Charlie joins a new reading club, from which he receives books to read. Suppose that books arrive as a Poisson process at a rate λ of books per week. For each book, the time it takes for Charlie to read it is exponentially distributed with parameter μ; i.e., on average it takes Charlie 1/μ weeks to finish one book. Assume that the reading times for different books are independent, and also independent from the book arrival process.
The problem with Charlie is that he is easily distracted. If he is reading a book when a new book arrives, he immediately turns to read the new one, and only comes back to the older book when he finishes the new book.
For all of the parts below, give your answers in terms of μ and λ, using standard notation. Enter μ as "mu" and λ as "lambda".
Hint: When Charlie starts reading a book, the total time he spends on reading it can be viewed as the first arrival from a Poisson process of rate μ, and you can then think about merging or splitting of Poisson processes.
When Charlie starts a new book, what is the probability that he can finish this book without being interrupted by a new book?
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Given that Charlie receives a new book while reading a book, what is the probability that he can finish both books, the new one and the interrupted one, without further interruption?
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What is the expected reading time of a book given that it is not interrupted?
Any ideas?
Anyone got this....?
" I found the answers to be
1.) mu/(mu+lambda)
2.) (mu/(mu+lambda))^2
3.) 1/(mu+lambda).
Sourced from course solutions found on CourseHero for Fall 2016 "
To solve this problem, we can use the concept of renewal theory. Let's break down each part of the problem and explain how to get the answers.
1. When Charlie starts a new book, what is the probability that he can finish this book without being interrupted by a new book?
First, we need to determine the distribution of Charlie's reading times before being interrupted by a new book. We know that the reading time follows an exponential distribution with parameter μ.
Let's define T as the time Charlie spends on reading a book before being interrupted. The probability that T > t (where t is a positive number) can be calculated using the exponential distribution:
P(T > t) = e^(-μ*t)
Therefore, the probability that Charlie can finish the book without being interrupted can be calculated as the complement of this probability:
P(Finish without interruption) = 1 - P(T > t)
2. Given that Charlie receives a new book while reading a book, what is the probability that he can finish both books, the new one and the interrupted one, without further interruption?
In this case, we need to consider the time Charlie spends on both books. Let's define T1 as the time spent on the interrupted book (the one he was reading when the new book arrived) and T2 as the time spent on the new book.
We know that T1 and T2 are both exponentially distributed with parameter μ. To find the probability that both books can be finished without further interruption, we need to calculate the joint probability:
P(Finish both without interruption) = P(T1 > t, T2 > t)
Using the exponential distribution, we can calculate this joint probability as:
P(T1 > t, T2 > t) = P(T1 > t) * P(T2 > t) = e^(-μ*t) * e^(-μ*t) = e^(-2μ*t)
3. What is the expected reading time of a book given that it is not interrupted?
To find the expected reading time of a book without interruption, we need to calculate the mean of the exponential distribution. The mean of an exponential distribution with parameter μ is given by:
Mean = 1 / μ
Therefore, the expected reading time of a book without interruption is 1 / μ.
In summary:
1. The probability that Charlie can finish a new book without interruption is 1 - e^(-μ*t).
2. The probability that Charlie can finish both the interrupted book and the new book without interruption is e^(-2μ*t).
3. The expected reading time of a book without interruption is 1 / μ.