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Probability and conditional probability, with a focus on case studies and examples. It covers topics such as calculating observed proportions, estimating differences between population probabilities, and testing if population proportions are different. The document also introduces notions of conditional probability and independence. The case study discussed in the document involves parasitic fish and predation. likely useful as study notes or lecture notes for university courses in statistics or probability theory.
Typology: Lecture notes
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Bret Hanlon and Bret Larget
Department of Statistics University of Wisconsin—Madison
September 27–29, 2011
Probability 1 / 33
Case Study Example 9.3 beginning on page 213 of the text describes an experiment in which fish are placed in a large tank for a period of time and some are eaten by large birds of prey. The fish are categorized by their level of parasitic infection, either uninfected, lightly infected, or highly infected. It is to the parasites advantage to be in a fish that is eaten, as this provides an opportunity to infect the bird in the parasites next stage of life. The observed proportions of fish eaten are quite different among the categories.
Uninfected Lightly Infected Highly Infected Total Eaten 1 10 37 48 Not eaten 49 35 9 93 Total 50 45 46 141
The proportions of eaten fish are, respectively, 1/50 = 0.02, 10/45 = 0.222, and 37/46 = 0.804.
Probability Case Studies Infected Fish and Predation 2 / 33
Case Study Example 9.4 on page 220 describes an experiment. In Costa Rica, the vampire bat Desmodus rotundus feeds on the blood of domestic cattle. If the bats respond to a hormonal signal, cows in estrous (in heat) may be bitten with a different probability than cows not in estrous. (The researcher could tell the difference by harnessing painted sponges to the undersides of bulls who would leave their mark during the night.)
In estrous Not in estrous Total Bitten by a bat 15 6 21 Not bitten by a bat 7 322 329 Total 22 328 350
The proportion of bitten cows among those in estrous is 15/22 = 0.682 while the proportion of bitten cows among those not in estrous is 6/328 = 0.018.
Probability Case Studies Infected Fish and Predation 4 / 33
Are the probabilities of being bitten different for cows in estrous or not? How do we estimate the difference in probabilities of being bitten? How do we estimate the odds ratio? Here, the odds of a cow in estrous being bitten are roughly 2 to 1, while the odds of a cow not in estrous being bitten are roughly 2 to 100, so the odds ratio is about 100 times larger to be bitten for cows in estrous compared to those not. How do we quantify uncertainty in this estimate?
Probability Case Studies Infected Fish and Predation 5 / 33
To understand the methods for comparing probabilities in different populations and analyzing categorical data, we need to develop notions of: I (^) conditional probability; and I (^) independence; We will also more formally introduce some probability ideas we have been using informally.
Probability The Big Picture 7 / 33
Bucket 1 contains colored balls in the following proportions: 10% red; 60% white; and 30% black. Bucket 2 has colored balls in different proportions: 10% red; 40% white; and 50% black. A bucket is selected at random with equal probabilities and a single ball is selected at random from that bucket.
Think of the buckets as two biological populations and the colors as traits.
Probability Probability Running Example 8 / 33
A probability is a number between 0 and 1 that represents the chance of an outcome. Each elementary outcome has an associated probability. The sum of probabilities over all outcomes in the outcome space is 1.
Probability Probability Probability 10 / 33
An event is a subset (possible empty, possibly complete) of elementary outcomes from the outcome space. The probability of an event is the sum of probabilities of the outcomes it contains.
P(Bucket 1) = P({(1, R), (1, W ), (1, B)}) = 0.05 + 0.30 + 0.15 = 0.5. P(Red Ball) = P({(1, R), (2, R)}) = 0.05 + 0.05 = 0.1. P(Bucket 1 and Red Ball) = P({(1, R)}) = 0.05. P(Ω) = 1.
Probability Probability Events 11 / 33
Events are mutually exclusive if they have no outcomes in common. This is the same as saying their intersection is empty. The symbol for the empty set (the set with no elementary outcomes) is ∅.
The events A = {choose Bucket 1} and B = {Choose Bucket 2} are mutually exclusive because there are no elementary outcomes in which both Bucket 1 and Bucket 2 are selected. Any event E is always mutually exclusive with its complement, E c^.
Probability Probability Events 13 / 33
If events A and B are mutually exclusive, then P(A or B) = P(A) + P(B). With more formal notation,
P(A ∪ B) = P(A) + P(B) if A ∩ B = ∅.
The probability of a red ball is 0.1 because
P
(1, R) or (2, R)
Probability Probability Events 14 / 33
The probability that an event does not happen is 1 minus the probability that it does. P(not A) = 1 − P(A). With more formal notation,
P(Ac^ ) = 1 − P(A)
Let A = {Ball is Red}. Earlier, we found that P(A) = 0.1. The probability of not getting a red ball is then
P(Ac^ ) = 1 − P(A) = 1 − 0 .1 = 0. 9
Probability Probability Events 16 / 33
A random variable is a rule that attaches a number to each elementary outcome. As each elementary outcome has a probability, the random variable specifies how the total probability of one in Ω should be distributed on the real line, which is called distribution of the random variable. For a discrete random variable, all of the probability is distributed in discrete chunks along the real line. A full description of the distribution of a discrete random variable is: I (^) a list of all possible values of the random variable, and I (^) the probability of each possible value.
Probability Probability Random Variables 17 / 33
The conditional probability of an event given another is the probability of the event given that the other event has occurred. If P(B) > 0, P(A | B) =
P(A and B) P(B) With more formal notation,
, if P(B) > 0.
The vertical bar | represents conditioning and is read given. P(A | B) is read The probability of A given B.
Probability Probability Conditional Probability 19 / 33
Define events B 1 and B 2 to mean that Bucket 1 or 2 was selected and let events R, W , and B indicate if the color of the ball is red, white, or black. By the description of the problem, P(R | B 1 ) = 0.1, for example. Using the formula,
Probability Probability Conditional Probability 20 / 33