Example_class_2_slides

Example_class_2_slides

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Unformatted text preview: · · P (Ck ) = P (C1 ∪ C2 ∪ · · · ∪ Ck ) 1 − (1 − p1 )(1 − p2 ) · · · (1 − pk ) 2. (a) P (C1 ∪ C2 ∪ C3 ) = 1 − (1 − P (C1 ))(1 − P (C2 ))(1 − P (C3 )) = 1 − 3 123 ××= 234 4 (b) c c P [(C1 ∩ C2 ) ∪ C3 ] = c c 1 − [1 − P (C1 ∩ C2 )](1 − P (C3 )) = c c 1 − [1 − P (C1 )P (C2 )](1 − P (C3 )) 12 1 1 1 − (1 − × )(1 − ) = 23 4 2 = Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 Problem 5 Problem 5 In the following figure of pipes, assume that the probability of each valves being opened is p and that the statuses of all the valves are mutually independent. Find the probability that water can flow from L to R. Chan Chi Ho, Chan Tsz Hin & Shi Yun Example class 2 Solution to Problem 5 Solution From above diagram it is easily seen that current can flow from L to R through either one of the follow paths: AB, DE, ACE, DCB. Denote AB as the event that both valves A and B are opened, ABC the event that valves A,B,C are all opened, and so on. Then using inclusion-exclusion formula, P (AB ∪ DE ∪ ACE ∪ DCB ) = P (AB ) + P (DE ) + P (ACE ) + P (DCB ) −P (AB ∩...
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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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