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Unformatted text preview: STAT 230 Test 1 Short Solutions 1. Four students, two in STAT and two in ACTSC, get on board the elevator at the first floor of the MC building which has six floors. Each student will get off at one of the five upper level floors with equal probability. Find the probabilities of the following events. [2] (a) A = “They all get off at the same floor” Soln: P ( A ) = 5 / 5 4 = 1 / 125. [2] (b) B = “They get off at four different floors” Soln: P ( B ) = 5 (4) / 5 4 = 24 / 125. [2] (c) C = “None of them gets off at the second floor” Soln: P ( C ) = 4 4 / 5 4 . [2] (d) D = “At least one of them gets off at the second floor” Soln: P ( D ) = 1 P ( C ). [2] (e) E = “The two STAT students get off together at one floor and the two ACTSC students get off together at another floor” Soln: Choose two floors first; and then put STAT group in one floor and the ACTSC group in another. P ( E ) = ( 5 2 )( 2 1 ) / 5 4 = 4 / 125. [2] (f) F = “The two STAT students get off (not necessarily together) after both ACTSC students have already departed at lower level floors” Soln: The two STAT students could get off together or separate, the two ACTSC students could get off together or separate, resulting in four possible scenarios. In the scenario of both ACTSC students getting off together at a lower level floor but the two STAT students getting off at two different higher level floors, the number of possible arrangements is (...
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This note was uploaded on 03/03/2010 for the course MATH 235 taught by Professor Celmin during the Winter '08 term at Waterloo.
 Winter '08
 CELMIN
 Linear Algebra, Algebra, Probability

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