =he has type A, A 2 =he has type B, A 3 =he has type AB, A 4 =he has type O and E=he is typed as type A. P(A1)=.41, P(A2)=.09, P(A3)=.04, P(A4)=.46 P(E A1) =.88 P(E A2) =.04 P(E A3) =.10 P(E A4) = P(A1 E) = = If a person was typed with type A blood, there was a 93.3% chance that his true type was in fact type A. 30/1