This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. This is a case of the Binomial probability law. P [ k errors] = parenleftBigg n k parenrightBigg 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] = parenleftBigg n k 1 parenrightBigg ( 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] = parenleftBigg n k 2 parenrightBigg ( p (1 a )) k 2 (1 p (1 a )) n k 2 d. This can be modelled as a multinomial distribution with three possible outcomes: no errors, type 1 errors and type 2 errors. These three events occur with probabilities 1 p , pa and p (1 a ) respectively. We are interested in the case where out of n operations, we have k 1 type 1 errors, k 2 type 2 errors and thus n k 1 k 2 operations without any errors. Then,operations without any errors....
View
Full
Document
 Spring '03
 WOODS

Click to edit the document details