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Probability & Statistical Inference Lecture 3 MSc in Computing (Data Analytics)

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1 Probability & Statistical Inference Lecture 3 MSc in Computing (Data Analytics)

2 Lecture Outline  Introduction  Introduction to Probability Theory  Discrete Probability Distributions  There is more than one way to skin a cat!

3 Introduction

4 Probability & Statistics  We want to make decisions based on evidence from a sample i.e. extrapolate from sample evidence to a general population  To make such decisions we need to be able to quantify our (un)certainty about how good or bad our sample information is. Population Representative Sample Sample Statistic Describe Make Inference

5 Example: How many voters will give F.F. a first preference in the next general election ? -researcher A takes a sample of size 10 and find 4 people who say they will -researcher B takes a sample of size 100 and find 25 people who say they will Researcher A => 40% Researcher B => 25% Who would you believe? Probability & Statistics - Example

6 Intuitively the bigger sample would get more credence but how much better is it, and are either of the samples any good?  probability helps Descriptive Statistics are helpful but still lead to decision making by 'intuition‘ Probability helps to quantify (un)certainty which is a more powerful aid to the decision maker Probability & Statistics - Example

7 Probability & Statistics  Using probability theory we can measure the amount of uncertainty/certainty in our statistics.

8 Intuitions and Probability – Lotto example  If you had an Irish lotto ticket which of these sets of numbers is more likely to win: Odds of winning are 1 in Odds of winning are 1 in

9 Intuitions and Probability – Disease example  Suppose we have a diagnostic test for a disease which is 99% accurate.  A person is picked at random and tested for the disease  The test gives a positive result. What is the probability that the person actually has the disease?  99% ?

10 Disease example Test Results Those that have the don’t have/do have the disease If you take a population of 1,000,000 1,000,000999,900989,9919, No!! IT depends on how common or rare the disease is. Suppose the disease affects 1 person in 10,000 Of those who test positive only have the disease

11 Introduction to Probability Theory

12 Some Definitions  An experiment is the process of obtaining or taking a measurement.  A simple event is a basic outcome of an experiment; it cannot be decomposed into simpler outcomes.  A sample space of an experiment is the collection of all its simple events.  Example:  Experiment: Toss two coins and observe the up face on each  Sample Space: 1. Observe HH 2. Observe HT 3. Observe TH 4. Observe TT S: {HH,HT,TH,TT}

13 Probability  The probability of a simple event is a number that measures the likelihood that the event will occur when the experiment is preformed, denoted by P(E).  Some rules for probabilities:  Let E 1, E 2, E 3, ,E k 1. All simple event probabilities must lie between 0 and 1: 0 <= P(E i ) <= 1 for i=1,2, ,k 2. The sum of the probabilities of all the simple events within a sample space must be equal to 1:  Example:  Experiment: Toss two coins and observe the up face on each  Probability of each event: 1. E = HH => P(HH) = 1/4 2. E = HT => P(HT) = 1/4 3. E = TH => P(TH) = 1/4 4. E = TT => P(HH) = 1/4

14 Probability  The probability of an event A is equal to the sum of all the probabilities in event A:  Example:  Experiment: Toss two coins and observe the up face on each  Event A: {Observe exactly one head} P(A) = P(HT) + P(TH) = ¼ + ¼ = ½  Event B : {Observe at least one head} P(B) = P(HH) + P(HT) + P(TH) = ¼ + ¼ + ¼ = ¾

15 Complementary Event  The complementary of an event A is the event that A does not occur denoted by A´  Note that AU A` = S, the sample space  P(A) + P(A`) =1 => P(A) = 1 – P(A`)

16 Questions 1. What is the sample space when a coin is tossed 3 times? 2. What is the sample space for the number of aces in a hand of 13 playing cards? 3. If a fair die is thrown what is the probability of throwing a prime number? 4. If two fair die are thrown what is the probability that at least on score is a prime number? 1. What is the compliment of the event. 2. `What is its probability?

