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Lesson 6 – 2b

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

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  1. Lesson 6 – 2b Hyper-Geometric Probability Distribution

  2. 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

  3. 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

  4. 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.

  5. 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.

  6. 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)

  7. 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

  8. 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

  9. 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

  10. 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

  11. Summary and Homework • Summary • Small samples without replacement • Computer applet required for pdf or cdf • Not on AP • Homework • pg 343; 55

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