Question

The occurrence of random events (such as phone calls or e-mail messages) is often idealized using an exponential distribution. If \(\lambda\) is the average rate of occurrence of such an event, assumed to be constant over time, then the average time between occurrences is \(\lambda^{-1}\) (for example, if phone calls arrive at a rate of \(\lambda=2 /\) min, then the mean time between phone calls is \(\lambda^{-1}=\frac{1}{2} \mathrm{min}\) ). The exponential distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\) a. Suppose you work at a customer service desk and phone calls arrive at an average rate of \(\lambda_{1}=0.8 /\) min (meaning the average time between phone calls is \(1 / 0.8=1.25 \mathrm{min}\) ). The probability that a phone call arrives during the interval \([0, T]\) is \(p(T)=\int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} d t .\) Find the probability that a phone call arrives during the first 45 s \((0.75\) min) that you work at the desk. b. Now suppose that walk-in customers also arrive at your desk at an average rate of \(\lambda_{2}=0.1 /\) min. The probability that a phone $$p(T)=\int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} x} d t d s$$ Find the probability that a phone call and a customer arrive during the first 45 s that you work at the desk. c. E-mail messages also arrive at your desk at an average rate of \(\lambda_{3}=0.05 /\) min. The probability that a phone call and a customer and an e-mail message arrive during the interval \([0, T]\) is $$p(T)=\int_{0}^{T} \int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} s} \lambda_{3} e^{-\lambda_{3} u} d t d s d u$$ Find the probability that a phone call and a customer and an e-mail message arrive during the first 45 s that you work at the desk.

Short answer

Answer: The probability of receiving a phone call, a customer, and an e-mail message during the first 45 seconds is approximately 0.94%.

Step-by-step solution

Part a: Probability of a phone call during the first 45 seconds

To find the probability that a phone call arrives during the first 45 seconds (0.75 minutes), we need to evaluate the given integral: \(p(T)=\int_{0}^{T}\lambda_{1}e^{-\lambda_{1}t}dt\) with \(T=0.75\) minutes and \(\lambda_{1} = 0.8\) calls per minute. Plug in the given values: \(p(0.75)=\int_{0}^{0.75}0.8e^{-0.8t}dt\) To solve the integral, we can use integration by parts or look up the antiderivative of \(xe^{-x}\), which is \(-(x+1)e^{-x}+C\). \(p(0.75) = -\left.\frac{t+1}{0.8}e^{-0.8t}\right|_{0}^{0.75} = \left(1-\frac{1.75}{0.8}e^{-0.6}\right) \approx 0.475\) So, the probability that a phone call arrives during the first 45 seconds is approximately 47.5%.

Part b: Probability of a phone call and a customer during the first 45 seconds

The probability of independent events happening in the time interval is given as a double integral: \(p(T)=\int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} s} dt ds\) Plug in the given values, with \(\lambda_{2}=0.1\) customer per minute and \(T=0.75\) minutes: \(p(0.75)=\int_{0}^{0.75}\int_{0}^{0.75}(0.8e^{-0.8t})(0.1e^{-0.1s})\,dt\,ds\) Since the two variables \(t\) and \(s\) are independent, we can rewrite the double integral as a product of single integrals: \(p(0.75)=\left(\int_{0}^{0.75}0.8e^{-0.8t}\,dt\right)\left(\int_{0}^{0.75}0.1e^{-0.1s}\,ds\right)\) We already found the first integral in part a, and found the second integral similarly: \(p(0.75) \approx 0.475\left(1-\frac{1.75}{0.1}e^{-0.075}\right) \approx 0.068\) So, the probability that a phone call and a customer arrive during the first 45 seconds is approximately 6.8%.

