Example_class_2_slides

And since p c1 c2 p c1 p c2 p c1 c2 then

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Unformatted text preview: ely. Compute 1 2 P (C1 ∪ C2 ∪ C3 ) c c P [(C1 ∩ C2 ) ∪ C3 ] Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 111 2, 3, 4, Solution to Problem 4 Solution: 1 We first consider C1 and C2 , if the probability of at least one of them happens is P (C1 ∪ C2 ). c c P (C1 ∪ C2 ) = 1 − P ((C1 ∪ C2 )c ) = 1 − P (C1 ∩ C2 ) Since C1 and C2 are independent,P(C1 ∩ C2 ) = P (C1 )P (C2 ). And since P (C1 ∪ C2 ) = P (C1 ) + P (C2 ) − P (C1 ∩ C2 ),then we get c c P (C1 ∩ C2 ) = P ((C1 ∪ C2 )c ) = 1 − P (C1 ∪ C2 ) = 1 − P (C1 ) − P (C2 ) + P (C1 ∩ C2 ) = 1 − P (C1 ) − P (C2 ) + P (C1 )P (C2 ) = (1 − P (C1 ))(1 − P (C2 )) = c c P (C1 )P (C2 ) c c Therefore C1 and C2 are independent. Then we can calculate c c P (C1 ∪ C2 ) = 1 − P (C1 )P (C2 ) = 1 − (1 − p1 )(1 − p2 ) Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 Based on mathematical induction we can get similar result, which is that c c c C1 , C2 , · · ·, Ck are independent, c c c c c c P (C1 ∩ C2 ∩ · · · ∩ Ck ) = P (C1 )P (C2 ) · · · P (Ck ). Therefore the probability of at least one of C1 , C2 , · · ·, Ck happens is P (C1 ∪ C2 ∪ · · · ∪ Ck ), which is equal to = 1 − P [(C1 ∪ C2 ∪ · · · ∪ Ck )c ] = c c c 1 − P (C1 ∩ C2 ∩ · · · ∩ Ck ) = c c c 1 − P (C1 )P (C2 ) ·...
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This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

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