Chapter 4 Sections 4.1 & 4.7 in Rosner September 23, 2008 Tree Diagrams Genetics 001 Permutations Combinations

Transcription

1 Chapter 4 Sections 4.1 & 4.7 in Rosner September 23, 2008 Tree Diagrams Genetics 001 Permutations Combinations Homework 2 (Due Thursday, Oct 2) is at the back of this handout

2 Goal for Chapter 4: To introduce discrete probability distributions including the binomial and Poisson distributions. Skill set: You should be able to construct the density function and cumulative density function given the probability of success and enough information to construct the appropriate tree diagram. Table of contents for Sections 4.1 and 4.7 Tree diagrams Pages 1-10 Example 1 - hemophilia Page 1 Genetics 001 Page 4 Example 2 - eye color Page 4 Example 3 - flower color Page 6 Example 4 - skin and ear lobes Page 7 Multiplication rule applied to tree diagrams Page 8 Permutations and combinations Pages Definitions Page 10 Example 5 - RN nucleotides Page 11 Guidelines for calculations Page 11 Example 6 - DN-RN code Page 11 Permutations of distinct objects Page 12 Example 7 - permutations Page 12 Definition of factorial Page 13 Example 8 Page 14 Definition of n P k Page 14 Theorem Page Permutation of Indistinguishable Objects Example 9 - selection of patients Page 16 Definition of combination Page 16 Example 10 - how to select 2 alike from 5 Page 17 Theorem: Combinations Page 17 Example 11 - blood bank problem Page 18 Stata commands: comb(5,2) lnfactorial(8) exp(lnfactorial(8)) round(exp(lnfactorial(8)),1)

3 Tree diagrams: The probability problems we have dealt with to date were fairly simple because the number of possibilities in each case was relatively small. s the experiments become more complex, it is helpful to have a systematic method for listing the possible outcomes. One such method is a tree diagram. This technique works well when the experiment can be visualized as taking place in a small number of distinct steps or stages. Example 1: woman is a carrier for classic hemophilia. This means that although the woman does not herself have hemophilia, she can pass the disease on to her son. She gives birth to three sons. What are the possibilities for this experiment? Let D+ = son has the disease (i.e. receives the hemophilia allele from Mom) n D- = son n does not have the disease and for n = 1, 2, 3 Let us assume that it is the case that the carrier is just as likely to pass the disease on to a given son as to not pass on the disease to that son. ssume that the Dad doesn t have the disease. That is, Pr(son n is D+) = Pr(son n is D-) = ½ for n = 1, 2 or 3. We will assume that whether or not the second son has hemophilia is independent of the status of the first son. Similarly the status of the third son is independent of the status of the first two sons. So, using the definition of independence, we have (note this is only a partial listing of the possibilities): Pr(son 1 is D+ and son 2 is D- ) = Pr(son 1 is D+) Pr(son 2 is D- ) = ½ ½ = ¼ Pr(son 1 is D+ and son 3 is D+) = Pr(son 1 is D+) Pr(son 3 is D+) = ½ ½ = ¼ Pr(son 2 is D- and son 3 is D- ) = Pr(son 2 is D- ) Pr(son 3 is D- ) = ½ ½ = ¼ Page -1-

4 We have similar results for the other combinations for sons 1 and 2, sons 1 and 3, and sons 2 and 3. The tree below gives the set of all possibilities for the three sons. Tree for Example #1 (Hemophilia) D+ D+ 1 D+ D+ D+ D- 2 D+ D+ D- D+ D- D+ 3 D+ D- D+ D- 4 D+ D- D- D+ D+ 5 D- D+ D+ D- 6 D- D+ D- D- D- D+ 7 D- D- D+ D- 8 D- D- D- First Second Third Path Path Son Son Son Number Notice that our tree above for Example 1 is such that each possibility for the disease status of the combination of all 3 sons is listed as one of the 8 branches of the tree. There are no other possibilities and no possibility is listed more than once. This is a characteristic of all properly drawn trees. This means you can add the probabilities associated with entire branches because there is no overlap. So if and represent two different branches, then Pr ( or ) = Pr(c) = Pr() + Pr() because Pr(1) = 0. Page -2-

