Consider a sample space S consisting of a countable number of outcomes. Suppose

Question

Consider a sample space S consisting of a countable number of outcomes. Suppose that a probability function P has been defined that satisfies the three Axioms of Probability. Let E be an event with non-zero probability, so that P(E)>0. For an arbitrary event F, consider the conditional probability defined as P(FE) P(FEyP(E). I probability is also a probability. That is, prove that it satisfies the Axioms of Probability. Show all work. Prove every step unless it is obvious (even then a reason is recommended).

Explanation / Answer

First axiom: Since P(F E) >= 0 and P(E) >= 0,

P(F|E) = P(F E) / P(E) >= 0.

Second axiom: Since S is countable, P(S) = 1.

Note that S is also the sample space for P(F|E).

Third axiom: If F|E G|E = ,

P(F|E) + P(G|E) = P(F E) / P(E) + P(G E) / P(E)

= (P(F E) + P(G E)) / P(E)

= (P(F U E) + P(G U E) - P((F E) (G E))) / P(E)

= (P(F U E) + P(G U E) - P((F G E))) / P(E)

= P((F U E) U P(G U E)) / P(E)

= P(F|E) U P(G|E).

Therefore conditional probability is also a probability.

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