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Chapter 4

# Answers to exercises in Chapter 4.docx

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University of Toronto St. George

Philosophy

PHL246H1

Colin Howson

Winter

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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

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