5 We already have the multiplication rule for two events but it also works for 3 or more. If, and C are independent events, then Pr( and and C) = Pr() Pr() Pr(C) Pr(each of the sons has the disease) = Pr(son 1 is D+ and son 2 is D+ and son 3 is D+) where D+ = has hemophilia = Pr(son 1 is D+) Pr(son 2 is D+) Pr(son 3 is D+) we can multiply the 3 probabilities because the disease status of each son is independent of the disease status of the other sons Since Pr(son 1 is D+) = Pr(son 2 is D+) = Pr(son 3 is D+) = ½, Pr(each of the sons has the disease) = (½) (½) (½) = 1/8 (path 1) Pr(exactly two sons has the disease) = 3/8 (paths 2, 3 and 5) [i.e. can add branches] The problem that we just went through above shows us that because of the independence of the three sons with respect to disease status (has or does not have hemophilia), we can get the probability that goes with each branch by multiplying the individual probabilities for each of the three sons together. Since Pr(son n is D+) = Pr(son n is D-) = ½ for n = 1,2 or 3 the probability for each branch is (½) (½) (½) = 1/8. Pr(branch 2 or branch 3 or branch 5) = Pr(branch 2) + Pr(branch 3) + Pr(branch 5) because Pr(branch n and branch m) = 0 for n = 1,2,...,8 and m = 1,2,...,8 = (1/8) + (1/8) + (1/8) = 3/8 Note that the sum of the probabilities of all 8 branches is one. Page -3-

6 Genetics 001 (i.e. genetics grossly oversimplified): The hereditary characteristics of an organism are determined by units called genes. gene can be defined as a region of DN that controls a hereditary characteristic. It usually corresponds to a sequence used in the production of a specific protein or RN. gene carries biological information in a form that must be copied and transmitted from each cell to all its progeny. n allele is one of two or more alternative forms of a gene at a given position (locus) on a chromosome, caused by a difference in the sequence of DN For example, consider the gene that determines the height of a pea plant. This gene has two alleles, T for tallness and t for dwarfism. Thus there are three possible genetic compositions or genotypes with respect to this trait. They are TT, Tt and tt. (These are Mendel s famous peas.) When the two genes are of the same form, we say that the organism is homozygous (TT and tt) for a given trait; otherwise it is heterozygous (Tt). trait that will appear when the allele for the trait is present is called a dominant trait, and the allele is the dominant allele. Its contrasting trait or allele is said to be recessive. In the case of pea plants, the allele for tallness is dominant. Thus each of the genotypes TT and Tt will result in a tall plant, while the genotype tt will result in a dwarfed plant. The resulting characteristic of the pea plant is called the phenotype. Thus the phenotype for the plants with genotypes TT and Tt is tall and that for the plant with genotype tt is dwarfed. For each trait the offspring inherits one gene randomly from each parent. Example 2: ssume each member of a couple has alleles for both brown and blue eyes. In genetic terms they are each heterozygous for eye color. In the case of eye color, the allele for brown eyes, which we will denote by, is dominant over that for blue eyes, b. That is, anyone with the allele will have brown eyes. t conception each parent contributes one allele for eye color. Hence we can view the experiment of determining the eye color of a child as a two-stage process. Stage 1 represents the inheritance of an allele from the mother ( or b since Mom is heterozygous); stage 2 represents the inheritance from the father ( or b since Dad is also heterozygous). The tree for this example is shown below. Page -4-

7 Tree for Example #2 (Eye Color) 1 = brown b 2 b = brown 3 b = brown b b 4 bb = blue Mother Father Stage 1 Stage 2 Path Number Path Since the alleles are inherited at random, at each step we are just as likely to inherit a as we are a b allele. This means and Pr( from Mom) = Pr(b from Mom) = ½ Pr( from Dad) = Pr(b from Dad) = ½ Since what the child inherits from Mom is independent of what the child inherits from Dad, we can get the probability for each branch by multiplying the probability related to Mom by the probability related to Dad. So the probability for each of the four branches of the tree is ¼. Since three of the four paths result in a brown-eyed child we conclude Pr(brown-eyed child) = ¼ + ¼ + ¼ = 3/4 Notice that each branch being equally likely is an important trait of this particular tree but is not true of all trees. It is true of all correctly constructed trees that you can add the branches of the tree without worrying about overlap - i.e. the intersection is zero for any 2 branches. Page -5-

