This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 2 solution (Fall 2008) 2.3.1 2. i. In both cities: ii. In neither of the cities: iii. In exactly one of the cities: 5. i. P ( A ∪ B ) = 0 . 6 ii. P ( A ∩ B ) = P ( A ) + P ( B ) P ( A ∪ B ) = 0 . 4 + 0 . 3 . 6 = 0 . 1 iii. P ( A B ) = p ( A ) p ( A ∩ B ) = 0 . 4 . 1 = 0 . 3 1 2.4.5 1. P ( A ) = P (Machine A produce a batch with no defectives) = 0 . 95 P ( B ) = P (Machine A produce a batch with no defectives) = 0 . 92 P ( A ∩ B ) = P (Both machine A and machine B produce a batch with no defectives) = 0 . 88 i. P ( A ∪ B ) = P ( A ) + P ( B ) P ( A ∩ B ) = 0 . 95 + 0 . 92 . 88 = 0 . 99 ii. P ( A ∪ B ) P ( A ∩ B ) = 0 . 99 . 88 = 0 . 11 5. i. P = 6 36 = 1 6 . ii. P = 15 36 = 5 12 . 7. i. 1 3 since originally the player chooses 1 out of 3. ii. 2 3 . Let A be the event that we select big prize for the first trial. Let B be the event that we select big prize when we change the original selection. P ( B ) = P ( B  A ) · P ( A ) + P ( B  A c ) · P ( A c...
View
Full
Document
This note was uploaded on 04/16/2009 for the course STAT 401 taught by Professor Akritas during the Spring '00 term at Penn State.
 Spring '00
 Akritas

Click to edit the document details