Part c: Probability of a phone call, a customer, and an e-mail during the first 45 seconds

The probability of these three independent events happening in the time interval is given as a triple integral: \(p(T)=\int_{0}^{T} \int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} s} \lambda_{3} e^{-\lambda_{3} u}\, dt\, ds\, du\) Plug in the given values, with \(\lambda_{3} = 0.05\) e-mails per minute: \(p(0.75)=\int_{0}^{0.75}\int_{0}^{0.75}\int_{0}^{0.75}(0.8e^{-0.8t})(0.1e^{-0.1s})(0.05e^{-0.05u})\,dt\,ds\,du\) The variables \(t\), \(s\), and \(u\) are independent, so we can rewrite the triple integral as a product of single integrals: \(p(0.75)=\left(\int_{0}^{0.75}0.8e^{-0.8t}\,dt\right)\left(\int_{0}^{0.75}0.1e^{-0.1s}\,ds\right)\left(\int_{0}^{0.75}0.05e^{-0.05u}\,du\right)\) For the third integral, \(p(0.75) \approx 0.475\left(1-\frac{1.75}{0.1}e^{-0.075}\right)\left(1-\frac{1.75}{0.05}e^{-0.0375}\right) \approx 0.0094\) So, the probability that a phone call, a customer, and an e-mail message arrive during the first 45 seconds is approximately 0.94%.

Key concepts

Understanding Exponential Distribution

The exponential distribution is a key concept in understanding the timing of random events, such as phone calls, email messages, or customer arrivals. At its core, the exponential distribution models the time between independent events that occur continuously and randomly at a constant average rate. This rate of occurrence is denoted by the symbol \(\lambda\), which is known as the rate parameter.

Events modeled by the exponential distribution have a memoryless property, meaning the probability of an event occurring in the future is independent of when the last event happened. For instance, if the average rate of phone calls is \(\lambda = 2\) calls per minute, it does not matter when the last call was made; the probability of the next call occurring in any given minute remains the same.

The probability density function for the exponential distribution is \(f(t) = \lambda e^{-\lambda t}\), where \(t\) represents time. Understanding this formula is crucial when working out probabilities related to timing such as 'How likely is an event to happen within the next minute?'

Integral Calculus in Probability

Integral calculus is a powerful tool for computing probabilities, especially when dealing with continuous probability distributions like the exponential distribution. The integral of the probability density function over a given interval provides the probability that a random event will occur within that interval.

For example, to find the probability that a customer service agent receives a phone call in the first 45 seconds (\(0.75\) minutes) of work, one must integrate the probability density function \(f(t) = \lambda e^{-\lambda t}\) from \(t = 0\) to \(t = 0.75\). The integral \(\int_{0}^{T}\lambda e^{-\lambda t}dt\) gives the total probability of an event occurring within the time interval up to \(T\).

The solution to this integral involves finding the antiderivative of \(\lambda e^{-\lambda t}\), often through techniques like integration by parts or by recognizing specific integral forms. Once the antiderivative is determined, you can calculate the probability by evaluating it at the ends of the interval, which provides the area under the curve of the exponential function over that range.

Rate of Occurrence in Multiple Event Scenarios

When analyzing scenarios involving multiple independent events, such as receiving phone calls, walk-in customers, and emails, the rate of occurrence becomes a multi-dimensional consideration. Each type of event has its own rate parameter, such as \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\), indicating how frequently phone calls, customers, and emails arrive, respectively.

To evaluate the joint probability that all such events happen within a certain time frame, one can use multiple integrals. For example, the probability that both a phone call and a walk-in customer arrive within the first 45 seconds is represented by the double integral \(\int_{0}^{T} \int_{0}^{T} \lambda_1 e^{-\lambda_1 t} \lambda_2 e^{-\lambda_2 s} dt ds\), which involves integrating the product of their individual probabilities over the time interval.

Similarly, add another event like an email arrival, and you'll end up with a triple integral. Here, understanding the concept of independence is crucial: the probability of simultaneous occurrences is the product of their individual probabilities because each event does not affect the occurrence of the others. Thus, integrating each event's probability density function in a multi-variable scenario gives a measure of the joint probability for all events within the specified time frame.