8 There also exist alleles such that neither is dominant over the other. Example 3 below illustrates this situation. Example 3: The plant known as the four-o clock can have red, white or pink flowers. The allele for redness is denoted by R and that for white by r. red flower has two R alleles and is said to be homozygous for color; a white flower is homozygous with genotype rr. When pure white plants are bred to pure red ones, the resulting flower is pink since neither allele is dominant. When two of the heterozygous plants (i.e. one allele is R and the other is r) are bred, the outcomes are as given in the tree diagram below. Each of the four paths of the tree is equally likely. The assumption is that Pr(R) = Pr(r) = ½ Pr(a red flower) = Pr(RR) = Pr(R) Pr(R) = ¼ (path 1) Pr(a white flower) = Pr(rr) = Pr(r) Pr(r) = ¼ (path 4) Pr(a pink flower) = Pr(Rr or rr) = Pr(Rr) + Pr(rR) - Pr(Rr and rr) = ¼ + ¼ - 0 = ½ (paths 2 and 3) Tree for Example #3 (Four-o clocks) R R 1 RR = red r 2 Rr = pink r R 3 rr = pink r 4 rr = white llele from Plant #1 llele from Plant #2 Path Number Path Trees can be used in a genetic setting to study more than one trait simultaneously. Page -6-

9 Example 4: In humans, the allele for normal skin pigmentation (S) is dominant over that for albinism (s). The allele for free ear lobes (F) is dominant over that for attached ear lobes(f). Suppose a woman has genotype SsFF and her husband has genotype ssff. Hence the woman is heterozygous for skin pigmentation and therefore has normal skin pigmentation. She is homozygous for ear lobes and has free ear lobes. Her husband is albino with free ear lobes (because F is dominant). What are the possible outcomes for their offspring? We visualize this as a four- stage experiment with the following stages: 1) inherit an allele for skin pigmentation from the mother 2) inherit an allele for skin pigmentation from the father 3) inherit an allele for ear formation from the mother 4) inherit an allele for ear formation from the father The tree diagram below gives the possibilities. Tree for Example #4 (Skin and Earlobes) S s F F 1 normal skin & free lobes f 2 normal skin & free lobes s s F F 3 albino & free lobes f 4 albino & free lobes Skin Skin Earlobe Earlobe Path Path Mom Dad Mom Dad Number Stage 1 Stage 2 Stage 3 Stage 4 Pr(branch 1) = Pr(branch 2) = Pr(branch 3) = Pr(branch 4) = (½) 1 1 (½) = 1/4 because Pr(S from Mom) = Pr(s from Mom) = ½ but Pr(S from Dad) = 0 and Pr(s from Dad) = 1 and Pr(F from Mom) = 1 and Pr(f from Mom) = 0 but Pr(F from Dad) = Pr(f from Dad) = 1/2. Pr(normal skin and free lobes) = Pr(branch 1) + Pr(branch 2) = (1/4) + (1/4) = ½ ll of the trees we have considered so far were such that the stages or twigs were independent. This meant that we could obtain the probability for the branch by multiplying the probabilities for each of the twigs on that branch. Page -7-

10 Let s go back to Example 1 and show that the product of the probabilities on the branches gives us the probability of the branch even if the stages or twigs are NOT independent. Tree for Example #1 (Hemophilia) Revisited D+ D+ 1 D+ D+ D+ D- 2 D+ D+ D- Pr() D+ Pr( ) Pr(C and ) D+ 3 D+ D- D+ D- D- 4 D+ D- D- D+ D+ 5 D -D+ D+ D- 6 D- D+ D- D- D- D+ 7 D- D- D+ D- 8 D- D- D- First Second Third Path Path Son Son Son Number Let = the 1 st son has hemophilia = the 2 nd son does not have hemophilia C = the 3 rd son has hemophilia Let s consider path #3 in the above tree. Note that because we are not assuming independence of the events, the probabilities on the twigs are not Pr() and Pr(C) but are instead conditional probabilities based on what has happened on the earlier twigs on the branch. Page -8-

