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2006-12-21 23:35:03 · 5 answers · asked by charan r 1 in Science & Mathematics Mathematics

5 answers

let A1,A2,A3......An are mutually exclusive and exhaustive events in a sample space and A is another event such tht p(A) not equal to zero then,
P(Aj/A) = P(Aj)P(A/Aj)
----------------
sigma P(Aj)P(A/Aj)

hope this helps..

2006-12-22 03:50:38 · answer #1 · answered by For peace 3 · 0 0

Bayes' theorem (also known as Bayes' rule or Bayes' law) is a result in probability theory, which relates the conditional and marginal probability distributions of random variables. In some interpretations of probability, Bayes' theorem tells how to update or revise beliefs in light of new evidence: a posteriori.

The probability of an event A conditional on another event B is generally different from the probability of B conditional on A. However, there is a definite relationship between the two, and Bayes' theorem is the statement of that relationship.

As a formal theorem, Bayes' theorem is valid in all interpretations of probability. However, frequentist and Bayesian interpretations disagree about the kinds of things to which probabilities should be assigned in applications: frequentists assign probabilities to random events according to their frequencies of occurrence or to subsets of populations as proportions of the whole; Bayesians assign probabilities to propositions that are uncertain. A consequence is that Bayesians have more frequent occasion to use Bayes' theorem. The articles on Bayesian probability and frequentist probability discuss these debates at greater length.
http://en.wikipedia.org/wiki/Bayes'_theorem

2006-12-21 23:39:20 · answer #2 · answered by cajadman 3 · 0 0

@ cidyah. Your answer looks super different than for errors in circumstances 2 and 5. In case 2 the wonderful calculation would desire to be 0.031494. In case 5, you will desire to have that the risk of B picking 0 white and 5 black balls as C(a million,0)*C(5,5) / C(6,5) = a million/6, no longer 5/6. thus for case 5 the risk would desire to be (25/252)*(a million/6) = 25/1512 = 0.0165343. as a consequence i'm getting an exceedingly final risk of 0.212585 / 0.4499715 = 0.472441. i will delete my submit once you have made the corrections. :) (i could have despatched you an digital mail yet that wasn't accessible.)

2016-12-11 14:13:36 · answer #3 · answered by chaplean 4 · 0 0

http://en.wikipedia.org/wiki/Bayes%27_theorem

2006-12-21 23:42:02 · answer #4 · answered by Anonymous · 0 0

P(H|D) = P(H intersect D) / P(D)

2006-12-21 23:45:40 · answer #5 · answered by the DoEr 3 · 0 1

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