BIOL2300 Biostatistics Chapter 4 Counting and Probability

Transcription

1 BIOL2300 Biostatistics Chapter 4 Counting and Probability

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3 Event, sample space sample space (generally denoted Ω, pronounced omega ): set of outcomes of a random experiment {H,T} set of coin flips {1,2,3,4,5,6} set of outcomes of die roll {HH,HT,TH,TT} set of 2 consecutive coin flips simple event: element of sample space

4 Events Event A is subset of Ω, i.e. A µ Ω. Examples: {1,3,5} set of odd numbered outcomes of die roll {1,3,5,...,35} = reds in roulette {2,4,...,36} = blacks in roulette {A,G} = purines in i-th position of a nucleotide seq {A,I,L,F,V,P,G} = hydrophobic amino acids A: Alanine - Ala I: Isoleucine - Ile L: Leucine - Leu F: Phenylalanine - Phe V: Valine - Val P: Proline - Pro G: Glycine - Gly

5 P(A) denotes prob of event A P(A) can be approximated by relative frequency #times A occurs in N independent experiments N

6 Law of large numbers In the limit as N approaches infinity, the relative frequency of occurrence of event A equals the probability P(A) Flip a coin 1000 times. The probability of heads is approximated by number of heads divided by 1000.

7 Histogram of relative frequency of number of times a 1 was rolled in 1000 die rolls, repeated 10,000 times.

8 Counting If each simple event has equal probability (uniform distribution), then probability P(A) equals # simple events in A / size of sample space

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14 Answer 1-26/54 = For events (or sets) A,B whose intersection is non-empty, we do NOT double count the items in the intersection. Thus answer is P(A union B) and NOT P(A)+P(B) The addition rule is ONLY valid if events (sets) are disjoint. If events not disjoint, then rule is P(A)+P(B)-P(A Å B). We ll often write P(A Å B) instead as P(A,B).

15 Sampling: selection with and without replacement

16 Independent Events

17 Conditional probability and independent Events Conditional probability P(A B) defined by P(A B) = P(A,B)/P(B) Events A,B µ Ω are independent if P(A,B) = P(A) P(B) Equivalently, A,B are independent if P(A) = P(A B) = P(A,B)/P(B)

18 Intersection of sets

19 Union of sets

20 Set difference

21 A,B,C are events, hence subsets of sample space Since A and B are subsets of the sample space, their intersection is denoted by A B. However, in probability theory, the intersection of events A and B Is usually denoted by A,B or less commonly by AB (not multiplication!). The probability that events A and B simultaneously occur is called the joint probability P(A,B).

22 Venn Diagrams The sample space, Ω, is also called the universe, domain of discourse, etc. and is represented by a large rectangle. Events A µ Ω are represented by circles within Ω, and P(A)= A /Ω Conditional probability P(A B) is visually represented by considering the universe to be B, rather than Ω.

23 Simple form of Bayes

24 Contingency Table

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27 Bayes theorem Bayes theorem allows one to reverse causality. If we know probability P(A) and we know conditional probability P(B A) then we can compute P(A B). This is used to compute the probability of a model L(Model Data) given knowledge of P(Data Model) and the prior probability P(Model) of a model

28 Application of Bayes rule Sally (blindfolded) draws a ball randomly from one of two urns: urn 1: 5 red balls and 5 black balls urn 2: 9 red balls and 1 black ball Sally draws a red ball. What is the probability that the ball comes from urn 2?

29 Easy solution using contingency table P(urn2 R) = P(urn2,R)/P(R) = (9/20) / (14/20) = 9/14=0.6429

30 Computation P( red urn 1) = 0.5 P( red urn 2) = 0.9 P( urn2 red) = P(urn2, red)/p(red) = P(red urn2)*p(urn2)/p(red) = 0.9*0.5/P(red) = 0.45/0.7 = To compute P(red), use total prob form P(red) = P(red urn1)p(urn1) + P(red urn2)p(urn2) P(red) = 0.5* *0.5 = 0.7

31 Prior and posterior probabilities prior probability: probability (estimate) prior to an experiment. For instance, prior probability of urn2 before Sally draws a ball, P(urn2) = 0.5 posterior probability: revised probability (estimate) after performing an experiment. For instance, posterior probability of urn2 after Sally draws red ball, P(urn2 red) = 0.7

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33 Answer Prob that 4 randomly selected persons have same birthday on Jan 1 is 1/ Similarly for Jan 2 and all 365 days. Thus prob that 4 randomly selected persons have same birthday is 365 1/365 4 = 1/365 3 =

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35 Answer (With replacement): (89/979) 2 = (Without replacement): (89 88)/( )

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37 Answer P( RB 40) = P(RB,40)/P(40). Now P(RB,40) = 30/( ) and P(40) = 85/( ), so answer is 30/85 Using table, what is P(40 RB)? Using Bayes, what is P(40 RB)?

