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Rule 2: P(f ) = 0
Proof: S¢=f . The from Rule 1, P(f)=P(S¢)=1-P(S) = 1-1 (by Axiom 2) = 0.

Rule 3: 0£P(E)£ 1.
Proof: First, 0£ P(E) by Axiom 1.  Consider P(E¢): since it is a probability, P(E¢) ³ 0 by Axiom 1, i.e., -P(E¢)£ 0 Þ 1-P(E¢) £ 1. But from Rule 1, P(E) = 1-P(E¢ ). Therefore P(E) £ 1.

Rule 4: For any two events E₁ and E₂, P(E₁ÈE₂) = P(E₁)+P(E₂)-P(E₁ÇE₂). As a special case, if E₁ and E₂ are mutually exclusive (so that P(E₁ÇE₂)=0), then P(E₁ÈE₂) = P(E₁)+P(E₂).

As an extension, P(E₁ÈE₂ÈE₃) = P(E₁)+P(E₂)+P(E₃) - P(E₁ÇE₂) - P(E₂ÇE₃) - P(E₁Ç E₃) + P(E₁ÇE₂ÇE₃)

Rule 5: Suppose that the compound event E consists of a total of k simple events (say E₁,E₂,...,E_k). Then P(E)= P(E₁)+P(E₂)+...P(E_k). Note that each simple event is an outcome. This rule is especially useful when the number of possible outcomes is very large so that there are many compound events, but it is relatively easy to compute the probability of each simple event (outcome).

Rule 6: Suppose the sample space S has a total of N outcomes and each outcome is equally likely, i.e., each has a probability equal to 1/N. Then if an event E consists of a total of N(E) outcomes, it has probability P(E) = N(E)/N. This rule is very common in practice - in many experiments each of the simple events in S are equally likely.

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