17 Questions 4. A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space For where (S,P) denotes that the first assembly line is shut down and the second one is operating at partial capacity. What is the probability that: a) Both assembly lines are shut down? b) Neither assembly lines are shut down c) At least one assembly line is on full capacity d) Exactly one assembly line is at full capacity Event AP(A)Event AP(A)Event AP(A) (S,S)0.02(S,P)0.06(S,F)0.05 (P,S)0.07(P,P)0.14(P,F)0.2 (F,S)0.06(F,P)0.21(F,F)0.19

18 Compound Events  The union of two event A and B is the event that occurs if either A or B, or both, occur on a single performance of the experiment denoted by A U B (A or B)  The intersection of two events A and B is the event that occurs if both A and B occur on a single performance of an experiment denoted by A B or (A and B)  Example: Consider a die tossing experiment with equally likely simple events {1,2,3,4,5,6}. Define the events A, B and C.  A:{Toss an even number} = {2,4,6}  B:{Toss a less than or equal to 3} = {1,2,3}  C:{Toss a number greater than 1} = {2,3,4,5,6}  Find:

19 Conditional Probability  The conditional probability of event A conditional on event B is for P(B)>0. It measures the probability that event A occurs when it is known that event B occur.  Example: A = odd result on die = {1,3,5}  B = result > 3 = {4,5,6}

20 Conditional Probability Example  Example: A study was carried out to investigate the link between people’s lifestyles and cancer. One of the areas looked at was the link between lung cancer and smoking. 10,000 people over the age of 55 were studied over a 10 year period. In that time 277 developed lung cancer. What is the likelihood of somebody developing lung cancer given that they smoke? CancerNo CancerTotal Smoker2413,325 3,566 Non-Smoker366,398 6,434 Total2779,72310,000

21 Conditional Probability Example  Event A: A person develops lung cancer Event B: A person is a smoker P(A) = 277/10,000 = P(B) = 3,566/10,000 = 0.356

22 Exercises 1. A ball is chosen at random from a bag containing 150 balls that are either red or blue and either dull or shinny. There are 36 red, shiny balls and 54 blue balls. 1. What is the probability of a chosen ball being shiny conditional on it being red? 2. What is the probability of a chosen ball being dull conditional on it being red?

23 Mutually Exclusive Events  Two events, A and B, are mutually exclusive given that if A happens then B can’t also happen.  Example: Roll of a die A = less than 2 B = even result There is no way that A and B can happen at the same time therefore they are mutually exclusive events

24 Rules for Unions  Additive Rule:  Additive Rule for Mutually Exclusive Events

25 Example  Recodes at an industrial plant show that 12% 0f all injured workers are admitted to hospital for treatment, 16% are back on the job the next day, and 2% are both admitted to a hospital for treatment and back to work the next day. If a worker is injured that is the probability that the worker will be either admitted to hospital or back on the job the next day or both?

26 Independent Events  Events A and B are independent if it is the case that A happening does not alter the probability that B happens.  Example :A = even result on die B = result > 2  Then, let us say we are told the result on the die (which someone has observed but not us) is even so knowing this, what is the probability that the event B has happened? Sample space: {2, 4, 6} B = 4 or 6 => P(B) = 2/3

27 Independent Events  But if we didn’t know about the even result we would get: Sample space: {1, 2, 3, 4, 5, 6} B = 3 or 4 or 5 or 6 => P(B) = 4/6 = 2/3 so knowledge about event A has in no way changed out probability assessment concerning event B

28 Rules for Intersection  Multiplicative Rule of Probability  If events A and B are independent then

29 Bayes Theorem  One of a number of very useful results: - here is simplest definition:  Suppose: You have two events which are ME and exhaustive – i.e. account for all the sample space –  Call these events A and event (read ‘not A’).  Further suppose there is another event B, such that  P(B|A) > 0 and P(A|B) > 0.  Then Bayes theorem states:

30 Discrete Probability Distributions

31 Discrete Random Variable  A Random Variable (RV) is obtained by assigning a numerical value to each outcome of a particular experiment.  Probability Distribution: A table or formula that specifies the probability of each possible value for the Discrete Random Variable (DRV)  DRV: a RV that takes a whole number value only

