Hypergeometric probability distribution

Hypergeometric probability distribution

• 1.
  NADEEM UDDIN ASSOCIATE PROFESSOR OFSTATISTICS HYPERGEOMETRIC PROBABILITY DISTRIBUTION
• 2.
  Hypergeometric Experiments: A hypergeometricexperiment is a random experiment that has the following properties: 1-The experiment consists of n repeated trials. 2-Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. 3-The probability of success, in each trial is not constant. 4-The trials are dependent.
• 3.
  Experiment-1 A recent studyfound that four out of nine houses were insured. If we select three houses from the nine without replacement and all the three are insured. Explanation: 1-The experiment consists of 3 repeated trials. 2-Each trial can result in just two possible outcomes. We call one of these outcomes a success (insured) and the other, a failure (not insured). 3-The probability of success, in each trial is not constant. 1 st trial probability would be 4 9 . 2 nd trial probability would be 3 8 . 3 rd trial probability would be 2 7 . 4-The trials are dependent.
• 4.
  Experiment-2 You have anurn of 15 balls - 5 red and 10 green. You randomly select 2 balls without replacement and count the number of red balls you have selected. Explanation: This would be a hypergeometric experiment. Note that it would not be a binomial experiment. A binomial experiment requires that the probability of success be constant on every trial. With the above experiment, the probability of a success changes on every trial. In the beginning, the probability of selecting a red ball is 5/15. If you select a red ball on the first trial, the probability of selecting a red ball on the second trial is 4/14. And if you select a green ball on the first trial, the probability of selecting a red ball on the second trial is 5/14. Note further that if you selected the balls with replacement, the probability of success would not change. It would be 5/15 on every trial. Then, this would be a binomial experiment.
• 5.
  Hypergeometric Random Variable: Ahypergeometric random variable is the number of successes that result from a hypergeometric experiment. Hypergeometric Distribution: The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Given x, N, n, and k, we can compute the hypergeometric probability based on the following formula: 𝑃 𝑥 = 𝑥 = 𝐾 𝑥 𝑁 − 𝐾 𝑛 − 𝑥 𝑁 𝑛 ; 𝑓𝑜𝑟𝑥 = 0, 1, 2, 3, … … … . , 𝑛 The mean of the distribution is equal to n × k / N . The variance is n × k × ( N - k ) × ( N - n ) / [ N 2 × ( N - 1 ) ] .
• 6.
  Where N = Sizeof population. n = Size of sample. K = Number of successes. N – K = Number of failures. x = Number of successes in the sample.
• 7.
  Use of HyperGeometricProbability Distribution 𝑝 𝑥 = 𝑥 = 𝐾 𝑥 𝑁−𝐾 𝑛−𝑥 𝑁 𝑛 Example-1 If 5 cards are dealt from a deck of 52 playing cards, what is the probability that 3 will be hearts? Solution: By using hypergeometric distribution 𝑝 𝑥 = 𝑥 = 𝐾 𝑥 𝑁−𝐾 𝑛−𝑥 𝑁 𝑛 Where N = 52 n = 5 K = 13 P(x=3) = ? 𝑃 𝑥 = 3 = 13 3 52−13 5 − 3 52 5 = 13 𝐶3×39 𝐶2 52 𝐶5 = 0.0815
• 8.
  Example-2 From 15 kidneytransplant operation,3 are fail within a year .consider a sample of 2 patients, find the probability that (a)Only 1 of the kidney transplant operations result in failure within a year. (b)All 2 of the kidney transplant operations result in failure within a year. (c)At least 1 of the kidney transplant operations result in failure within a year.
• 9.
  Solution: By using hypergeometricdistribution 𝑝 𝑥 = 𝑥 = 𝐾 𝑥 𝑁−𝐾 𝑛−𝑥 𝑁 𝑛 Where N =1 5 n = 3 K = 3 (a) P(x=1) = ? 𝑃 𝑥 = 1 = 3 1 15−3 3 − 1 15 3 = 3 𝐶1×12 𝐶2 15 𝐶3 = 0.4351 (b) P(x=2) = ? 𝑃 𝑥 = 2 = 3 2 15−3 3 − 2 15 3 = 3 𝐶2×12 𝐶1 15 𝐶3 = 0.0791
• 10.
  (c) P( x≥ 1) = ? 𝑃 𝑥 ≥ 1 = 𝑃 𝑥 = 1 + 𝑃(𝑥 = 2) P(x ≥ 1) = 0.4351 + 0.0791 (from part a and b) P(x ≥ 1) = 0.5142
• 11.
  Example-3 A small votingdistrict has 1000 female voters and 4000 male voters. A random sample of 10 voters is drawn. What is the probability exactly 7 of the voters will be male? Solution: Since the population size is large relative to the sample size, We shall approximate the hypergeometric distribution to binomial distribution. Therefore n = 10 𝑝 = 4000 5000 = 0.8(𝑚𝑎𝑙𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦) q = 0.2(female probability) 𝑃 𝑥 = 𝑥 = 𝑛 𝑐 𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑃 𝑥 = 7 = 10 𝑐7 0.8 7 0.2 10−7 = 0.2013