Assignmnt #2

# Assignmnt #2 - S tatistics Assignment#2 1 P.134 3.42 a...

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Statistics- Assignment #2 1. P.134, 3.42 a. Sample points are the most basic outcome of an experiment. A:{{1,6}, {6,1}, {2,5}, {5,2}, {3,4}, {4,3}} B: {{1,4}, {2,4}, {3,4}, {4,4}, {5,4}, {6,4}, {4,1}, {4,2}, {4,3}, {4,5}, {4,6}} The intersection of A and B is the event that occurs if we roll both a 7 and at least one of the two dice is showing a 4. Therefore the sample points for the intersection are: A B: {{3,4}, {4,3}} The union of A and B is the event that occurs if we observe a sum of 7, at least one of the two dice showing a 4 or both on a single throw of the die. The sample points in the event A B are those for which A occurs, B occurs, or both A and B occur. The sample points for the union are: A B: {{1,6}, {6,1}, {2,5}, {5,2}, {3,4}, {4,3}, {1,4}, {2,4}, {4,4}, {5,4}, {6,4}, {4,1}, {4,2}, {4,5}, {4,6}} We know that A :{Observe a sum of 7} and the compliment of A is defined as the event that occurs when A does not occur. Therefore, A c :{Observe no sum of 7}={{1,1}, {1,2}, {1,3}, {1,4}, {1,5}, {2,1}, {2,2}, {2,3}, {2,4}, {2,6} {3,1}, {3,2}, {3,3}, {3,5}, {3,6}, {4,1}, {4,2}, {4,4}, {4,5}, {4,6}, {5,1}, {5,3}, {5,4}, {5,5}, {5,6},

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{6,2} , {6,3} , {6,4} , {6,5} , {6,6}} b. We can assume that all pairs of outcomes have probability 1/36. Since the event A contains 6 sample points- all with probability 1/36 we reason that the probability of A is the sum of the probabilities of the sample points in A. P(A)= 1/36+1/36+1/36+1/36+1/36+1/36
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## This note was uploaded on 02/13/2010 for the course MATH MATH 203 taught by Professor Drjosecorrea during the Winter '08 term at McGill.

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Assignmnt #2 - S tatistics Assignment#2 1 P.134 3.42 a...

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