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MIT6_041F10_assn07_sol

# MIT6_041F10_assn07_sol - Massachusetts Institute of...

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) Problem Set 7: Solutions 1. (a) The event of the i th success occuring before the j th failure is equivalent to the i th success occurring within the first ( i + j 1) trials (since the i th success must occur no later than the trial right before the j th failure). This is equivalent to event that i or more successes occur in the first ( i + j 1) trials (where we can have, at most, ( i + j 1) successes). Let S i be the time of the i th success, F j be the time of the j th failure, and N k be the number of successes in the first k trials (so N k is a binomial random variable over k trials). So we have: i + j 1 P ( S i < F j ) = P ( N i + j 1 i ) = i + j 1 p k (1 p ) i + j 1 k k k = i (b) Let K be the number of successes which occur before the j th failure, and L be the number of trials to get to the j th failure. L is simply a j th order Pascal, with probability of 1 p (since we are now interested in the failures, not the successes.) Plugging into the formula for j th order Pascal random variable, E [ L ] = j , σ 2 p j 1 p K = (1 p ) 2 Since K = L j , p p E [ K ] = K = j, σ 2 j 1 p (1 p ) 2 (c) This expression is the same as saying we need at least 42 trials to get the 17th success. Therefore, it can be rephrased as having a maximum of 16 successes in the first 41 trials. Hence b = 41, a = 16. 2. A successful call occurs with probability p = 3 4 2 3 = 1 2 . · (a) Fred will give away his first sample on the third call if the first two calls are failures and the third is a success. Since the trials are independent, the probability of this sequence of events is simply 1 1 1 1 (1 p )(1 p ) p = · · = 2 2 2 8 (b) The event of interest requires failures on the ninth and tenth trials and a success on the eleventh trial. For a Bernoulli process, the outcomes of these three trials are independent of the results of any other trials and again our answer is 1 1 1 1 (1 p )(1 p ) p = · · = 2 2 2 8 (c) We desire the probability that L 2 , the time to the second arrival is equal to five trials. We know that p L 2 ( ) is a Pascal PMF of order 2, and we have 5 1 1 5 1 p L 2 (5) = p 2 (1 p ) 5 2 = 4 · = 2 1 2 8 Page 1 of 6

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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Fall 2010) (d) Here we require the conditional probability that the experimental value of L 2 is equal to 5, given that it is greater than 2. p L 2 (5) p L 2 (5) P ( L 2 = 5 | L 2 > 2) = = P ( L 2 > 2) 1 p L 2 (2) � � 5 5 1 p 2 (1 p ) 5 2 4 · 1 2 1 2 1 = = = 1 2 1 p 2 (1 p ) 0 1 2 6 2 1 1 2 (e) The probability that Fred will complete at least five calls before he needs a new supply is equal to the probability that the experimental value of L 2 is greater than or equal to 5.
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