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: York University MATH 2030 3.0AF (Elementary Probability) Assignment 5 Solutions November 2008 (postponed to February 2009) Salisbury 3.2 No. 8 E [( X + Y ) 2 ] = E [ X 2 +2 XY + Y 2 ] = E [ X 2 ]+2 E [ XY ]+ E [ Y 2 ] = 3+2 2+4 = 11 3.2 No. 14 Let X be the number of floors stopped at. Because X counts something, we can use the method of indicators rather than trying to figure out the exact distribution of X . We write X = 1 A 1 + 1 A 2 + + 1 A 10 where A i is the event that the elevator stops at the i th floor (we add up the above terms by counting the number of 1s, but counting 1s is the same as counting floors). So E [ X ] = E [1 A i ] = P ( A i ) = 10 P ( A 1 ) (since all the floors have the same probability of being stopped at). A c 1 is the event that we dont stop at the first floor, or in other words, that none of the 12 passengers chooses the 1st floor....
View Full
Document
 Fall '08
 Salisbury
 Probability

Click to edit the document details