Example_class_2_slides

P ac b 1 p a b c p ac p b p ac b 3 2 p

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Unformatted text preview: er. Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 Solution to Problem 3 Solution: 1 P (A ∩ B c ) = P (B c |A)P (A) = P (B c |A)[1 − P (Ac )] = 2 P (Ac ∩ B ) = P (Ac |B )P (B ) = Chan Chi Ho, Chan Tsz Hin & Shi Yun 3 p 5 Example class 2 2 9 3 [1 − ] = 4 5 20 3. P (Ac ∪ B ) = 1 − P (A ∩ B c ) P (Ac ) + P (B ) − P (Ac ∩ B ) 3 2 +p− p 5 5 2 p 5 = p = 1 − P (A ∩ B c ) 9 1− 20 3 20 3 8 = = 4. P (A ∩ B ) + P (Ac ∩ B ) 33 P (A ∩ B ) + × 58 = P (A ∩ B ) = = P (B ) 3 8 3 =0 20 therefore A and B are not mutually exclusive. Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 Problem 4 Problem 4 1 Say that C1 , C2 , · · ·, Ck are (mutually) independent events that have respective probabilities p1 , p2 , · · ·, pk . Argue that the probability of at least one of C1 , C2 , · · ·, Ck happens is equal to 1 − (1 − p1 )(1 − p2 ) · · · (1 − pk ) c c c HINT:(C1 ∪ C2 ∪ · · · ∪ Ck )c = C1 ∩ C2 ∩ · · · ∩ Ck 2 Let C1 , C2 , C3 be independent events with probabilities respectiv...
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