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Chapter 2: Q. 2.13 (page 53)

Prove thatP(EFc)=P(E)−P(EF).

Short Answer

Expert verified

Apply Axiom 3for mutually exclusive events role="math" localid="1649247228736" EFandEFc.

Step by step solution

01

Given Information.

P(EFc)=P(E)−P(EF)

02

Explanation.

For Eand Fevents within some outcome space:

PEFc=P(E)-P(EF)

Events EF& localid="1649247667317" EFcare mutually exclusive because they Xwould be an element of that intersection x∈EF⇒x∈Fand localid="1649247675420" x∈EFc⇒x∈Fcthat is a contradiction.

EF∪EFc=Ethis is the first Theoretical exercise in this section

These two facts lead to the use of the Axiom3:

PEF∪EFc=PEFc+P(EF)⇒P(E)=PEFc+P(EF)

And the right to equality is easily transformed into the wanted identity.

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