This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: P [ C ∪ D c ] = P [ C ] + P [ D ] − P [ C ∩ D c ] = 1 / 2 + (1 − 2 / 3) − 1 / 6 = 2 / 3 . (8) (e) By Defnition 1.7, events C and D c are independent because P [ C ∩ D c ] = 1 / 6 = (1 / 2)(1 / 3) = P [ C ] P [ D c ] . (9) 1 Problem 1.6.5 Solution For a sample space S = { 1 , 2 , 3 , 4 } with equiprobable outcomes, consider the events A 1 = { 1 , 2 } A 2 = { 2 , 3 } A 3 = { 3 , 1 } . (1) Each event A i has probability 1 / 2. Moreover, each pair of events is independent since P [ A 1 A 2 ] = P [ A 2 A 3 ] = P [ A 3 A 1 ] = 1 / 4 . (2) However, the three events A 1 ,A 2 ,A 3 are not independent since P [ A 1 A 2 A 3 ] = 0 n = P [ A 1 ] P [ A 2 ] P [ A 3 ] . (3) 2...
View
Full
Document
This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.
 Spring '08
 Staff

Click to edit the document details