# solution04 - Solutions to Homework 04 1 a P A ∪ B = P A P...

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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 = 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....
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solution04 - Solutions to Homework 04 1 a P A ∪ B = P A P...

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