11 The probability that goes with branch #3 is Pr( and and C). I claim we can get this answer by multiplying together the three probabilities listed on the branch. Multiplying together the first two twigs on the branch we get the following: Pr( ) Pr( ) = Pr( ) Pr( and ) Pr( ) Pr( ) Pr( and ) = Pr( and ) The two s cancel out and so Pr( ) Pr( ) = Pr( and ) (Multiplication principle, Equation 1) Note that the multiplication principle is true regardless of whether or not and are independent. Now what happens when we consider the 3 rd twig on branch or path 3. First note that Pr( C and ) = Pr( C and[ and ]) Pr( and ) = Pr([ and ] and C) Pr( and ) = Pr( and and C) Pr( and ) Equation 2 So multiplying together the three probabilities on ranch 3 we get Page -9-

12 Pr( ) Pr( ) Pr( C and ) = Pr( and ) Pr( C and ) = Pr( and ) = Pr( and and C) Pr( and and C) Pr( and ) See Equation 1 See Equation 2 So the product of the probabilities on the branch really does give us the probability for the branch regardless of whether or not the stages are independent. Permutations and Combinations: The tree diagrams that we have been considering give us the ability to count the number of outcomes for an event and the total number of possible outcomes N, so that we can calculate the probability that will occur as the number of ways in which occurs divided by the total number of possibilities. s experiments become more complex, we will need other techniques in order to obtain the desired probability. One such technique involves permutations and combinations. Definition: permutation is an arrangement of objects in a definite order. Definition: combination is a selection of objects without regard to order. Notice that the key to distinguishing between these two definitions is order. We ll consider permutations first. Once a problem has been identified as being one in which order is important, the next question to be answered is: How many permutations or arrangements of the given objects are possible? This question is frequently answered by means of the multiplication principle. Multiplication Principle Pr( and ) = Pr() Pr( ) Page -10-

13 Example 5: iologists are interested in the order in which the four ribonucleotides adenine (), uracil (U), guanine (G) and cytosine (C) combine to form small chains. These nucleotides provide the principle subunits of RN, the intermediate information-carrying molecule involved in translating the DN genetic code. How many chains, each consisting of two different nucleotides, can be formed. Notice that C and C are considered to be different (i.e. order is important). nswer = 4@3 = 12 (any of the 4 letters can be used in the first slot but the letter chosen for the first slot cannot be used in the second position, so there are only 3 choices for the second position). Guidelines to consider when applying the multiplication principle: 1) Watch out for repetition versus non-repetition. 2) Watch for subtraction. 3) If there is a stage in the experiment with a special restriction, then you should worry about the restriction first. Example 6: The DN-RN code is a molecular code in which the sequence of molecules provides significant genetic information. Each segment of RN is composed of words. Each word specifies a particular amino acid and is composed of a chain of three ribonucleotides that are not necessarily all different. For example, the word UUU corresponds to the amino acid phenylalanine, whereas UG identifies methionine. ) Consider the experiment of forming an RN word. How many words can be formed? Remember from Example 5 above that we have four choices (, U, G and C) for each letter of each of the three-letter words. nswer = = 64 Page -11-

14 ) How many of the 64 possible words above have at least two identical nucleotides? This is a problem that lends itself to subtraction. First find the number of words that have no repeats. This is = 24. So at least two identical is = 40. Pr(word will contain some repeats) = 40/64 = C) Consider event that a randomly formed word ends in U (uracil) and involves no repetitions. Find Pr(). Start by looking at the last letter of the word because that letter has the restriction. Number of ways can select the last letter is 1. Number of ways you can select the first letter given that the last letter is U and there can be no repetitions is 3. Number of ways you can select the middle letter given that the first and last letters have been selected and there are no repetitions is 2. Pr() = (3 2 1)/64 = 6/64 = Permutations of distinct objects: Example 7: Suppose we have three distinct objects (, and C) and we wish to know how many permutations we can have with these 3 objects. Remember that permutation means that order counts. So the first choice can be any of, or C. If the first choice is, then the second choice has to be either or C. If the first choice is and the second choice is C, then the only possibility for the third choice is. The tree below gives the 6 possible arrangements for the three distinct letters. Page -12-