32 Example: What is the probability distribution for the experiment to assess the no of tails from tossing 2 coins;  Sample Space Coin 1Coin 2 TT TH HT HH x = no. of tails is the RV x P(x) 0= P(HH) = = P(TH) + P(HT) = = P(TT)= 0.25 P( any other value )= 0 N.B.  P(x) = 1 0  P(x)  1 for all values of x

33 Mean of a Discrete Random Variable  Mean of a DRV =  = Σ x * p(x)  Example: Throw a fair die x P(x)x * P(x) P(any other value) = 0 0 Mean =  = Σ x * p(x) = 3.5

34 Simulated Sample size = mean = 3.7 S.D. = 2.1 Simulated Sample size = mean = 3.54 S.D. = 1.67 Some Examples

35 Simulated Sample size = Mean = 3.49 S.D. = 1.73

36 Note:  The largest simulation had the mean closest to that predicted by the probability distribution  As the simulations got bigger the mean approached 3.5  Mean of DRV is the mean of a large number of independent experiments (trials).

37 Standard Deviation of a DRV

38 x P(x)x 2 * P(x) P(any other value) = 0 0 = (3.5) 2 = = 2.92 => S.D. = 1.71 Simulations: N=10 => S.D. = 2.1 N=100 => S.D. = 1.67 N=1,000 => S.D. = 1.73 Example: Rolling one die

39 Binomial (Probability) Distribution  Many experiments lead to dichotomous responses (i.e. either success/failure, yes/no etc.)  Often a number of independent trials make up the experiment  Example: number of people in a survey who agree with a particular statement? Survey 100 people => 100 independent trials of Yes/No The random variable of interest is the no. of successes (however defined)  These are Binomial Random Variables

40 4 people tested for the presence of a particular gene. success = presence of gene P(gene present / success) = 0.55 P(gene absent / failure) = 0.45 P(3 randomly tested people from 4 have gene)? Assume trials are independent - e.g. the people are not related There is 4 ways of getting 3 successes Binomial Distribution Example

41  Using Independence rule we can calculate the probability of each outcome:  Outcome 1: 0.55  0.55  0.55  0.45 = Outcome 2: 0.55  0.55  0.45  0.55 = Outcome 3: 0.55  0.45  0.55  0.55 = Outcome 4: 0.45  0.55  0.55  0.55 = ways of getting result each with P= => 4  = => P(3 randomly tested people have gene) =

42 Binomial Distribution Example  A more convenient way of mathematically writing the same result is as follows:  the number of ways you can get three successes from 4 trials can be written mathematically as 4 C 3, this is known as a combination:

43 Binomial Distribution – General Formula  This all leads to a very general rule for calculating binomial probabilities: In General Binomial (n,p) n = no. of trials p = probability of a success x = RV (no. of successes)  Where P(X=x) is read as the probability of seeing x successes.

44

45 Binomial Distribution  For all binomials the mean is given by the simple formula;  = n  p  Example: from previous example  = 4  0.55 = 2.2  Standard deviation also has simple formula for all Binomials  Example: from previous example = 0.995

46 Binomial Distribution  What is P(< 3 people have gene) from a group of four people tested at random?  Use the fact that the possible outcome are mutually exclusive (ME) = P(0) + P(1) + P(2) = = [ to 3 decimal places ]  We can write this probability like this;  P(X>3)=?

47 Binomial Question  There are two hospitals in a town. In Hospital A, 10 babies are born each day, in Hospital B there are 30 babies born each day. If the hospitals only count those days on which over 70% of babies born are girls, and assuming the probability that a girl is born is ½, which of the two hospitals will record more such days?  Hospital A: Binomial (n=10, p=0.5)  Hospital B: Binomial (n=30, p=0.5)

48 Answer  Hospital 1:  Calculate :  Hospital 2 :  Calculate : There is a higher probability of getting 70% of babies born being girl from hospital 1.

