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Unformatted text preview: University of Illinois Fall 2010 ECE 313: Problem Set 3: Solutions Conditional probabilities, independence, and the binomial distribution 1. [Conditional probability] Let X 1 and X 2 denote the numbers showing on the dice. P (at least one six  different) = P { at least one six and different } P { different } = P { X 1 = 6 ,X 2 6 = 6 } + P { X 1 6 = 6 ,X 2 = 6 } 5 6 = 2 · 1 6 · 5 6 5 6 = 1 3 . (Here’s an alternative solution, that makes explicit reference to the underlying probability space. Let Ω = { ( i,j ) : 1 ≤ i ≤ 6 , 1 ≤ j ≤ 6 } , where for a given outcome ( i,j ) , i denotes the number show ing on the first die and j denotes the number showing on the second die. We assume that all 36 elements of Ω have equal probability. The event (at least one six and different) can be written as { (1 , 6) , (2 , 6) , (3 , 6) , (4 , 6) , (5 , 6) , (6 , 1) , (6 , 2) , (6 , 3) , (6 , 4) , (6 , 5) } ; it has ten elements, and hence proba bility 10 36 . The event (different) has 30 elements, and hence probability 30 36 . The ratio of these probabilities...
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This note was uploaded on 02/09/2011 for the course ECE 313 taught by Professor Milenkovic,o during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Milenkovic,O

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