Textbook Notes (363,001)
Philosophy (115)
PHL246H1 (3)
Chapter 4

# Answers to exercises in Chapter 4.docx

1 Page
66 Views

School
University of Toronto St. George
Department
Philosophy
Course
PHL246H1
Professor
Colin Howson
Semester
Winter

Description
Answers to exercises in Chapter 4 (I’m using keyboard symbols only). Ex. 1 i.That the additivity rule IV and rule III hold for Prob is Essentially in Ex. 5, Chapter 2. To see that rule I holds for Prob noEe 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) = Prob (HE) and so E rule I holds for Prob .EThat rule II is holds for Prob is Erivial. ii.a.By rule V, Prob EH|E’) = Prob (H&E)/Prob (E’) =EProb(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’) = Prob E&E’(H). ii.b.By a. Prob (H|E) = Prob Prob(H). Therefore Prob (H|E’) = Prob (H|E). E’ E’&E E E’ Ex.2 is now not on the syllabus. Ex. 3. Assume Prob(E|H) = 1 and Prob(E)>0. By Bayes’s Theorem Prob(H|E) = Prob(E|H)Prob(H)/Prob(E) = Prob(H)/Prob(E). Ex.4 i. No. It would only be a contradiction if both Prob(H|E) and Prob(H’|E) > ½, but t
More Less

Related notes for PHL246H1

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.