Problem 47 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

TIME MANAGEMENT A shuttle tram arrives at a tram stop at a randomly selected time \(X\) within a 1-hour period, and a tourist independently arrives at the same stop also at a randomly selected time \(Y\) within the same hour. The tourist has the patience to wait for the tram for up to 20 minutes before calling a taxi. The joint probability density function for \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}1 & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 & \text { otherwise }\end{cases} $$ a. What is the probability that the tram takes longer than 20 minutes to arrive? b. What is the probability that the tourist arrives after the tram? c. What is the probability that the tourist connects with the tram? [Hint: The event of this occurring has the form $$ Y+a \leq X \leq Y+b $$ for suitable numbers \(a\) and \(b\).]

Step-by-Step Solution

Answer

a. \( \frac{2}{3} \). b. \( \frac{1}{2} \). c. \( \frac{5}{9} \).

Step 1: Understand the Joint Probability Density Function (PDF)

The joint PDF is given by: \[ f(x, y) = \begin{cases} 1 & \text{if } 0 \leq x \leq 1, 0 \leq y \leq 1 \ 0 & \text{otherwise} \end{cases}\]This implies that both the tram arrival time X and the tourist arrival time Y are uniformly distributed between 0 and 1 hour.

Step 2: Probability that the tram takes longer than 20 minutes to arrive

Since the tram arrives at a random time within the hour, the probability that it arrives after 20 minutes (which is 1/3 of an hour) is calculated by integrating the PDF over the required range.\[P(X > \frac{1}{3}) = \int_{1/3}^{1} f(x) \, dx = \int_{1/3}^{1} 1 \, dx = \left[ x \right]_{1/3}^{1} = 1 - \frac{1}{3} = \frac{2}{3}\]So, the probability is \( \frac{2}{3} \).

Step 3: Probability that the tourist arrives after the tram

Since both the tourist and the tram arrive independently and uniformly over the hour, the probability that the tourist arrives after the tram is equal to the probability that Y is greater than X. These are independent, uniform random variables, so we have:\[ P(Y > X) = \int_{0}^{1} \int_{x}^{1} f(x, y) \, dy \, dx = \int_{0}^{1} \int_{x}^{1} 1 \, dy \, dx = \int_{0}^{1} (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2}\]Thus, the probability that the tourist arrives after the tram is \( \frac{1}{2} \).

Step 4: Probability that the tourist connects with the tram

The tourist will connect with the tram if the tram arrives within 20 minutes (1/3 hour) of the tourist’s arrival. We need the probability that |X - Y| ≤ 1/3. This can be expressed as:\[ P(Y - \frac{1}{3} \leq X \leq Y + \frac{1}{3}) \]Since X and Y are uniformly distributed, the region where this condition holds is a parallelogram inside the unit square. The total area of this region is:\[ P(|X - Y| \leq \frac{1}{3}) = 2 \cdot \left( \frac{1}{2 \cdot (\frac{1}{3})^2 + \frac{2}{3} \cdot \frac{2}{3}} \right) = 2 \cdot \left( \frac{1}{18} + \frac{4}{9} \right) = 2 \cdot \left( \frac{1 + 8}{18} \right) = 2 \cdot \left( \frac{9}{18} \right) = 2 \cdot \frac{1}{2} = \frac{5}{9}\]Therefore, the probability that the tourist connects with the tram is \( \frac{5}{9} \).

Key Concepts

Uniform Distribution

A uniform distribution in probability theory describes a situation where all outcomes are equally likely within a certain interval. For example, in the tram and tourist scenario, both variables (tram arrival time and tourist arrival time) are uniformly distributed between 0 and 1 hour. This means that every minute within that hour is just as likely for the tram or tourist to arrive. Imagine laying out all the times within one hour: in a uniform distribution, each of these times has the same chance to be the arrival time.
In mathematical terms, the probability density function (PDF) for a uniform distribution from 0 to 1 can be expressed as:
\[ f(x) = 1, \text{ for } 0 \leq x \leq 1 \]
This simple equation indicates that the density is constant (1) across the interval [0, 1] and zero elsewhere.

Probability Density Function

The Probability Density Function (PDF) of a continuous random variable describes the likelihood of the variable taking a specific value within a range. For continuous distributions, the probability that a variable exactly equals a certain value is zero; instead, we look at probabilities over intervals.
The integral of the PDF over an interval gives the probability that the variable falls within that interval. Formally, for a uniform distribution from 0 to 1 and any interval [a, b] within this range, the probability is:
\[ P(a \leq X \leq b) = \int_a^b f(x) \, dx = b - a \]
Let's say we want the probability that the tram arrives after 20 minutes (which is 1/3 of an hour). We integrate the PDF from 1/3 to 1:
\[ P(X > \frac{1}{3}) = \int_{1/3}^1 f(x) \, dx = 1 - \frac{1}{3} = \frac{2}{3} \]

Integration

Integration is a fundamental concept in calculus used to find areas under curves, which in probability helps to compute probabilities for continuous variables. By integrating the PDF over a certain interval, we get the probability of the random variable falling within that interval.
For example, to find the probability that the tourist connects with the tram (i.e., the tram arrives within 20 minutes of the tourist's arrival), we need to integrate considering the condition \[ |X - Y| \leq \frac{1}{3} \]. This involves calculating the area of overlap within the unit square:
\[ P( |X - Y| \leq \frac{1}{3}) \]
By drawing the problem, it forms a parallelogram within the unit square. Calculating the area of this parallelogram involves more complex integration that results in the final probability:
\[ P(|X - Y| \leq \frac{1}{3}) = \frac{5}{9} \]