ASSIGNMENT 3

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

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Unformatted text preview: 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 ) = . 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 ) = . 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 ....
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This note was uploaded on 06/23/2010 for the course MATH 203 taught by Professor Dr.josecorrea during the Spring '08 term at McGill.

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

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