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 = φ ⇒ 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
 Probability, Probability theory, Binomial distribution, k2, 0.9k, 1 0k

Click to edit the document details