49 Binomial Question  A flu virus hits a company employing 180 people. Independent of other employees, there is a probability p=0.35 that each person needs to take sick leave. What is the expectation and variance of the proportion of the workforce who needs to take sick leave. In general what is the value of the sick rate p that produces the largest variance for this proportion.

50 Poisson Probability  Many experiments don't have a simple success/failure response  Responses can be the number of events occurring over time, area, volume etc.  We don't know the number of 'failures' just the number of successes. Example: The number of calls to a telesales company - we know how many calls got through (successes) - but don't know how many failed (lines busy etc.)  Knowledge of the mean number of events over time etc => Poisson Random Variable  Events must occur randomly

51 Poisson Probability Distribution  Probability Distribution for Poisson Where  is the known mean:  x is the value of the RV with possible values 0,1,2,3,…. e = irrational constant (like  ) with value …  The standard deviation, , is given by the simple relationship;   =

52 Example: Bombing of London WW2  1944 German V1 rockets feel on London  Were they aimed at specific targets or falling randomly?  Important in AA strategy & Civil Defence  Divide London into a 24  24 grid of equal sizes (576 equal square areas).

53 Example: Bombing of London WW2  If rockets are random => should fall according to Poisson random variable per square   (mean) = No. of Bombs/ No of squares = 535/576 =  So, for a particular square (assuming randomness)  Where x is the number of bombs landing in the square on the map grid.

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55 Example: Bombing of London WW2  Prediction from Poisson so good => British concluded rockets were not being aimed at specific targets - were falling randomly on London X = no. of rockets P(x) 576  p(x) > 4 (i.e. 5+) Actual no. of squares Hit

56 Other Basic Discrete Probability Distributions  Geometric – No. of independent trials to first success.  Negative Binomial - No. of independent trials to first, second, third fourth… success.  Hypergeometric – lottery type experiments.  many others….

57 Question  The number of cracks in a ceramic tile has a Poisson distribution with a mean µ = 2.4.  What is the probability that a tile has no cracks?  What is the probability that a tile has four or more cracks?

58 There is more than one way to skin a cat!

59 Questions 1. If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)? 2. What is the compliment of the event? 3. What is its probability? There are three ways (at least) that we can approach this problem

60 Solution 1: Brute Force Approach  Enumerate the sample space and select those outcomes that satisfy the desired conditions  36 possible combinations of 2 die

61 Solution 1: Brute Force Approach  Enumerate the sample space and select those outcomes that satisfy the desired conditions  36 possible combinations of 2 die  27 combinations include a prime number

62 Solution 1: Brute Force Approach  Enumerate the sample space and select those outcomes that satisfy the desired conditions  36 possible combinations of 2 die  27 combinations include a prime number  Probability of at least one prime is 27/36 = 0.75

63 The Compliment  What is the compliment of the event?  That neither score is a prime number (2, 3, 5) when two fair dice are thrown  What is its probability?  Let the event be E and its probability be P(E)  Then the compliment of E is E’ and the probability of E`, P(E`), is equal to 1 – P(E)  In our case P(E) = 0.75 => P(E`) = 1 – 0.75 = 0.25

64 The Compliment  What is the compliment of the event?  That neither score is a prime number (2, 3, 5) when two fair dice are thrown

65 The Compliment  What is the compliment of the event?  That neither score is a prime number (2, 3, 5) when two fair dice are thrown  What is its probability?

66 The Compliment  What is the compliment of the event?  That neither score is a prime number (2, 3, 5) when two fair dice are thrown  What is its probability?  Let the event be E and its probability be P(E)  Then the compliment of E is E’ and the probability of E`, P(E`), is equal to 1 – P(E)  In our case P(E) = 0.75 => P(E`) = 1 – 0.75 = 0.25

67 Solution 2: Find the Probability of the Compliment  The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!  Our question can be slightly rearranged to reveal a possible solution  If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?

68 Solution 2: Find the Probability of the Compliment  The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!  Our question can be slightly rearranged to reveal a possible solution  If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?

69 Solution 2: Find the Probability of the Compliment  The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!  Our question can be slightly rearranged to reveal a possible solution  If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?  What is the probability of one or more primes from two dice throws?

