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Unformatted text preview: Solutions to Homework 04 1. a. P [ A B ] = P [ A ] + P [ B ] P [ A B ] = P [ A ] + P [ B ] P [ A ] P [ B ]. b. P [ A B ] = P [ A ] + P [ B ] P [ A B ] = P [ A ] + P [ B ]. (since A B = P [ A B ] = 0) 2. a. Using the theorem on total probability: P [ A ] = P [ A  A ] P [ A ] + P [ A  A ] P [ A ] = P [ A  A ](1 P [ A ]) + P [ A  A ] P [ A ] = . 3(1 P [ A ]) + 0 . 5 P [ A ] P [ A ] = 0 . 3750 b. P [ A ] = 1 . 3750 = 0 . 6250 3. a. This is a case of the Binomial probability law. P [ k errors] = n k ! p k (1 p ) n k b. Type 1 errors occur with probability pa and do not occur with probability 1 pa . Each operation can now be considered a Bernoulli trial with probability of success as pa . Then, P [ k 1 type 1 errors] = n k 1 ! ( pa ) k 1 (1 pa ) n k 1 c. Similarly, type 2 errors occur with probability p (1 a ) and do not occur with probability 1 p (1 a ). Then, P [ k 2 type 2 errors] = n k 2 !...
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This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 GELFAND

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