15 Tree for Example 7 Path C C 1 2 C C 3 4 C choices X 2 choices X 1 choice = 3! Definition: Let n be a positive integer. The product n(n - 1)(n - 3@2@1 is called n factorial an is denoted n! 3! = 3@2@1 = 6 Definition: 0! = 1 (zero factorial = 1) Stage 1 Stage 2 Stage 3 So we have that the number of permutations of n distinct objects is n! Stata does not have a direct way to get the factorial of a number. It uses the natural log of the factorial (instead of a product it switches to a sum). This is to avoid overflow problems. We know that the exponential function and the natural log function are inverses, so we back to the factorial by taking exp(lnfactorial(n)). Sometimes exp(lnfactorial(n)) doesn t result in an integer because of the way things are stored. So we use the round function to make sure we get an integer. di round(exp(lnfactorial(3)),1) 6. di round(exp(lnfactorial(0)),1) 1 Stata help/functions/mathematical functions Page -13-

16 Example 8: Let s go to a larger tree. Suppose we want the number of permutations of 5 different objects. If you were going to draw a tree 1) the first stage would have 5 twigs 2) the 2 nd stage is formed by adding 4 twigs to each of those 5 twigs 3) the 3 rd stage is formed by adding 3 twigs to each of the twigs on level 2 4) the 4 th stage is formed by adding 2 twigs to each of the twigs on level 3 5) the 5 th stage is formed by adding 1 twig to each of the twigs on level 5 So we have = 5! = 120 Now what would happen if from 5 different objects, we decided to choose 2 and order is important. The tree would then be just the first two stages of Example 8 above. So there would be 5 4 = 20 ways (the number of choices in Stage 1 times the number of choices in Stage 2). In other words we cut off the last three stages. When we arranged all 5 distinct objects there were 5! distinct patterns. ut now that we select only two, we have 20 distinct patterns. 20= 5 4 = = 5! = 5! = ! (5 2)! 5 P 2 Note: I wouldn t expect you to think of this. So we have the definition of the number of permutations of 5 things taken 2 at a time. Definition: The number of permutations of n distinct objects taken k at a time is n P k = n ( n! k)!. Page -14-

17 Permutations of Indistinguishable Objects: How many distinct arrangements of n objects are possible if some of the objects are identical and therefore cannot be distinguished? Let us go back to the tree for Example 7. Instead of 3 different objects, we will change the s to s so that we still have 3 objects but 2 of them are alike. C C & $ C C & $ C % % Notice that paths 1 and 3 (&) are now alike, paths 2 and 4 ($) are alike and paths 5 and 6 (%) are alike. So we have not 6 but 3 distinct patterns because we have lost the distinction between and since they are both now (i.e. we lost a 2!). 3 = ( 2 1) 1 = 3! 21!! Which is the number of permutations of 3 objects 2 of which are alike (which leaves 1 or you could say 1 alike). The more general form is the theorem on the next page. Page -15-

18 Theorem: Permutation of indistinguishable objects. n n 1 n 2 n k n objects can be arranged is Consider objects where are of type 1, are of type 2,..., are of type k. The number of ways in which the n! n! n! n!... n! k Given that n= n + n + n n k Remember n 1 were alike; n 2 were alike and so on to n k were alike. Example 9: Fifteen patients are to be part of an experiment to test a standard drug, an experimental drug, and a placebo. Each patient is to be randomly assigned a treatment. In how many different ways can the three treatments be assigned to the 15 patients? What is the probability that a random assignment of treatments would result in 10 patients receiving the placebo, 3 receiving the experiment drug and 2 the standard drug? We will consider this a permutation problem because we will assume it makes a difference which drug is given to a given participant. Total number of assignments: 3@3@3@...@3 = 3 15 = 14,348,907 placebo 10, experimental drug 3, standard drug 2 = 15!/(10!3!2!) = 30,030 Pr(placebo 10, experimental 3, standard 2) = 30030/ = Please note that this is not a design that anyone (I hope) would consider reasonable. Combinations: Definition: combination is a selection of objects without regard to order. Page -16-

