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Binomial and Poisson Distribution. Topic 7. Binomial Probability Distribution. Consider a sequence of independent events with only two possible outcomes called success (S) and failure (F) Example: outcome of treatment (cured/not cured) opinion (yes/no) S=yes, F=no
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Binomial Probability Distribution • Consider a sequence of independent events with only two possible outcomes called success (S) and failure (F) • Example: outcome of treatment (cured/not cured) • opinion (yes/no) S=yes, F=no • gender (boy/girl) S=boy, F=girl • Let p be the probability of S • Consider n number of such independent events. • Then the total no of success out of n such events is a random variable called Binomial r.v.
Binomial Probability Distribution Let p = probability of having a boy q = probability of having a girl Assume the outcome of the first child (boy or girl) does not affect the outcome of the second and subsequent children i.e. events are independent , we can multiply the probabilities together to get Pr(BBB) = p x p x p = Pr(3 boys) Pr(GGG) = q x q x q = Pr(No boy)
Binomial Probability For a 3-child family, 1 boy = BGG or GBG or GGB Pr(1 boy) = Pr(BGG)+Pr(GBG)+Pr(GGB) = pqq + qpq +qqp = 3pqq Similarly, Pr(2 boys) = Pr(BBG)+Pr(BGB)+Pr(GBB) = 3ppq
Example 1 Choose a group of 7 old individuals randomly from the population of 65-74 years old in US. Suppose 12.5% of the population in that age is diabetic. The total no. of persons out of the 7 selected who suffers from Diabetes has a binomial distribution. Let X = # diabetic Binomial( n = 7, p = 0.125)
Example 2 Past records indicate that 70% of patients responded to treatment. What is the probability that 16 out of the next 20 patients will respond to treatment? # patients responding ~ Bin( n = 20, p = 0.7) P(16 out of 20 responded) = 0.1304
The number of deaths attributable to typhoid fever follows a Poisson distribution at a rate of 4.6 deaths per year X = # deaths in 1 year ~ Poisson(4.6) Y = # deaths in 6 months ~ Poisson(2.3)