/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 146 A target for a bomb is in the ce... [FREE SOLUTION] | 91Ó°ÊÓ

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A target for a bomb is in the center of a circle with radius of 1 mile. A bomb falls at a randomly selected point inside that circle. If the bomb destroys everything within \(1 / 2\) mile of its landing point, what is the probability that the target is destroyed?

Short Answer

Expert verified
The probability that the target is destroyed is \( \frac{1}{4} \).

Step by step solution

01

Understand the Problem

We need to find the probability that a bomb, landing anywhere inside a circle of radius 1 mile, destroys a target located at the circle's center. The bomb destroys everything within 1/2 mile of its landing point.
02

Define the Event and Area of Interest

The target will be destroyed if the bomb falls within 1/2 mile of the center. The area where the target will be destroyed is a smaller circle with radius 1/2 mile centered at the target.
03

Calculate the Area of the Destruction Zone

The area of a circle is given by the formula \( A = \pi r^2 \). For the destruction zone with radius 1/2 mile, the area is \( A = \pi (1/2)^2 = \pi/4 \) square miles.
04

Calculate the Area of the Circle

The area of the entire circle with radius 1 mile is given by \( A = \pi (1)^2 = \pi \) square miles.
05

Calculate the Probability

The probability that the target is destroyed is the ratio of the area of the destruction zone to the area of the circle. Thus, the probability is \( \frac{\pi/4}{\pi} = \frac{1}{4} \).

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Key Concepts

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

Geometric Probability
Geometric probability is the study of probability related to geometric figures and measurements. In simple terms, it is the likelihood of an event happening in a given geometric setting. This concept is based on comparing the sizes of different areas, lengths, or volumes within a given space.

Let's consider a practical scenario: a bomb dropped into a circular field. To find the probability that the bomb hits a specific smaller area within that field, we need geometric probability. We assess the ratio of the area in which the event is successful (destruction zone) to the total area where the event could occur (entire circle).

For this exercise, we understand the event space as the entire area of the circle with a radius of 1 mile. The success space is the smaller circle with a radius of 0.5 miles, representing the area destroyed if the bomb hits. Hence, we calculate probability by determining these two areas and forming a ratio. It's like comparing a piece of the cake (destruction zone) to the whole cake (entire circle) to gauge the likelihood of hitting the target.
Circle Area Calculation
Knowing how to calculate the area of a circle is crucial for this problem. The formula is straightforward: the area of a circle is given by \[ A = \pi r^2 \],where \( A \) is the area, \( \pi \) is the mathematical constant approximately equal to 3.14159, and \( r \) is the radius of the circle.

For the bomb problem, there are two circles we need to consider:
  • The larger circle, representing the entire area where the bomb can land, has a radius of 1 mile. Its area is \( \pi (1)^2 = \pi \) square miles.
  • The smaller circle, representing the danger zone for the target destruction, has a radius of 0.5 miles. Its area is \( \pi (0.5)^2 = \frac{\pi}{4} \) square miles.
These calculations show the importance of understanding circle area computation, enabling us to accurately determine the probability.
Random Point Selection
In probability theory, random point selection is the process of choosing a location in a space in a way that gives every point the same chance of being chosen. This concept is important to ensure fairness and uniformity in probability calculations.

In our context of a bomb randomly landing within a circle, each point inside the circle is equally likely to be the bomb's landing site. This randomness means we can use geometric probability to determine where the bomb might land with relation to the target.

Random point selection ensures that we are fair in our calculations. We're assuming there isn't any bias on where the bomb could fall, making each point inside the radius equally probable. This is why we only need the areas to calculate our probability. Each small possible "landing site" within that entire circle has an equal probability of being hit by the bomb.

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Most popular questions from this chapter

The length of life \(Y\) for fuses of a certain type is modeled by the exponential distribution, with $$f(y)=\left\\{\begin{array}{ll}(1 / 3) e^{-y / 3}, & y>0 \\\0, & \text { elsewhere }\end{array}\right.$$ (The measurements are in hundreds of hours.) a. If two such fuses have independent lengths of life \(Y_{1}\) and \(Y_{2}\), find the joint probability density function for \(Y_{1}\) and \(Y_{2}\). b. One fuse in part (a) is in a primary system, and the other is in a backup system that comes into use only if the primary system fails. The total effective length of life of the two fuses is then \(Y_{1}+Y_{2} .\) Find \(P\left(Y_{1}+Y_{2} \leq 1\right)\).

Let \(Y_{1}\) and \(Y_{2}\) have the joint probability density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} k\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1, \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(k\) that makes this a probability density function. b. Find \(P\left(Y_{1} \leq 3 / 4, Y_{2} \geq 1 / 2\right)\)

In Exercise 5.1 we determined that the joint distribution of \(Y_{1}\), the number of contracts awarded to firm \(\mathrm{A},\) and \(Y_{2},\) the number of contracts awarded to firm \(\mathrm{B}\), is given by the entries in the following table. $$\begin{array}{cccc}\hline & \multicolumn{3}{c} {y_{1}} \\\\\\)\cline { 2 - 4 }\\( y_{2} & 0 & 1 & 2 \\\\\hline 0 & 1 / 9 & 2 / 9 & 1 / 9 \\\1 & 2 / 9 & 2 / 9 & 0 \\\2 & 1 / 9 & 0 & 0 \\\\\hline\end{array}$$ The marginal probability function of \(Y_{1}\) was derived in Exercise 5.19 to be binomial with \(n=2\) and \(p=1 / 3 .\) Are \(Y_{1}\) and \(Y_{2}\) independent? Why?

In Exercise 5.8 , we derived the fact that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find a. the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. \(P\left(Y_{1} \leq 1 / 2 | Y_{2} \geq 3 / 4\right)\) c. the conditional density function of \(Y_{1}\) given \(Y_{2}=y_{2}\) d. the conditional density function of \(Y_{2}\) given \(Y_{1}=y_{1}\) e. \(P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)\)

In Exercise 5.38 , we determined that the joint density function for \(Y_{1}\), the weight in tons of a bulk item stocked by a supplier, and \(Y_{2}\), the weight of the item sold by the supplier, has joint density $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1 / y_{1}, & 0 \leq y_{2} \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ In this case, the random variable \(Y_{1}-Y_{2}\) measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find \(E\left(Y_{1}-Y_{2}\right)\) In this case, the random variable \(Y_{1}-Y_{2}\) measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find \(E\left(Y_{1}-Y_{2}\right)\)

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