{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Example_class_2_slides

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ) ·...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online