{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ASSIGNMENT 3

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

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

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

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

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

{[ snackBarMessage ]}