PHL246H1 Chapter 4: Answers to exercises in Chapter 4.docx
Document Summary
Answers to exercises in chapter 4 (i"m using keyboard symbols only). 1 i. that the additivity rule iv and rule iii hold for probe is essentially in ex. To see that rule i holds for probe note that if h <=> h" then h&e <=> h"&e and so by rule i prob(h&e) = prob(h"&e) and so. Prob(h&e)/prob(e) = prob(h"&e)/prob(e), i. e. prob(h|e) = prob(h"|e), i. e. Prob(h&e)/prob(e) = prob(h"&e)/prob(e), i. e. prob(h|e) = prob(h"|e), i. e. probe(h) = probe(h") and so rule i holds for probe. That rule ii is holds for probe is trivial. ii. a. by rule v, probe(h|e") = probe(h&e)/probe(e") = prob(h&e|e)/prob(e"|e) = prob(h&e&e")/prob(e) divided by prob(e&e")/prob(e) = prob(h&e&e")/prob(e&e") = prob(h|e&e") = probe&e"(h). ii. b. by a. probe"(h|e) = probe"&eprob(h). It would only be a contradiction if both prob(h|e) and prob(h"|e) > , but this is impossible because if h and h" are exclusive then by the additivity rule iv prob(h or h"|e) would be greater than 1.
