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Unformatted text preview: Introduction to Information Theory (67548) November 27, 2008 Assignment 0: Solution Lecturer: Prof. Michael Werman Due: Problem 1 Probability and Conditional Probability Part (a) Leting A be the event that there is a girl in the family, and B be the event that there is a boy in the family, the answer is simply Pr( A  B ) = Pr( A ∩ B ) / Pr( B ), which is equal to (1 / 2) / (3 / 4) = 2 / 3. Part (b) Given that the boy is the first child, the second child is either a girl or a boy with equal probability, so the answer is 1 / 2. The same applies if the boy is the second child. Part (c) The problem is that this statistical experiment is not well defined. For instance, one scenario is that all families decide that boys, when available, should open the door. In this case, the fact that a boy opens the door only rules out the possibility of two girls in the family, and the answer is as in Part (a). If one of the two childs opens the door ar random, we can think of the sample space as Ω = ( { bb,bg,gb,gg }×{ 1 , 2 } , where { 1 , 2 } signifies whether the first or the second child opens the door. All 8 outcomes are equally likely. Letting A be the event that there is a girl in the family, and B the event that a boy opens the door, we have that Pr( A  B ) = P ( A ∩ B ) P ( B ) = P ( { ( bg, 1) , ( gb, 2) } ) P ( { ( bb, 1) , ( bb, 2) , ( bg, 1) , ( gb, 2) } ) = 2 / 8 4 / 8 = 1 2 . Problem 2 Geometric Distribution Part (a) By the definition of a geometric distribution, Pr( X = k ) = (1 p ) k 1 p . The expectation is equal to E [ X ] = ∞ X k =1 k Pr( X = k ) = ∞ X k =1 k (1 p ) k 1 p = p ∞ X k =1 k (1 p ) k 1 ....
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This note was uploaded on 12/10/2009 for the course CS 67543 taught by Professor Michaelwerman during the Spring '08 term at Hebrew University of Jerusalem.
 Spring '08
 MichaelWerman

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