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Chapter 4. Conditional Probability

# Chapter 4. Conditional Probability - 4 4.1 Conditional...

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§ 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 | B ) = P 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

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(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 ) . Proof: Note that P ( A B j ) = P ( A | B j ) P ( B j ) . For (i), P ( A ) = P ( A ( B 1 ∪ · · · ∪ B n )) = P (( A B 1 ) ∪ · · · ∪ ( A B n )) = n X j =1 P ( A B j ) = n X j =1 P ( A | B j ) P ( B j ) .
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