19 Example 10: Five people have volunteered to take part in an experimental program. Only two are needed to complete the study. In how many ways can two people be selected from five? Label the subjects,, C, D, E. Order is not important so {,C} is the same as {C,}., C, D, E, C, D, E, CD, CE, DE There are 10 such sets. Notice that 5 P 2 = 5! ( 5 2)! ut for our problem above we have lost the ordering of the 2 we pick or we have lost a 2!. So the combination (i.e. no order) of 5 things chosen 2 at a time is 5 C 2 5P 2 5! 5! 5 4 = = = = = 5 2= 10 2! 2!(5 2)! 2!3! 2! Note that selecting 2 of the 5 means 3 of the 5 are not selected. Theorem: The number of combinations of distinct objects selected at a time, denoted by n C r, is given by the equation n C r n! = r!(n r)! n also denoted n r r. di comb(5,2) = comb(n,k) returns the combinatorial function -- "k out of n". The arguments must be nonnegative integers with n $ k $ 0. di round(exp(lnfactorial(2)),1)*comb(5,2) = 5 P 2 20 Page -17-

20 Example 11: blood bank has available 10 units of positive blood. Four happen to be contaminated with serum hepatitis. Three units are randomly selected for use in three patients. What is the probability that all three patients will be exposed to hepatitis from this source? Total number of ways to select 3 units from the 10 is 10 C 10 10! = = = !7! In order for all three patients to be exposed to hepatitis, all three units must be selected from the 4 contaminated units. The number of ways to make this selection 4 C 4 4! = = = !1! So Pr(all three exposed to risk) = 4/120 = Examples from Statistical Methods in the iological and Health Sciences, 2 nd edition by J. Susan Milton. Where have we been and where are we going: We have been considering some of the methods currently used to describe data sets (means, histograms, etc.). Some of the data sets we have considered were the entire population under consideration, while others were a sample drawn from a larger population. To date the conclusions we have drawn from these samples have been pretty crude since they have been based solely on our opinions. We have also discussed probability theory and we have seen that it is possible to answer some fairly complex questions through the use of probability theory. What we have not yet done is to consider the implications of probability theory to data analysis. That is, we have not yet shown how probability theory can be used to draw conclusions about a population based on a sample drawn from that population. To do this we must turn our attention to a topic that provides the link between probability theory and applied statistics, namely the notion of a random variable. Page -18-

21 Homework 2 This homework is due at the beginning of class on October 1, 2008 You may work in small groups Using multiple tests to make a diagnosis: We have seen that usually when we use a single test for diagnosis, as we increase the sensitivity we decrease the specificity. nd as we decrease the sensitivity, we increase the specificity. (The MI/non-MI study is a good example of this.) nd we did this by changing the referent value. ut it is frequently the case that multiple tests are used and under these circumstances we can increase the sensitivity or the specificity (but not both at the same time) above that of either test alone without changing the referent value. Two independent tests in a screening or diagnostic situation can be used in three different ways: 1) Test is applied first and all those with a positive result are retested with test. This is called the series approach and (first + and then +) = T+ otherwise T Made Up Data Disease Disease Present bsent (D+) (D-) , , ,250 1,000 99, ,990 4,890 84, ,000 2) Test is applied first and all those with a positive result are retested with test. This is also the series approach and (first + and then +) = T+ otherwise T-. Page -19-

22 3) Tests and can be used together and all those individuals with positive results on either or both tests are considered to be positive. This is called the parallel approach and T+ = {(++), (+-), (-+)} and T- = (--) Please calculate each of the following. You might want to give the formula you are using, just in case your arithmetic is incorrect we ll at least know you started out right. For the parallel and series questions, you might find the little graph below helpful. sensitivity of test = sensitivity of = sensitivity of the series combination = sensitivity of the parallel combination = specificity of = specificity of = specificity of the series combination = specificity of the parallel combination = D+ D- See next page. Page -20-

23 Please fill in the table below. Explain what, if anything, you gain by using the series and parallel combinations. Under which conditions would you use the tests in parallel and under which conditions would you use the tests in series? Tests and are independent Only Only and in Series and in Parallel Sensitivity Specificity Page -21-