Charlie joins a new reading club, from which he receives books to read. Suppose that books arrive as a Poisson process with rate λ=1/2 books per week. For each book, suppose that the time it takes for Charlie to finish reading is exponentially distributed with parameter μ=1/3 i.e., on average it takes Charlie 1/μ=3 weeks to finish one book. Assume that the reading times of different books are independent.
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.
Hint: when Charlie starts reading a book, the finishing time can be viewed as the first arrival from a Poisson process of rate μ=1/3, and you can apply merging and splitting of this Poisson process with other processes.
When Charlie starts a new book, what is the probability that he will finish this book without being interrupted?
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Given that Charlie receives a new book while reading another 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 average reading time of a book given that it is not interrupted? Hint: The answer is not 1/μ=3.
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When Charlie starts a new book, what is the probability that he will finish this book without being interrupted?
1) 0.4
2) 0.16
3) 1.2
1. When Charlie starts a new book, the probability that he will finish this book without being interrupted can be calculated as the probability that no new books arrive while he is reading.
Since books arrive as a Poisson process with rate λ=1/2 books per week, the time between book arrivals follows an exponential distribution with parameter λ. The probability that no new books arrive in a given time period is given by the probability density function (pdf) of the exponential distribution, which is e^(-λt).
Therefore, the probability that Charlie finishes the new book without being interrupted is given by the complement of the probability that a new book arrives while he is reading. This can be calculated as 1 - e^(-λt), where t is the average time it takes Charlie to finish reading a book (1/μ = 3 weeks).
2. Given that Charlie receives a new book while reading another book, the probability that he can finish both books without further interruption can be calculated by considering the time it takes for Charlie to finish reading the interrupted book and the time until the next arrival of a new book.
The time it takes for Charlie to finish reading the interrupted book is exponentially distributed with parameter μ = 1/3. The time until the next arrival of a new book is exponentially distributed with parameter λ = 1/2.
We can calculate the probability of finishing both books without further interruption by considering the probability that the time it takes for Charlie to finish the interrupted book is less than the time until the next arrival of a new book. This can be calculated using the cumulative distribution function (cdf) of the exponential distribution.
3. The average reading time of a book given that it is not interrupted can be calculated using the concept of conditional expectation.
The reading time of Charlie is exponentially distributed with parameter μ = 1/3. However, if the book is not interrupted, then Charlie will take exactly 3 weeks to finish it. Therefore, the average reading time of a book given that it is not interrupted is 3 weeks.
To answer these questions, we can use the concepts of Poisson processes and exponential distributions.
1. When Charlie starts a new book, what is the probability that he will finish this book without being interrupted?
Let's denote the time it takes for Charlie to finish a book as T. Since T is exponentially distributed with parameter μ=1/3, the probability distribution function (pdf) of T can be expressed as f(T) = μ * exp(-μT).
Since books arrive as a Poisson process with rate λ=1/2 per week, the arrival time of a new book can be modeled as a Poisson process with rate λ. The inter-arrival time between two consecutive books, denoted as R, follows an exponential distribution with parameter λ.
The probability that there are no interruption events during the reading of a book is equal to the probability that no new books arrive during the time it takes Charlie to finish the book. Let's denote this probability as P(no interruption).
Mathematically, P(no interruption) = P(R > T). Since R and T are both exponentially distributed, we can calculate P(no interruption) as follows:
P(no interruption) = ∫[T,∞] ∫[0,∞] f(T) * f(R) dR dT
= ∫[T,∞] μ * exp(-μT) * λ * exp(-λR) dR dT
= μ * λ * ∫[T,∞] exp(-μT) * exp(-λR) dR dT
Simplifying the above expression, we have:
P(no interruption) = μ * λ / (λ + μ)
Substituting the given values of μ=1/3 and λ=1/2, we can find the probability that Charlie will finish a book without being interrupted.
