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solution04

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

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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. 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 ] = 0 . 3(1 - P [ A ]) + 0 . 5 P [ A ] P [ A ] = 0 . 3750 b. P [ ¯ A ] = 1 - 0 . 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 ! ( p (1 - a )) k 2 (1 - p (1 - a )) n - k 2 d. This can be modeled 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

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

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