Downloaded from https://www.everyspec.com AMCP 706.19 Example No. 29 (Cont'd) 6. New exgerimental result H (test C) 6. .Pr{H I Ell = 0 wad 1-head, 2-tails. Calculate (J) X (J)2 Pri{H IZ: row 7 in Tablc 3-10. PrHI E 3 } - 0. 7. Calculate { lE}. Use Eq. 3-3 8 7. Pr{E4IHI ox 0 x=0 first. The probabilities after test F•{E1H , 0.7562 X '3/8) 0.2836 B are now the prior probabilities I for test C. PrE I HV c 0.2,•,1 x 0 Total 0.2836 Use Eq. 3-39 next as in step 5. PrB I HI - 0/02836 =0 Pr-E21 )' 0. 2836/0.2836 1 1 "P,{ 3 1 HI- 0/0. 2830 -- 0. This simple illustration shows how Bayes formula cau be applied repeatedly to a prob- lem. A more test data become available.
Once th - probability of any Ej is zero, it can never again he anything but zero. Thus, it is important to allow at least some initial prior probability for any possible outenmne. A ,,