PS05_sol - University of Illinois Spring 2010 ECE 313:...

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University of Illinois Spring 2010 ECE 313: Problem Set 5: Solutions Conditional Probability, Law of Total Probability, Bayes’ Formula 1. [Definition of Conditional Probability] P (red apple) = P (red) P (apple | red) = 0 . 15 P (apple) = P (red apple) P (red | apple) = 0 . 25 2. [Conditional Probability and Total Probability] (a) 1 (b) 0 (c) P ( BCD | E ) (d) P ( AC ) 3. [Bernoulli Bus Lines] (a) i. What is the probability that you will wait exactly k minutes for your bus? Third-order waiting time is a negative binomial random variable, p NB 3 ( k ) = ± k - 1 2 ² P 3 (1 - P ) k - 3 ii. What is your expected waiting time? E [ NB 3] = 3 P iii. What is the variance of your waiting time? Var( NB 3) = 3 ± 1 - P P 2 ² (b) The waiting time in this part is five minutes less than the waiting time in the previous part, i.e., Y = X - 5, therefore p Y ( m ) = p NB 3 ( m + 5) = ± m + 4 2 ² P 3 (1 - P ) m +2 (c) Since the Turquoise bus has come and gone, you only have to wait for two buses to arrive. The waiting time for the second arrival in a Bernoulli process is a second-order negative binomial: p NB 2 ( m ) = ± m - 1 2 ² P 2 (1 - P ) m - 2 The fact that the Turquoise bus arrived at 5:03 is irrelevant; the only thing that matters is that it has already arrived. (d) i. What is the probability that you will wait m minutes for your bus to arrive? With probability 0.25, you must wait for two buses to arrive; with probability 0.75, you only need to wait for one bus to arrive. The probability mass function for your waiting time is therefore p X ( m ) = 0 . 25 p NB 2 ( m ) + 0 . 75 p NB 1 ( m ) p X ( m ) = 0 . 25 ± m - 1 2 ² P 2 (1 - P ) m - 2 + 0 . 75 P (1 - P ) m - 1
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ii. What is your expected waiting time? E
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PS05_sol - University of Illinois Spring 2010 ECE 313:...

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