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Chapter 6: Problem 102

Two sentries are sent to patrol a road 1 mile long. The sentries are sent to points chosen independently and at random along the road. Find the probability that the sentries will be less than 1/2 mile apart when they reach their assigned posts.

Short Answer

Expert verified
The probability is 3/4.

Step by step solution

01

Understand the Problem

The problem involves two sentries positioned randomly along a 1-mile road, and we need to determine the probability that the distance between them is less than 1/2 mile.
02

Define the Variables

Let the positions of the two sentries on the road be denoted by coordinates \((x_1, y_1)\). Since they are positioned randomly, \(x_1\) and \(y_1\) are two independent uniform random variables each ranging from 0 to 1.
03

Set Up the Inequality

We need to find the probability that the absolute difference between their positions, \(|x_1 - y_1|\), is less than 1/2 mile.
04

Visualize the Problem

Visualize a square with sides of length 1, representing all possible positions \((x_1, y_1)\) of the sentries, where \(0 \leq x_1, y_1 \leq 1\). The area within this square represents all possible configurations of the two sentries.
05

Identify the Desired Region

The condition \(|x_1 - y_1| < 1/2\) describes a band around the line \(x_1 = y_1\). This corresponds to the region within the lines \(x_1 - y_1 = -1/2\) and \(x_1 - y_1 = 1/2\) inside the square.
06

Calculate the Area of the Desired Region

The band described by \(|x_1 - y_1| < 1/2\) has a width of 1/2. This band forms a parallelogram within the square with vertices at \((0, 1/2), (1/2, 0), (1, 1/2), (1/2, 1)\). The area of this parallelogram is \(1 - (1/2)^2 = 3/4\).
07

Calculate the Probability

Since the total area of the square is 1, the probability that the two sentries are less than 1/2 mile apart is the area of the band, which is \(3/4\).

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Key Concepts

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

Random Variables
In probability and statistics, a **random variable** is essentially a numerical outcome of a random process. It can take on different values, each with a certain probability, and is used to describe phenomena whose outcomes are uncertain.
Here, in the context of the sentry problem, the positions of the two sentries along the road are considered as random variables. These positions are chosen independently, meaning the position of one sentry does not affect the position of the other.
  • Let's denote these random variables as \(x_1\) and \(y_1\), where \(x_1\) and \(y_1\) represent the positions of the first and second sentries, respectively.
  • The interval from which these positions can be chosen is between 0 and 1 mile, as the road is 1 mile long.
Random variables allow us to express complex real-world problems using mathematical and probabilistic models.
Uniform Distribution
A **uniform distribution** is a type of probability distribution where all outcomes are equally likely. For a continuous uniform distribution over an interval, each point in the interval is equally probable.
In the problem of the sentries, both \(x_1\) and \(y_1\) are uniformly distributed over the interval [0, 1]. This means each position along the 1-mile road is equally likely for each sentry.
  • Think of the interval [0, 1] as the entire length of the road, where each position (or length on the road) has the same chance of being chosen.
  • This equality of probability across the interval simplifies calculations since it creates a predictable pattern of how the random variables behave.
Uniform distribution is a fundamental concept in probability theory, making it easier to model random processes like this.
Distance Calculation
**Distance calculation** between two points is crucial in problems involving position and movement. In this exercise, we want to find the probability that the distance between the two sentries is less than 1/2 mile.
The absolute difference \(|x_1 - y_1|\) gives us the distance between their respective positions.
  • The challenge is to understand this distance conceptually since absolute values transform negative distances into positive ones.
  • Imagine both sentries walking on the road, and we're only interested in how far apart they are, not which one is ahead.
This concept is easier when visualizing the line they meet on and the conditions needed for that distance to be less than 1/2 mile.
Geometric Probability
**Geometric probability** involves finding probabilities by examining geometric shapes and areas. In scenarios like the sentry problem, where the outcomes can be represented in a geometric space, it's useful.
For the sentries, we represent their positions as points \((x_1, y_1)\) in a 2D plane, forming a square between 0 and 1 on both axes.
  • The condition \(|x_1 - y_1| < 1/2\) forms a band around the line \(x_1 = y_1\) within this square, representing all the possible configurations where sentries are within the desired distance.
  • Geometrically, this means finding the area of the parallelogram defined by this band and comparing it to that of the entire square.
      • Thus, geometric probability connects spatial and probability concepts to solve complex problems using visual aids and calculations.

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Most popular questions from this chapter

If a random variable \(U\) has a gamma distribution with parameters \(\alpha>0\) and \(\beta>0\), then \(Y=e^{U}\) [equivalently, \(U=\ln (U)]\) is said to have a log-gamma distribution. The log-gamma distribution is used by actuaries as part of an important model for the distribution of insurance claims. Let \(U\) and \(Y\) be as stated. a. Show that the density function for \(Y\) is $$f(y)=\left\\{\begin{array}{ll} {\left[\frac{1}{\Gamma(\alpha) \beta^{\alpha}}\right] y^{-(1+\beta) / \beta}(\ln y)^{\alpha-1},} & y>1 \\ 0, & \text { elsewhere } \end{array}\right.$$b. If \(\beta<1,\) show that \(E(Y)=(1-\beta)^{-\alpha}\) $$\text { c. If } \beta<.5, \text { show that } V(Y)=(1-2 \beta)^{-\alpha}-(1-\beta)^{-2 \alpha}$$

Suppose that the number of occurrences of a certain event in time interval \((0, t)\) has a Poisson distribution. If we know that \(n\) such events have occurred in \((0, t),\) then the actual times, measured from \(0,\) for the occurrences of the event in question form an ordered set of random variables, which we denote by \(W_{(1)} \leq W_{(2)} \leq \cdots \leq W_{(n)} .\left[W_{(i)} \text { actually is the waiting time from } 0\) until the \right. occurrence of the \(i \text { th event. }]\) It can be shown that the joint density function for \(W_{(1)}, W_{(2)}, \ldots, W_{(n)}\) is given by $$f\left(w_{1}, w_{2}, \ldots, w_{n}\right)=\left\\{\begin{array}{ll} \frac{n !}{t^{n}}, & w_{1} \leq w_{2} \leq \cdots \leq w_{n} \\ 0, & \text { elsewhere } \end{array}\right.$$ [This is the density function for an ordered sample of size \(n\) from a uniform distribution on the interval (0,t).] Suppose that telephone calls coming into a switchboard follow a Poisson distribution with a mean of ten calls per minute. A slow period of two minutes' duration had only four calls. Find the a. probability that all four calls came in during the first minute; that is, find \(P(W_{(4)} \leq 1\) ). b. expected waiting time from the start of the two-minute period until the fourth call.

Let \(Y_{1}\) and \(Y_{2}\) be independent Poisson random variables with means \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Find the a. probability function of \(Y_{1}+Y_{2}\). b. conditional probability function of \(Y_{1}\), given that \(Y_{1}+Y_{2}=m\).

Let \(Y\) have a uniform ( 0,1 ) distribution. Show that \(U=-2 \ln (Y)\) has an exponential distribution with mean 2.

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random variables such that each \(Y_{i}\) has a gamma distribution with parameters \(\alpha_{i}\) and \(\beta\). That is, the distributions of the \(Y\) 's might have different \(\alpha\) 's, but all have the same value for \(\beta\). Prove that \(U=Y_{1}+Y_{2}+\cdots+Y_{n}\) has a gamma distribution with parameters \(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\) and \(\beta\).
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