38 Mortality rate

39 Mortality rate, continued

40 Fertility rate

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42 Answer Neonatal: 5000 Fetal deaths: 33,500 Live births: 4,052,000 perinatal mortality rate is thus

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46 Some simple counting problems

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48 Answer 26 letters, 10 digits number of good configurations: 26*10 number of all configurations: (26*10) 3 ratio of good/all is 1/260 2

49 Birthday paradox In a room of 23 persons, what is the probability that 2 persons have the same birthday?

50 Birthday paradox Question: In a room of 23 persons, what is the probability that 2 persons have the same birthday? Answer: (365-0)/365 * (365-1)/365 *... * (365-22)/365 = Use complement rule: P(at least 2 persons with same birthday) = 1 - P(no 2 persons have same birthday) = = > 1/2

51 Permutations and combinations Number of ordered k-tuples of distinct elements drawn from set of n elements is npk, defined to be n*(n-1)*...*(n-(k-1)), consisting of k factors Number of unordered k-tuples of distinct elements drawn from set of n elements is nck, defined to be number npk/ k! Here 0! = 1 and for k>0, k! = k*(k-1)*...*1

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56 Risk Absolute risk reduction P(event occurs in treatment group) - P(event occurs in control group) Relative risk (risk ratio) P(event occurs in treatment group) P(event occurs in control group)

57 Number to treat Number needed to treat equals absolute risk reduction

58 Disease No Disease Treatment a b No treatment c d

59 Absolute & relative risk reduction absolute risk reduction a/(a+b) - c/(c+d) relative risk reduction (a/a+b) / (c/(c+d)

60 Odds Odds against event A: P(not A)/P(A) Odds in favor of event A: P(A)/P(not A) Odds ratio (OR) = (a/b)/(c/d) odds in favor of event for treatment odds in favor of event for control

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62 Answer biker rate: 0.2 nonbiker rate: 0.8 odds against A is ratio of P(not A)/P(A) = 0.8/0.2 = 4 to 1

63 Card counting problems 5-card poker hands

64 Flush Probability of a flush (5 cards, all same suit). which is , or about a 2 in 1000 chance.

65 Straight Probability of a straight (5 cards in order, ace high or low, suit does not matter). which is , or about a 4 in 1000 chance.

66 Probability of a straight (5 cards in consecutive order, suit does not matter, ace counts as low )

67 Note that the numerator of the first expression counts the number of straights (ace low), where the player has sorted the cards in increasing order: choose starting card among 1,2,...,9 (36 possibilities since 4 suits), then 4 choices for each successive card (since suit does not matter).

68 The denominator of the first expression is the number of unordered hands in a deck of 52 cards. Since 52 C 5 = (52 P 5)/5! we see that the results of first and second expression must be identical

69 Royal straight Probability of a straight of form 10,jack,queen,king,ace (5 cards in consecutive order beginning with 10, suit does not matter, ace counts as high )

70 Probability of straight (ace low or high) Probability of a straight (5 cards in consecutive order, suit does not matter, ace counts either low or high) which equals or , around 4 in 1000 chance.

71 Straight flush Probability of a straight flush (5 cards in consecutive order with same suit, ace counts as low or high ) which equals or around a one in 10,000 chance.

72 Problem of finding the right key to open the door A woman is given n keys, of which only one fits the front door. She randomly tries the keys successively (sampling without replacement) until she can open the front door. This procedure may require 1,2,...,n trials. Prove that each of these n outcomes has probability 1/n.

73 Sketch of solution Define the events A1,A2,...,An as follows: A1: first key fits A2: first key does not fit, second key fits A3: first two keys do not fit, third key fits etc. An: first n-1 keys do not fit, nth key fits

74 Clearly P(A1) = 1/n P(A2) = (n-1)/n x 1/(n-1) = 1/m P(A3) = (n-1)/n x (n-2)/(n-1) x 1/(n-2) = 1/n Etc.

75 Coin tossing problem

76 Coin tossing problem

77 Solution

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80 Acceptance sampling

81 Solution of problem on acceptance sampling General procedure: select sample of size k out of n reject lot if any items in sample defective 3% of 5000 blood monitors defective. Find prob of rejecting batch when selecting 10 w/o replacement

82 HIV test accuracy

83 Solution of problem on HIV test accuracy Given facts from problem: HIV test is 95% accurate 10% of high risk group is HIV positive X tested positively. What is prob that X has HIV? X has HIV. What is prob that X tested positively?

84 Interpretation of test accuracy true positive (TP): tests positively, has disease false positive (FP): tests positively, does not have disease true negative (TN): tests negatively, does not have disease false positive (FN): tests negatively, has disease

85 Medical test definitions accuracy: probability of being correct: P(correct) (TP + TN)/(TP + FP + TN + FN) sensitivity (true positive rate): conditional prob of correct positive prediction P(TP pos) TP /(TP + FN) = TP / positives specificity (true negative rate): conditional prob of correct negative prediction P(TN neg) TN/(TN + FP) = TN / negatives positive predictive value (PPV): TP /(TP + FP) = TP/ predicted positives

86 Solution to HIV problem Assume that 95% accuracy means that sensitivity and specificity are 95% Hence P( pos test HIV ) = 0.95 Use Bayes to compute P( HIV pos test) P( pos test HIV)*P(HIV)/P(pos test) = (0.95)(0.1)/ P(pos test)

87 P(pos test) = P(pos test HIV)*P(HIV) + P(pos test not HIV)*P(not HIV) = (0.95*0.1)+(0.05*0.9) = 0.14 It follows that P(HIV pos test) = 0.679

88 Sickle cell disease According to the CDC (Center for Disease Control), SCD (sickle cell disease) affects approximately 100,000 Assuming Hardy- Weinberg equilibrium, what percentage of Americans are protected from malaria? (Assume that both heterozygotic and homozygotic carriers of the sickle-cell allele are protected from malaria.)

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90 Multinomial coefficients

91 Scrabble tiles & Mississippi

92 Misspellings of Mississippi

93 Sample counting problems