{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ASSIGNMENT 3 - Assignment.3 solution February 12 2010 MATH...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment.3 solution February, 12, 2010 MATH 203 1 Solution : Defining the events as A i = { the ith draw is defective. } So we are looking for finding the probability of P ( A c 1 A c 2 A 3 A 4 ) = P ( A c 1 ) P ( A c 2 | A c 1 ) P ( A 3 | A c 1 A c 2 ) P ( A 4 | A c 1 A c 2 A 3 ) = 15 20 . 14 19 . 5 18 . 4 17 = 0 . 03611971 3.66 Solution : Since A, B, and C are all mutually exclusive, we know that P ( A B ) = P ( A C ) = P ( B C ) = 0 (a) P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = . 30 + . 55 - 0 = . 85 (b) P ( A B ) = 0 (c) P ( A | B ) = P ( A B ) P ( B ) = 0 . 55 = 0 (d) P ( B C ) = P ( B ) + P ( C ) - P ( B C ) = . 55 + . 15 - 0 = . 70 (e) No, B and C are not independent events. If B and C are independent events, then P ( B | C ) = P ( B ). From the problem, we know P(B) = .55. P ( B | C ) = P ( B C ) P ( C ) = 0 . 15 = 0 . Thus, since P ( B | C ) 6 = P ( B ), events B and C are not independent. 3.72 Solution : Define the following events, and from the start we can assume that everything is conditional on neuroblastoma so B: { Child undergoes surgery } B: { Surgery is successful in curing the disease } From the problem, we know P ( B ) = . 20 and P ( C | B ) = . 95 . We also know that P ( C | B ) = P ( C B ) P ( B ) for any events B and C. Thus, P ( C B ) = P ( C B ) P ( B ) = . 95( . 20) = . 19 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Assignment.2 solution February, 3, 2010 MATH 203 3.79 Solution : Define the following events: R i : { Fish i
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}