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hw1_sol

# hw1_sol - EE 351K Probability and Random Processes...

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EE 351K Probability and Random Processes FALL 2010 Instructor: Haris Vikalo [email protected] Homework 1 Solutions Problem 1 We are given that P ( A ) = 0 . 55 , P ( B c ) = 0 . 45 , and P ( A B ) = 0 . 25 . Determine P ( B ) and P ( A B ) . Solution : We have P ( B ) = 1 - P ( B c ) = 1 - 0 . 45 = 0 . 55 . Also, by rearranging the formula P ( A B ) = P ( A ) + P ( B ) - P ( A B ) , we obtain P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 55 + 0 . 55 - 0 . 25 = 0 . 85 . Problem 2 Let A and B be two sets. (a) Show that ( A c B c ) c = A B and ( A c B c ) c = A B . (b) Consider rolling a six-sided die once. Let A be the set of outcomes where an odd number comes up. Let B be the set of outcomes where a 1 or a 2 comes up. Calculate the sets on both sides of the equalities in part (a), and verify that the equalities hold. Solution : (a) Sketching Venn diagram or applying De Morgan’s law directly leads to the desired result. Let us here prove De Morgan’s law. That is, let us show ( E F ) c = E c F c for any two sets E and F . Given ω E c F c , ω must be either in E c or F c . As E F E and E F F , ω must be not in E F . Therefore, ω ( E F ) c . On the other hand, suppose ω ( E F ) c . As E F E , ω E c or ω E \ ( E F ) . Furthermore, E \ ( E F ) F c . Therefore, ω E c F c . Similarly, we can also prove

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hw1_sol - EE 351K Probability and Random Processes...

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