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Chapter 5: Problem 199

Two individuals agree to meet at a certain spot sometime between \(5: 00\) and 6:00 P.M. They will each wait 10 minutes starting from when they arrive. If the other person does not show up, they will leave. Assume the arrival times of the two individuals are independent and uniformly distributed over the hour-long interval, find the probability that the two will actually meet.

Short Answer

Expert verified
The probability that the two individuals will meet is \(\frac{11}{36}\).

Step by step solution

01

Identify and visualize time intervals on a graph

Since their arrival times are uniformly distributed within the hour (\(0\leq t_1, t_2 \leq 60\) minutes), we can represent their arrival times on a Cartesian graph, with \(t_1\) being the arrival time of person 1 and \(t_2\) being the arrival time of person 2. They will meet if \(|t_1-t_2|\leq 10\). We can create a graphical representation of this condition where the curve denotes the meeting condition.
02

Define the region where they meet on the graph

According to the condition \(|t_1-t_2|\leq 10\), they will meet if \(-10 \leq t_1-t_2 \leq 10\). We can represent this as two separate inequalities: 1. \(t_1-t_2 \geq -10\) or \(t_1 \geq t_2-10\) 2. \(t_1-t_2 \leq 10\) or \(t_1 \leq t_2+10\) Now, we'll find the intersection points where these inequalities are true.
03

Determine the intersection points

We have the following boundary conditions: - Intersection at \(t_1 = 0\) gives \(0 = t_2 - 10\), hence \(t_2 = 10\) - Intersection at \(t_2 = 0\) gives \(t_1 = 10\) - Intersection at \(t_1 = 60\) gives \(60 = t_2 + 10\), hence \(t_2 = 50\) - Intersection at \(t_2 = 60\) gives \(t_1 = 50\) Now we have all four intersection points: \((0, 10)\), \((10, 0)\), \((50, 60)\), and \((60, 50)\).
04

Calculate the area of the region where they meet

Now that we have identified the intersection points, we can calculate the area of the region where the two people will meet. The region is a square with side length \(60\), minus two right triangles with side lengths \(50\). The area of the square is: \(A_{square}= 60\times60 = 3600\) The area of one triangle is: \(A_{triangle} = \frac{1}{2} \times 50 \times 50 = 1250\) Since there are two triangles, the area of both triangles is: \(A_{triangles} = 2 \times 1250 = 2500\) The area of the meeting region is the area of the square minus the area of both triangles: \(A_{meeting} = A_{square} - A_{triangles} = 3600 - 2500 = 1100\)
05

Calculate the probability

Since the arrival times are uniformly distributed, the probability of them meeting is the ratio of the meeting area to the total area in the graph. The total area is the area of the square. The probability that they will meet is: \[P(meeting) = \frac{A_{meeting}}{A_{square}} = \frac{1100}{3600} = \frac{11}{36}\] Thus, the probability that the two individuals will actually meet is \(\frac{11}{36}\).

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Key Concepts

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

Uniform Distribution
When we discuss the probability of events, specifically the likelihood of two individuals meeting at a designated timeframe, understanding the concept of a uniform distribution is crucial. A uniform distribution is a type of probability distribution in which all outcomes are equally likely to occur. Imagine a flat line on a graph where every point on that line represents an event that is just as probable as any other point on the line.

In our exercise, the arrival times for both individuals are uniformly distributed between 5:00 P.M. and 6:00 P.M. This implies that each minute within this hour-long interval has an equal chance of being the arrival time for each individual. It serves as an underlying assumption that simplifies calculations, as the probability curve does not weigh certain time frames more heavily than others.
Probability Theory
Probability theory is the mathematical framework we use to analyze random phenomena and make sense of uncertainties. It consists of rules and principles that help us calculate the likelihood of different outcomes. In the textbook exercise, we are asked to determine the probability that two individuals will meet based on their independent and uniformly distributed arrival times.

Applying probability theory, we consider the entire set of possible arrival times (the sample space) and the event where their individual arrival times allow them a chance to meet. By visualizing these times on a graph and calculating areas, we use geometric probability鈥攖he area representing the event of meeting compared to the total area representing all possible events鈥攖o arrive at our final answer.
Independent Events
The concept of independent events is foundational in calculating probabilities for multiple occurrences. Independent events are those whose outcomes do not influence one another. In simpler terms, the result of one event has no bearing on the result of another. When dealing with independent events, we can analyze each event's probability without needing to consider other events.

In the problem we are examining, it is stated that the individuals' arrival times are independent. This means the time one person arrives has no impact on the time the other person arrives. Developing solutions based on this independence assumption allows us to employ straightforward computations, specifically when using the uniform distribution to calculate the probability of the two meeting.

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Most popular questions from this chapter

Two continuous random variables \(\mathrm{X}\) and \(\mathrm{Y}\) may also be jointly distributed. Suppose \((\mathrm{X}, \mathrm{Y})\) has a distribution which is uniform over a unit circle centered at \((0,0)\). Find the joint density of \((\mathrm{X}, \mathrm{Y})\) and the marginal densities of \(\mathrm{X}\) and \(\mathrm{Y}\). Are \(\mathrm{X}\) and \(\mathrm{Y}\) independent?

Conditional probabilities can often be used to calculate unconditional probabilities. Show that the marginal distribution of \(\mathrm{X}\) can be calculated from the conditional distribution of \(\mathrm{X}\) given \(\mathrm{Y}\) and the marginal distribution of \(\mathrm{Y}\).

Consider a bag containing two white and 4 black balls. If two balls are drawn at random without replacement from the bag, let \(\mathrm{X}\) and \(\mathrm{Y}\) be random variables representing the results of these two drawings. Let 0 correspond to drawing a black ball and 1 correspond to drawing a white ball. Find the joint, marginal, and conditional distributions of \(\mathrm{X}\) and \(\mathrm{Y}\).

Let a random variable Y represent the diameter of a shaft and a random variable \(\mathrm{X}\) represent the inside diameter of the housing that is intended to support the shaft. By design the shaft is to have diameter \(99.5\) units and the housing inside diameter 100 units. If the manufacturing process of each of the items is imperfect, so that in fact \(\mathrm{Y}\) is uniformly distributed over the interval \((98.5,100.5)\) and \(\mathrm{X}\) is uniformly distributed over \((99,101)\), what is the probability that a particular shaft can be successfully paired with a particular housing, when "successfully paired" is taken to mean that \(\mathrm{X}-\mathrm{h}<\mathrm{Y}<\mathrm{X}\) for some small positive quantity \(\mathrm{h}\) ? Assume that \(\mathrm{X}\) and \(\mathrm{Y}\) are independent.

Consider the experiment of tossing two tetrahedra (regular four sided polyhedron) each with sides labeled 1 to 4 . Let \(X\) denote the number on the downturned face of the first tetrahedron and \(\mathrm{Y}\) the larger of the two downturned numbers. Find the joint density of \(\mathrm{X}\) and \(\mathrm{Y}\).
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