HW1_solutions_b

HW1_solutions_b - HW#1, Solutions AMS 311 Probability...

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HW#1, Solutions AMS 311 Probability Theory (Spring 2011) (1). (8 points) (give the student 5 points if he/she shows a venn diagram to indicate it but no theoretical proof) Proof. Based on Proposition P(A B) = P(A) + P(B) – P(A B), we have P(A B) = P(A) + P(B) – P(A B). Besides, it is always true that P(A B) 1, so P(A B) P(A) + P(B) 1 . (2). (15 points) (a) There are different ways to define the sample space. For example, if label the 6 people and their laptop from 1 to 6, and the outcome set could be {123456, 135246, 321654,….}, which is a permutation of 1 to 6. (b) Based on notation in (a), P=#{123456}/ |S |=1/6! The question could be more interesting if I have asked “Find the probability that none of them gets her/his own laptop.” (c) Based on notation in (a), P=5!/6!=1/6 Another thought is “Joe randomly selects one of the 6 laptops, what’s the chance he gets his own”, of course it’s 1/6. (3). (20 points)
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This note was uploaded on 05/09/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

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HW1_solutions_b - HW#1, Solutions AMS 311 Probability...

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