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by: Favian Swaniawski

Populatn BIOL 464

Favian Swaniawski
GPA 3.82


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Class Notes
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This 27 page Class Notes was uploaded by Favian Swaniawski on Saturday September 12, 2015. The Class Notes belongs to BIOL 464 at West Virginia University taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/202742/biol-464-west-virginia-university in Biology at West Virginia University.


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Date Created: 09/12/15
Review of Probability Biol 464 Gen 535 82008 Probability Defined Probability relative frequency of an event in a long series of trials Can be calculated simply by the Sample Point Method SamplePoint Method Conce tual Ste s 1 Define your sample space S by listing all sample points 2 Assign probabilities PI of those sample points such that ZPz39zl 3 Determine which sample points comprise the event of interest and sum the probabilities of those points If all events equiprobable then PA Example Find the probability of getting exact 0 heads in three tosses of a coin Define Sample Space Outcome T055 1 T055 2 T055 3 1 Head Head Head 2 Head Head Tail 3 Head Tail Head 4 Tail Head Head 5 Tail Tail Head 6 Tail Head Tail 7 Head Tail Tail 8 Tail Tail Tail Example Find the probability of getting exact heads in three tosses of a coin Assign Pi to each outcome All outcomes are equally probable 18 0 Outcome T055 1 T055 2 T055 3 1 Head Head Head 2 Head Head Tail 3 Head Tail Head 4 Tail Head Head 5 Tail Tail Head 6 Tail Head Tail 7 Head Tail Tail 8 Tail Tail Tail Find the probability of getting exactly two heads in three tosses of a coin What sample points constitute the event of interest Where Heads occurs exactly twice ie Outcomes 23 and 4 Sum of the probability of these outcomes 18 18 18 38 or 0375 In other words In other words there is a 375 chance of obtaining exactly two heads in a series of three coin tosses Problem1 1015 minutes Use the SamplePoint Method to find the most probable sum from a single roll of a pair of balanced dice What is the probability of that sum Combinatorics There are simpler methods rather than generating a list of all possible outcomes that are useful when counting sample points mn rule if you have m elements from one group and n elements from another you can calculate the number of pairs of these elements by m X n Dice Example Die 1 b outcomes elementS Die 2 6 outcomes m X n 6 X 6 36 possible combinations Combinatorics As you add additional elements to the experiment the number of possible combinations increases markedly Dice Example 3 dice m n and z m X n X z 6X 6X 6 216 combinations Example Adding more elements to the experiment Calc the probability that everyone class of 14 will have a different birthday Student 1 365 different days on which birthday could fall Student 2 364 different days on which birthday could fall cant fall on bday of S1 Student 3 363 different days on which birthday could fall cant fall on bdays of S1 or 82 Student 14 352 different days Example Adding more elements to the experiment Can apply the SamplePoint equation as follows PA na 365x364x363352 N 36514 07769 So in a class of 14 there is a 7769 chance that you will not share the same birthday as another student In a class of 18 Prob 623Why would it decrease More on SamplePoints Sample points often represented as sequences of numberssymbols In some instances the total number of sam le points is equal to the number of distinct ways you can order these numbers or symbols Perm utation total number of ways to order n elements taken r at a time n n r n r Permutation example How many trinucleotide sequences 0a be formed Without repeating a nucleotide n number of elements 4 r number of samples 3 l l B quot39 439 24 rt r 4 3 With permutations the order of the elements in a sequence is important Des ite having the same three nucleotides the trinucleotide sequences below are all considered different each counted separately in a permutation because the order nucleotides varies among sequences ACG CGA AGC Combinations When the sequence of the elements is not important this is no longer a permutation Combination an unordered set ofr elements Chosen at random Without replacement can only be Chosen twice from n available elements n n r rn r Combinations Eve though the order nucleotides varies among the sequences below in a combination this set sequences would only be counted once because they contain the same combination of elements ie nucleotides ACG CGA AGC Combinations What would be the total number of trinucleotide combinations that could be generated without repeating a nucleotide n n 4 4 r rn r 34 3 Notice there are considerably rewer potential outcomes in this experiment Problems 2 and 3 20 min to complete Union of Events Union of events considering samplespaces blue circles Events A and B the Union of events A u B contains A or B or both A and B ie Events A or B are both true Union Example Union of events In a roll of a pair of dice Event A occurrence of a single odd number 1 3 5 Event B occurrence of a number less than 4 123 A u B will consist of all possible sample points such that A u B 1235 Intersection of Events The intersection of oven maximums consist of onythose sample points that lie within samplespaces A and B shaded area of circles Dice example Event A 135 Event B 123 A n B consists of 13 Intersection C9D Conditional Probability The occurrence of an event is sometimes dependent on the occurrence of another event The occurrence of Event A given the conditions of Event B can be CaICUIateCi by the equation PA m B PAiB 2 MB Where PAB is read as the probability ofA given B Example Conditional aility A person rolls a die and asks you to guess the outcome If you guess 3 You have a probability of 16 of being correct PA Ifthe person tells you that the roll is an odd number you now have a probability of 12 of being correct PB 1 PAlBwamp13 PltBgt Independent Events Tw events are said to be independent if The probability of event A given B is equal to the probability ofA PAB PA The probability of event B given A is equal to the probability of B PBA PB occurrence of event A is not dependent on the occurrence of B or vice versa Multiplicative law of Probability B rearranvinv the formula for conditional PAlB probability PAnB 133 the prob of the intersection of two events can be calculated PA m B PABPB If the two events are PAnBPAPB independent The probability that two events will both occur can be determined by mult their probabilities Mutually Exclusive Events If PltAmBgt 0 then events A and B are said to be mutually exclusive Ex flipping a Coin the outcome can be Heads A or Tails B but not both Additive Law of Probability PA u B PA PB PA m B If events A and B are mutually exclusive PA u B PA 103


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