Bayes Theorem (Special Case)

Bayes Theorem (Special Case)

P(F│E) = \frac{P(F)\cdot P(E\mid F)}{P(F)\cdot P(E\mid F)+P({F}')\cdot P(E\mid {F}')}.

 

Using Bayes’ Theorem

a.  Start a tree diagram with branches representing  F1,F2,………….Fn.   Label each

branch with its corresponding probability.

b.  From the end of each of these branches, draw a branch for event E. Label

this branch with the probability of getting to it, P(E│Fi).

c. You now have n different paths that result in event E. Next to each path,

put its probability—the product of the probabilities that the first branch

occurs,P(Fi) and that the second branch occurs, P(E│Fi)that is, the

product P(Fi)*P(E│Fi)which equals P(Fi ∩ E).

d.  P(Fi│E)is found by dividing the probability of the branch for Fi by the

sum of the probabilities of all the branches producing event E.

 

 

One Response to “Bayes Theorem (Special Case)”

  • Bas Knuppe says:

    I had to use this for my university bachelor’s degree. Although I’m incredibly smart with an IQ of 144(testen when I was 10 years old), I needed a clearer definition of this since my university obviously assumed that we already knew everything on this matter. Luckily my intelligence allows me to look at this definition once and never forget it anymore, but for the one use it had it prooved very useful. Thanks a lot!

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