probability assignment 5

# probability assignment 5 - York University MATH 2030...

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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 1’s, but counting 1’s 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 don’t stop at the first floor, or in other words, that none of the 12 passengers chooses the 1st floor. Every passenger has probability 9 / 10 = 0 . 9 of not choosing the first floor, and by independence the probability that none choose the 1st floor is P ( A c 1 ) = (0 . 9) 12 .

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