70 Solution 2: Find the Probability of the Compliment  The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!  Our question can be slightly rearranged to reveal a possible solution  If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?  What is the probability of one or more primes from two dice throws?  What is the probability of one or more of outcome O from X trials?

71 Solution 2: Find the Probability of the Compliment  The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!  Our question can be slightly rearranged to reveal a possible solution  If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?  What is the probability of one or more primes from two dice throws?  What is the probability of one or more of outcome O from X trials?  If questions are of this form we can work out the answer by working out the compliment first

72 Solution 2: Find the Probability of the Compliment  What is the probability that neither score is a prime number (2, 3, 5) when two fair dice are thrown?  This is an easier probability to calculate as we can consider throwing each dice as an independent event and combine the probabilities that neither results in a prime  It is the “one or more” in the previous problem that makes things tricky as we cannot consider each dice throw as an independent event 

73 Solution 2: Find the Probability of the Compliment  To start let’s work out, if we throw a single dice what is the probability of not getting a prime number? 

74 Solution 2: Find the Probability of the Compliment  To start let’s work out, if we throw a single dice what is the probability of not getting a prime number?  Sample space: {1, 2, 3, 4, 5, 6}  Primes: {2, 3, 5}  Non-primes: {1, 4, 6}  So, probability is 3/6

75 Solution 2: Find the Probability of the Compliment  If the probability of getting no prime if we throw one dice is 3/6, what is the probability of getting no primes if we throw two dice in a row?

76 Solution 2: Find the Probability of the Compliment  If the probability of getting no prime if we throw one dice is 3/6, what is the probability of getting no primes if we throw two dice in a row?  Dice rolls are independent events  Remember our intersection rule for independent events  So, the probability of getting no primes if we throw two dice in a row is:

77 Solution 2: Find the Probability of the Compliment  Our event, E, was that neither score is a prime number (2, 3, 5) when two fair dice are thrown  So the complement of this event, E`, is that at least one score is a prime number (2, 3, 5) when two fair dice are thrown  We know that given the probability of event E, P(E), we can work our the probability of the complement of this event, P(E`), as 1 – P(E)  So for our dice example

78 Solution 2: Find the Probability of the Compliment  The great thing is that this works for any number of dice  The probability, P(E), of getting no primes if we throw n dice in a row is:  So, for three dice the probability of getting no primes is  This means that the probability of getting at least one prime from 3 dice rolls is 1 – 1/8 = 7/8

79 Solution 3: Use the Binomial Distribution  Problems that can be stated as:  what is the probability of seeing x successes in n independent binary trials  can be solved using the Binomial distribution.  For example:   what is the probability of seeing 1 prime in 2 dice throws

80 Solution 3: Use the Binomial Distribution  The Binomial probability, P(X=x), (read as the probability of seeing x successes) is given by:   where n is the number of trials, p is the probability of a success and, known as a combination, is the number of ways of getting x successes from n trials

81 Solution 3: Use the Binomial Distribution  So, what is the probability of seeing 1 prime in 2 dice throws  n = 2 p = 1/2 x = 1

82 Solution 3: Use the Binomial Distribution  Exercise: What is the probability of seeing 2 primes in 2 dice throws

83 Solution 3: Use the Binomial Distribution  Exercise: What is the probability of seeing 2 primes in 2 dice throws

84 Solution 3: Use the Binomial Distribution  So, what is the probability of seeing one or more primes in 2 dice throws?  P(1 ≥ X ≤ 2) = P(X = 1) + P(X = 2) = ½ + ¼ = ¾  More generally then we can say that the probability of seeing one or more primes in n dice throws is:

85 Exercise  Consider a multiple choice test in which:  Each question has 4 possible answers of which only 1 is correct  The test is made up of 10 questions  The pass mark is 40%  How well could we do if we just guessed each answer? 1. What is the probability of guessing a single question correctly? 2. What is the probability of getting no answers correct in the test? 3. What is the probability of getting at least one question correct in the test? 4. What is the probability of getting a score of 40% in the test? 5. What is the probability of passing the test?


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