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elemprob-fall2010-page2

# elemprob-fall2010-page2 - (4 If A1 A2 are pairwise...

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Unformatted text preview: (4) If A1 , A2 , . . . are pairwise disjoint (Ai ∩ Aj = ∅ if i = j ), then ∞ P ∪∞ i=1 Ai = P(Ai ). i=1 Actually part (3) is a special case of part (4) where A3 = A4 = · · · = ∅. If ω ∈ Ω, we often write P(ω ) for P({ω }). For another example, suppose we toss a quarter, a dime, and a nickel, each a fair coin, where the result of one coin does not inﬂuence another. Here Ω = {HHH, HHT, . . . , T T T }), where the ﬁrst is the quarter, the second the dime, and the third the nickel. We have P(HHH ) = 1 and so on. 8 Proposition 1.1 P(Ac ) = 1 − P(A). Proof. This follows from 1 = P(Ω) = P(A) + P(Ac ). Proposition 1.2 P(A ∪ B ) = P(A) + P(B ) − P(A ∩ B ). A picture makes this clear. Proof. P(A) = P(A ∩ B ) + P(A ∩ B c ), P(B ) = P(A ∩ B ) + P(Ac ∩ B ), and P(A ∪ B ) = P(Ac ∩ B ) + P(A ∩ B ) + P(A ∩ B c ). Adding the ﬁrst two lines and subtracting P(A ∩ B ) from both sides gives the proof. 2 ...
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