# Bayesian analysis with a discrete distribution Source: Stattrek.com.

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Bayesian analysis with a discrete distribution Source: Stattrek.com

Bayes Theorem Probability of event A given event B depends not only on the relationship between A and B but on the absolute probability of A not concerning B and the absolute probability of B not concerning A. Start with --P(A k ∩B)=P(A k ).P(B|A k )

Bayes Theorem If A1, A2, A3…An are mutually exclusive events of sample space S and B is any event from the same sample space and P(B)>0. then, P(A k |B)= P(A k ). P(B|A k ) P(A 1 ). P(B|A 1 )+P(A 2 ). P(B|A 2 )+… P(A n ). P(B|A n )

Sufficient conditions for Bayes The sample space is partitioned into a set of mutually exclusive events { A1, A2,..., An }. Within the sample space, there exists an event B, for which P(B) > 0. The analytical goal is to compute a conditional probability of the form: P( A k | B ). We know at least one of the two sets of probabilities described below. 1.P( A k ∩ B ) for each A k 2.P( A k ) and P( B | A k ) for each A k

Eg. of Discrete Distribution Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year.

Eg. of Discrete Distribution The weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% P( B | A1 )of the time. When it doesn't rain, he incorrectly forecasts rain 10% P( B | A2 ) of the time. What is the probability that it will rain on the day of Marie's wedding (P(A1))? A2==does not rain on wedding.

Values for calculation P( A 1 ) = 5/365 =0.0136985 [It rains 5 days out of the year.] P( A 2 ) = 360/365 = 0.9863014 [It does not rain 360 days out of the year.] P( B | A 1 ) = 0.9 [When it rains, the weatherman predicts rain 90% of the time.] P( B | A 2 ) = 0.1 [When it does not rain, the weatherman predicts rain 10% of the time.]

Applying the Bayes theorem P(A k |B)= P(A k ). P(B|A k ) P(A 1 ). P(B|A 1 )+P(A 2 ). P(B|A 2 )+… P(A n ). P(B|A n ) P(A 1 |B)= P(A 1 ). P(B|A 1 ) P(A 1 ). P(B|A 1 )+P(A 2 ). P(B|A 2 ) P(A 1 |B)= 0.111