2. Given that Charlie receives a new book while reading another book, what is the probability that he can finish both books, the new one and the interrupted one, without further interruption?
Let's denote the time it takes for Charlie to finish the interrupted book as T_int and the time it takes for him to finish the new book as T_new. Both T_int and T_new are exponentially distributed with parameter μ=1/3.
The probability that Charlie can finish both books without further interruption is equal to the probability that no new books arrive during the time it takes him to finish both books. Let's denote this probability as P(no further interruption).
Mathematically, P(no further interruption) = P(R > T_int + T_new). Similar to the previous question, we can calculate P(no further interruption) as follows:
P(no further interruption) = μ * λ / (λ + μ)
Note that this probability is the same as the one calculated in the previous question, as the arrival of the new book does not affect the calculation.
3. What is the average reading time of a book given that it is not interrupted?
To calculate the average reading time of a book given no interruption, we need to find the conditional expectation of T, denoted as E(T | no interruption). This expectation can be calculated using the formula:
E(T | no interruption) = ∫[0,∞] T * f(T | no interruption) dT
Since the conditional distribution of T given no interruption is still exponential with parameter μ=1/3, the conditional expectation can be directly calculated as:
E(T | no interruption) = 1 / μ = 3
Therefore, the average reading time of a book given no interruption is 3 weeks.
To answer these questions, we can use the concept of conditional probability and the properties of Poisson and exponential distributions. Let's go through each question step by step.
1. When Charlie starts a new book, what is the probability that he will finish this book without being interrupted?
To find the probability that Charlie finishes a book without being interrupted, we need to calculate the probability that no new books arrive while he is reading. Since the arrival of books follows a Poisson process with rate λ = 1/2 books per week, we can model the number of new books arriving during his reading time as a Poisson distribution with parameter λ.
The time it takes for Charlie to finish reading a book is exponentially distributed with parameter μ = 1/3, which means the average time to finish a book is 1/μ = 3 weeks. So, the probability that no new books arrive during this time is given by the probability density function (PDF) of the exponential distribution: P(T > t), where T is the time taken to finish reading a book and t is the time duration.
The probability that Charlie finishes a book without being interrupted is therefore P(T > t), which can be calculated as:
P(T > t) = e^(-μt)
Where e is the base of the natural logarithm.
2. Given that Charlie receives a new book while reading another 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 two scenarios: Charlie receiving a new book and finishing it before the interrupted one, and Charlie receiving a new book but not finishing it before the interrupted one.
Let's denote the time it takes for Charlie to finish the interrupted book as T1 and the time it takes to finish the new book as T2.
The probability of finishing both books without further interruption can be calculated as:
P(T1 < T2) * P(T2 > T1)
Where P(T1 < T2) is the probability that Charlie finishes the interrupted book before the new book arrives, and P(T2 > T1) is the probability that he finishes the new book before finishing the interrupted book.
To calculate P(T1 < T2) and P(T2 > T1), we need to use the properties of exponential distributions. In this case, since both T1 and T2 are exponentially distributed with the same parameter μ = 1/3, we have:
P(T1 < T2) = μ / (μ + μ) = 1/2
P(T2 > T1) = μ / (2μ) = 1/2
So, the probability of finishing both books without further interruption is (1/2) * (1/2) = 1/4.
3. What is the average reading time of a book given that it is not interrupted?
To find the average reading time of a book given that it is not interrupted, we need to calculate the conditional expectation of the reading time.
The expectation (average) of an exponential distribution with parameter μ is given by 1/μ. However, in this case, we need to account for the fact that the book is not interrupted. Since the reading time of different books is independent, the average reading time of a book given that it is not interrupted is simply the reciprocal of the rate of the exponential distribution for the reading time, which is equal to μ = 1/3.
Therefore, the average reading time of a book given that it is not interrupted is 3 weeks.
Please note that in the above calculations, we assume that Charlie can only start reading a new book when the previous one is finished.