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Chapter 5: Problem 29

Two individuals, \(A\) and \(B\), both require kidney transplants. If she does not receive a new kidney, then \(A\) will die after an exponential time with rate \(\mu_{A}\), and \(B\) after an exponential time with rate \(\mu_{B}\). New kidneys arrive in accordance with a Poisson process having rate \(\lambda\). It has been decided that the first kidney will go to \(A\) (or to \(B\).if \(B\) is alive and \(A\) is not at that time) and the next one to \(B\) (if still living). (a) What is the probability that \(A\) obtains a new kidney? (b) What is the probability that \(B\) obtains a new kidney?

Short Answer

Expert verified
In summary, the probabilities are: (a) The probability that \(A\) obtains a new kidney is given by \(P(\text{A obtains a new kidney}) = \frac{\lambda}{\lambda + \mu_A}\). (b) The probability that \(B\) obtains a new kidney is given by \(P(\text{B obtains a new kidney}) = \frac{\lambda}{\lambda + \mu_A} \cdot \frac{\lambda}{\lambda + \mu_B}\).

Step by step solution

01

Part (a): Probability that A obtains a new kidney

To find the probability that \(A\) obtains a new kidney, we need to find the probability that a kidney arrives before \(A\) dies. We know that the time until \(A\) dies follows an exponential distribution with rate \(\mu_A\), and the time until new kidneys arrive follows a Poisson process with rate \(\lambda\). The probability that A obtains a new kidney is then the probability that the exponential random variable with rate \(\lambda\) (time to arrival of the first kidney) is less than the exponential random variable with rate \(\mu_A\) (time until A dies). Mathematically, we can write this as: \[P(\text{A obtains a new kidney}) = P(T_\lambda < T_{\mu_A})\] By considering the properties of exponential distributions, we can calculate this probability as: \[P(T_\lambda < T_{\mu_A}) = \frac{\lambda}{\lambda + \mu_A}\]
02

Part (b): Probability that B obtains a new kidney

Now, let's find the probability that \(B\) obtains a new kidney. For this to happen, the following events must occur: 1. \(A\) receives a new kidney before dying (which we found in part a). 2. The second kidney arrives before \(B\) dies. Using conditional probability, we can write this as: \[P(\text{B obtains a new kidney}) = P(T_{\mu_A} < T_\lambda < T_{\mu_A} + T_{\mu_B} | T_\lambda < T_{\mu_A})\] Considering the memoryless property of exponential distributions and using the result from part (a), we can calculate this probability as: \[P(\text{B obtains a new kidney}) = \frac{\lambda}{\lambda + \mu_A} \cdot \frac{\lambda}{\lambda + \mu_B}\] Thus, we have found the probability that both \(A\) and \(B\) obtain a new kidney. In summary, the probabilities are: (a) \(P(\text{A obtains a new kidney}) = \frac{\lambda}{\lambda + \mu_A}\) (b) \(P(\text{B obtains a new kidney}) = \frac{\lambda}{\lambda + \mu_A} \cdot \frac{\lambda}{\lambda + \mu_B}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson Process
The concept of the Poisson process is pivotal when dealing with events that occur randomly over time. It serves as a mathematical model for phenomena where events happen independently and at a constant average rate. In the kidney transplant scenario, the arrival of new kidneys represents such events, arriving in accordance with a Poisson process at rate \(\lambda\).
A key attribute of the Poisson process is its discrete count of occurrences over continuous time. This device is particularly useful for modeling scenarios where you need to predict the number of times an event will happen in a fixed interval, such as the arrival of organ donations over a month, or the number of cars passing a point on a road per hour.
In the exercise improvement context, understanding the Poisson process helps to set expectations regarding the availability of kidneys over time, improving the anticipation for when transplants may occur.
Exponential Distribution
The exponential distribution describes the time between consecutive events in a Poisson process. It is a continuous probability distribution that's often used to model waiting times or lifespans鈥攍ike in our example, where it's used to model the time until individuals \(A\) and \(B\) will die without a transplant. The parameter of the exponential distribution, often denoted as \(\mu\), is the rate at which the events (in this case, deaths) occur.
The primary characteristic to note about the exponential distribution is that it is memoryless, meaning the future probability of an event occurring is independent of the past. This has a direct impact on calculating probabilities in situations where timing is essential. For instance, if \(A\) waits a certain amount of time for a kidney, the probability that she will have to wait the same amount of time again remains unchanged.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. In our kidney transplant problem, we see this concept come into play when calculating the probability of \(B\) receiving a kidney after \(A\) has already received one. It's denoted as \(P(B|A)\), signifying the probability of event \(B\) occurring given event \(A\) has occurred.
Understanding conditional probability is imperative for gauging the likelihood of sequences of events. In the healthcare context, this tells us how to determine the probability of one patient receiving treatment based upon another patient's treatment outcome. This knowledge aids in strategic healthcare planning, such as allocating resources or predicting the need for patient care.
Memoryless Property
The memoryless property is a unique characteristic of the exponential distribution and relates directly to the lack of 'memory' in terms of past events affecting future probabilities. Simply put, in a memoryless distribution, the probability of an event occurring in the next interval is the same regardless of how much time has already passed.
In the context of the kidney transplant scenario, this property implies that the likelihood of a kidney arriving doesn't change based on how long individuals have been waiting already. Once \(A\) gets a kidney, the probability of \(B\) getting the next kidney is calculated as if \(B\) was just starting to wait for it, disregarding any time \(B\) has already waited. This insight is crucial for the students to understand how to apply the exponential distribution in real-life situations, ensuring they can accurately model time-to-event data without unnecessary complications.

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Most popular questions from this chapter

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu\). Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \quad \text { and } \quad X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(\dot{E}\left[X_{(1)}\right]\) (b) \(-\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[\dot{X}_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

The random variable whose probability density function is given by $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{2} \lambda e^{\lambda x}, & \text { if } x \leqslant 0 \\ \frac{1}{2} \lambda e^{-\lambda x}, & \text { if } x>0 \end{array}\right. $$ is said to have a Laplace, sometimes called a double exponential, distribution. (a) Verify that the preceding is a probability density function. (b) Show that the distribution function of a Laplace random variable is $$ F(x)=\left\\{\begin{array}{ll} \frac{1}{2} e^{\lambda x}, & \text { if } x \leqslant 0 \\ 1-\frac{1}{2} e^{-\lambda x}, & \text { if } x>0 \end{array}\right. $$ Let \(X\) and \(Y\) be independent exponential random variables with parameter \(\lambda\). Also, let \(I\) be independent of \(X\) and \(Y\) and let it be equally likely to be 1 or \(-1\). (c) Show that \(X-Y\) is a Laplace random variable. (d) Show that \(I X\) is a Laplace random variable. (e) Show that \(W\) is a Laplace random variable, where $$ W=\left\\{\begin{array}{ll} X, & \text { if } I=1 \\ -Y, & \text { if } I=-1 \end{array}\right. $$

An average of 500 people pass the California bar exam each year. A California lawyer practices law, on average, for 30 years. Assuming these numbers remain steady, how many lawyers would you expect California to have in \(2050 ?\)

The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?

A store opens at 8 A.M. From 8 until 10 customers arrive at a Poisson rate of four an hour. Between 10 and 12 they arrive at a Poisson rate of eight an hour. From 12 to 2 the arrival rate increases steadily from eight per hour at 12 to ten per hour at 2; and from 2 to 5 the arrival rate drops steadily from ten per hour at 2 to four per hour at \(5 .\) Determine the probability distribution of the number of customers that enter the store on a given day.
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