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Lesson 6 – 2b. Hyper-Geometric Probability Distribution. Objectives. Determine whether a probability experiment is a geometric, hypergeometric or negative binomial experiment Compute probabilities of geometric, hypergeometric and negative binomial experiments
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Lesson 6 – 2b Hyper-Geometric Probability Distribution
Objectives • Determine whether a probability experiment is a geometric, hypergeometric or negative binomial experiment • Compute probabilities of geometric, hypergeometric and negative binomial experiments • Compute the mean and standard deviation of a geometric, hypergeometric and negative binomial random variable • Construct geometric, hypergeometric and negative binomial probability histograms
Vocabulary • Trial – each repetition of an experiment • Success – one assigned result of a binomial experiment • Failure – the other result of a binomial experiment • PDF – probability distribution function • CDF – cumulative (probability) distribution function, computes probabilities less than or equal to a specified value
Criteria for a Hyper-Geometric Probability Experiment An experiment is said to be a hyper-geometric experiment provided: • The experiment is performed a fixed number of times. Each repetition is called a trial. • The trials are dependent • For each trial there are two mutually exclusive (disjoint) outcomes: success or failure One of the conditions of a binomial distribution was the independence of the trials so the probability of a success is the same for every trial. If successive trials are done without replacement and the sample size or population is small, the probability for each observation will vary.
Hyper-Geometric Example • If a sample has 10 stones, the probability of taking a particular stone out of the ten will be 1/10. If that stone is not replaced into the sample, the probability of taking another one will be 1/9. But if the stones are replaced each time, the probability of taking a particular one will remain the same, 1/10. • When the sampling is finite (relatively small and known) and the outcome changes from trial to trial, the Hyper-geometric distribution is used instead of the Binomial distribution.
Hyper-Geometric PDF The formula for the hyper-geometric distribution is: NpCxN(1-p)Cn-x P(x) = --------------------------- x = 0, 1, 2, 3, …, n NCn Where N is the size of the population, p is the proportion of the population with a certain attribute (success), x is the number of individuals from the population selected in the sample with the attribute and n is the number selected to be in the sample (n-x is the number selected who do not have the attribute)
Hyper-Geometric PDF Mean (or Expected Value) and Standard Deviation of a Hyper-Geometric Random Variable: A hyper-geometric experiment with probability of success p has a Mean μx = np Standard Deviation σx = np(1-p)(N-n)/N-1) Where N is the size of the population, p is the proportion of the population with a certain attribute (success), x is the number of individuals from the population selected in the sample with the attribute and n is the number selected to be in the sample
Examples of Hyper-Geometric PDF • Pulling the stings on a piñata • Pulling numbers out of a hat (without replacing) • Randomly assigning the lane assignments at a race • Deal or no deal
Example 1 Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)? • N = 52; since there are 52 cards in a deck. • p = 26/52 = 0.5; since there are 26 red cards in a deck. • n = 5; since we randomly select 5 cards from the deck. • x = 2; since 2 of the cards we select are red. NpCxN(1-p)Cn-x P(x) = --------------------------- NCn 26C226C3 325 ∙ 2600 P(x) = ------------------ = --------------- 52C5 2,598,960 = 0.32513
Example 2 Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts? • N = 52; since there are 52 cards in a deck. • p = 13/52 = 0.25; since there are 13 hearts in a deck. • n = 5; since we randomly select 5 cards from the deck. • x ≤ 2; since 2 or less of the cards we select are hearts. NpCxN(1-p)Cn-x P(x) = --------------------------- NCn P(x ≤ 2) = P(0) + P(1) + P(2) = 0.9072
Summary and Homework • Summary • Small samples without replacement • Computer applet required for pdf or cdf • Not on AP • Homework • pg 343; 55