91Ó°ÊÓ

Consider \(X\) and \(Y\) two random variables of probability densities \(p_{1}(x)\) and \(p_{2}(x),\) respectively. The random variables \(X\) and \(Y\) are said to be independent if their joint density function is given by \(p(x, y)=p_{1}(x) p_{2}(y) . \) At a drive-thru restaurant, customers spend, on average, 3 minutes placing their orders and an additional 5 minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events \(X\) and \(Y\) . If the waiting times are modeled by the exponential probability densities $$p_{1}(x)=\left\\{\begin{array}{ll}{\frac{1}{3} e^{-x / 3}} & {x \geq 0} \\\ {0} & {\text { otherwise }}\end{array}\right.\text { and } p_{2}(y)=\left\\{\begin{array}{ll}{\frac{1}{5} e^{-y / 5}} & {y \geq 0} \\\ {0} & {\text { otherwise }}\end{array}\right.$$ respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by \(P[X+Y \leq 6]=\iiint_{D} p(x, y) d x d y\) where \(D=\\{(x, y) \| x \geq 0, y \geq 0, x+y \leq 6\). Find \(P[X+Y \leq 6]\) and interpret the result.

Step by step solution

01

Define the Joint Density Function

Since the random variables \(X\) and \(Y\) are independent, the joint density function is given by the product of their individual density functions. So we have:\[ p(x, y) = p_1(x) \cdot p_2(y) = \frac{1}{3} e^{-x/3} \cdot \frac{1}{5} e^{-y/5} \] for \(x, y \geq 0\). Otherwise, it's zero.

02

Set Up the Probability Integral

The probability that \(X + Y \leq 6\) is given by the integral:\[ P[X + Y \leq 6] = \int_{0}^{6} \int_{0}^{6-x} \frac{1}{15} e^{-x/3} e^{-y/5} \, dy \, dx \] We express \(D\) as \(0 \leq y \leq 6 - x\) when \(0 \leq x \leq 6\). This sets up the limits for the integration.

03

Integrate with Respect to \(y\)

First, integrate \(p(x, y) = \frac{1}{15} e^{-x/3} e^{-y/5}\) with respect to \(y\):\[ \int_{0}^{6-x} \frac{1}{15} e^{-y/5} \, dy = \left[ -\frac{1}{3} e^{-y/5} \right]_{0}^{6-x} = \frac{1}{3}\left(1 - e^{-(6-x)/5}\right) \]

04

Integrate with Respect to \(x\)

Now integrate the result of the previous step with respect to \(x\):\[ \int_{0}^{6} \frac{1}{3} e^{-x/3} \left(1 - e^{-(6-x)/5}\right) \, dx \]This requires breaking into two parts, resulting in:\[ \int_{0}^{6} \frac{1}{3} e^{-x/3} \, dx - \int_{0}^{6} \frac{1}{3} e^{-x/3} e^{-(6-x)/5} \, dx \]

05

Evaluate Both Integrals

For the first integral:\[ \int_{0}^{6} \frac{1}{3} e^{-x/3} \, dx = \Big[-e^{-x/3} \Big]_{0}^{6} = 1 - e^{-2}\]For the second integral, using substitution:\[ \int_{0}^{6} \frac{1}{3} e^{-x/3} e^{-(6-x)/5} \, dx\] = \[- e^{-2} \int_{0}^{6} e^{x/15} \, dx\] = \[ -e^{-2}\Big[15e^{x/15}\Big]_{0}^{6} = -15e^{2}\left(e^{-2} - e^{-2.4}\right)\]

06

Calculate Final Probability

After evaluating and substituting back the result of the two integrals:\[ P[X + Y \leq 6] = 1 - e^{-2} - 15 \left(e^{-2} - e^{-2.4}\right)\]Leaves us with the final probability.

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Key Concepts

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

Exponential Distribution

The exponential distribution is a continuous probability distribution used to model the time between events in a Poisson process. It's characterized by its rate parameter, often denoted by \( \lambda \). In our exercise, we have two independent events that follow different exponential distributions. Each event has its own average time, which serves to determine the rate parameter of the exponential distribution. For a random variable \( X \) following an exponential distribution with rate \( \lambda = 1/3 \), its probability density function (PDF) is given by:\[p_1(x) = \frac{1}{3} e^{-x/3} \text{ for } x \geq 0.\]The same principle applies to \( Y \), with a rate parameter \( 1/5 \):\[p_2(y) = \frac{1}{5} e^{-y/5} \text{ for } y \geq 0.\]These equations describe how likely it is for the events (e.g., ordering and payment in the drive-thru) to take a specific amount of time. Understanding exponential distribution helps in planning and optimizing processes and resources where wait times are variable.

Independent Random Variables

Random variables are said to be independent when the occurrence of one does not affect the probability of occurrence of the other. Independence is a crucial concept that helps simplify the complexity of multi-variable problems.In our context, placing an order and picking it up at a drive-thru are treated as independent random events. This independence allows us to calculate the joint probability distribution by simply multiplying their individual probability density functions (PDFs).The joint density function for independent variables \( X \) and \( Y \) is:\[p(x, y) = p_1(x) \cdot p_2(y) = \frac{1}{3} e^{-x/3} \cdot \frac{1}{5} e^{-y/5}\] This joint function serves as a foundation for further calculations, like finding the probability that the total time is less than certain limits, such as 6 minutes. Understanding independency and joint distributions aids in breaking down complex processes into manageable calculations.

Probability Density Function

A probability density function (PDF) helps determine the likelihood of a continuous random variable taking on a specific value. The PDF is crucial in understanding distributions of any continuous random variables, like time or distance.The PDF for an exponential distribution, \( p(x) \), gives us the relative likelihood for the random variable to take on a given value. For instance, the probability density function for \( X \) was given by:\[p_1(x) = \begin{cases} \frac{1}{3} e^{-x/3} & \text{if } x \geq 0 \0 & \text{otherwise}\end{cases}\]This expression tells us that for non-negative \( x \), the probability decreases exponentially as \( x \) increases, meaning shorter times are more likely. A similar PDF pattern can be observed for \( Y \). Fully grasping how PDFs function allows us to solve integrations that predict probabilities over a range, like the ones in the exercise above. They bridge the idea between a theoretical continuous distribution and practical calculations of probabilities.