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Unformatted text preview: § 4 Conditional probability § 4.1 Introduction 4.1.1 Definition. Let B be an event such that P ( B ) > 0. Then the conditional probability P ( A  B ) of A given B is P ( A  B ) = P ( A ∩ B ) P ( B ) . 4.1.2 Physical interpretation: If the event B has been observed, what do we know about the probability of event A ? Example. Dial a mobile phone no. at random. It is natural to assume that P (respondent is male) = P (respondent is female) ≈ 1 / 2 . Ask the person whether he/she weighs over 70 kg and the answer is “yes” say. Then, naturally, P (respondent is male  over 70 kg) > P (respondent is female  over 70 kg) . Example. Toss 2 dice. Given first dice scores 1, P (2nd dice scores 6  1st dice scores 1) = P (2nd dice scores 6) , since the two events { 1st dice scores 1 } and { 2nd dice scores 6 } are independent (discussed later). 4.1.3 Note that (i) P (Ω  B ) = P (Ω ∩ B ) / P ( B ) = 1; (ii) if A ∩ C = ∅ , then P ( A ∪ C  B ) = P ( A  B ) + P ( C  B ); (iii) if A 1 ⊂ A 2 ⊂ ··· , then P ( A 1 ∪ A 2 ∪ ···  B ) = lim n →∞ P ( A n  B ) . 16 (i), (ii) and (iii) together confirm that for any fixed event B , P ( · B ) defines a probability for the events of Ω. 4.1.4 Example. (a) Toss a $1 coin and a $2 coin. You ask, “are they both heads?” I say “No”. What is the probability that they are both tails? (b) Further, you ask, “Does the $1 coin turn up a head?”. I say “No”. What is the probability that they are both tails? Solution: Take Ω = { HH,HT,TH,TT } with equally likely outcomes (each with probability 1 / 4 ). Set A = { TT } ,B 1 = { HH } ,B 2 = { HT,HH } . For (a), we have P ( A  B c 1 ) = P ( A ∩ B c 1 ) P ( B c 1 ) = P ( A ) 1 P ( B 1 ) = 1 / 4 3 / 4 = 1 / 3 . For (b) (in addition to (a)), P ( A  B c 1 ∩ B c 2 ) = P ( A ) 1 P ( B 1 ∪ B 2 ) = 1 / 4 1 / 2 = 1 / 2 . 4.1.5 Theorem. Suppose that A ⊂ B 1 ∪ B 2 ∪ ··· ∪ B n , where the events B 1 ,...,B n are mutually exclusive and have nonzero probabilities. Then (i) (Law of total probability) P ( A ) = P ( A  B 1 ) P ( B 1 ) + P ( A  B 2 ) P ( B 2 ) + ··· + P ( A  B n ) P ( B n ); (ii) (Bayes’ Theorem) P ( B j  A ) = P ( A  B j ) P ( B j ) P ( A  B 1 ) P ( B 1 ) + ··· + P ( A  B n ) P ( B n ) ....
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 Fall '08
 SMSLee
 Conditional Probability, Probability

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