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Unformatted text preview: Probability Theory and Statistics (EE/TE 3341) Homework 2 Solutions Problem Solutions : Yates and Goodman, 1.6.3 1.6.7 1.7.1 1.7.3 1.8.1 1.8.2 1.8.7 1.9.1 1.9.5 1.10.1 and 1.10.4 Problem 1.6.3 Solution (a) Since A and B are disjoint, P [ A ∩ B ] = 0. Since P [ A ∩ B ] = 0, P [ A ∪ B ] = P [ A ] + P [ B ] − P [ A ∩ B ] = 3 / 8 . (1) A Venn diagram should convince you that A ⊂ B c so that A ∩ B c = A . This implies P [ A ∩ B c ] = P [ A ] = 1 / 4 . (2) It also follows that P [ A ∪ B c ] = P [ B c ] = 1 − 1 / 8 = 7 / 8. (b) Events A and B are dependent since P [ AB ] negationslash = P [ A ] P [ B ]. (c) Since C and D are independent, P [ C ∩ D ] = P [ C ] P [ D ] = 15 / 64 . (3) The next few items are a little trickier. From Venn diagrams, we see P [ C ∩ D c ] = P [ C ] − P [ C ∩ D ] = 5 / 8 − 15 / 64 = 25 / 64 . (4) It follows that P [ C ∪ D c ] = P [ C ] + P [ D c ] − P [ C ∩ D c ] (5) = 5 / 8 + (1 − 3 / 8) − 25 / 64 = 55 / 64 . (6) Using DeMorgan’s law, we have P [ C c ∩ D c ] = P [( C ∪ D ) c ] = 1 − P [ C ∪ D ] = 15 / 64 . (7) (d) Since P [ C c D c ] = P [ C c ] P [ D c ], C c and D c are independent. Problem 1.6.7 Solution (a) For any events A and B , we can write the law of total probability in the form of P [ A ] = P [ AB ] + P [ AB c ] . (1) Since A and B are independent, P [ AB ] = P [ A ] P [ B ]. This implies P [ AB c ] = P [ A ] − P [ A ] P [ B ] = P [ A ](1 − P [ B ]) = P [ A ] P [ B c ] . (2) Thus A and B c are independent. 1 (b) Proving that A c and B are independent is not really necessary. Since A and B are arbitrary labels, it is really the same claim as in part (a). That is, simply reversing the labels of A and B proves the claim. Alternatively, one can construct exactly the same proof as in part (a) with the labels A and B reversed. (c) To prove that A c and B c are independent, we apply the result of part (a) to the sets A and B c . Since we know from part (a) that A and B c are independent, part (b) says that A c and B c are independent. Problem 1.7.1 Solution A sequential sample space for this experiment is a24 a24 a24 a24 a24 a24 H 1 1 / 4 a88 a88 a88 a88 a88 a88 T 1 3 / 4 a24 a24 a24 a24 a24 a24 H 2 1 / 4 T 2 3 / 4 H 2 1 / 4 a88 a88 a88 a88 a88 a88 T 2 3 / 4 • H 1 H 2 1 / 16 • H 1 T 2 3 / 16 • T 1 H 2 3 / 16 • T 1 T 2 9 / 16 (a) From the tree, we observe P [ H 2 ] = P [ H 1 H 2 ] + P [ T 1 H 2 ] = 1 / 4 . (1) This implies P [ H 1  H 2 ] = P [ H 1 H 2 ] P [ H 2 ] = 1 / 16 1 / 4 = 1 / 4 . (2) (b) The probability that the first flip is heads and the second flip is tails is P [ H 1 T 2 ] = 3 / 16....
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This note was uploaded on 01/30/2012 for the course CS 3341 taught by Professor Baron during the Fall '08 term at University of Texas at Dallas, Richardson.
 Fall '08
 baron

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