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Chapter 5: Problem 123

The National Fire Incident Reporting Service stated that, among residential fires, \(73 \%\) are in family homes, 20\% are in apartments, and 7\% are in other types of dwellings. If four residential fires are independently reported on a single day, what is the probability that two are in family homes, one is in an apartment, and one is in another type of dwelling?

Short Answer

Expert verified
The probability is approximately 0.0895.

Step by step solution

01

Understand the Problem

We are given three categories for residential fires: family homes (73%), apartments (20%), and other dwellings (7%). We're tasked with finding the probability that, out of four reported fires, two are in family homes, one is in an apartment, and one is in another type of dwelling.
02

Define the Probability Variables

Let the probability of a fire being in a family home be \( p_1 = 0.73 \), in an apartment be \( p_2 = 0.20 \), and in other dwelling types be \( p_3 = 0.07 \).
03

Set Up the Multinomial Probability Formula

The probability distribution for this situation can be modeled by a multinomial distribution. The probability of a specific combination of outcomes \((x_1, x_2, x_3)\) is given by \[ P(X_1 = x_1, X_2 = x_2, X_3 = x_3) = \frac{n!}{x_1! x_2! x_3!} p_1^{x_1} p_2^{x_2} p_3^{x_3} \] where \(n\) is the total number of events, and \(x_1\), \(x_2\), and \(x_3\) are the counts of events in each category.
04

Apply the Values to the Formula

Here, \(n = 4\), \(x_1 = 2\), \(x_2 = 1\), \(x_3 = 1\), \(p_1 = 0.73\), \(p_2 = 0.20\), and \(p_3 = 0.07\). Applying these values: \[ P(X_1 = 2, X_2 = 1, X_3 = 1) = \frac{4!}{2!1!1!} (0.73)^2 (0.20)^1 (0.07)^1 \]
05

Calculate the Factorial and Simplify

Calculate the factorial: \(4! = 24\), \(2! = 2\), and \(1! = 1\). Thus, the factorial part of the formula is \(\frac{24}{2} = 12\).
06

Calculate the Probability Components

Calculate each probability component:- \((0.73)^2 = 0.5329\)- \((0.20)^1 = 0.20\)- \((0.07)^1 = 0.07\)Multiply these together: \(0.5329 \times 0.20 \times 0.07 = 0.0074606\).
07

Final Calculation

Multiply the factorial value by the combined probability component: \[ 12 \times 0.0074606 = 0.0895272 \] Therefore, the probability of two fires in family homes, one in an apartment, and one in another type of dwelling is approximately 0.0895.

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Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur. It ranges from 0 (impossible) to 1 (certain). In the context of residential fires, it helps to determine the likelihood of various outcomes involving fire incidents in different types of dwellings. Let's break it down:
  • Probability of a single event occurring: This is the basic likelihood associated with one particular event. For example, the probability of a fire being reported in a family home is 0.73.

  • Independent events: When the probability of each fire occurring does not affect the others, these events are considered independent. In our exercise, each fire is reported independently.

  • Joint probability: This allows us to calculate the combined probability of two or more independent events happening simultaneously. In this case, we need to calculate the probability of having two fires in family homes, one in an apartment, and one in another dwelling.
Understanding probability is key to analyzing situations involving more than one possible outcome, as it allows us to calculate and predict the likelihood of each scenario happening.
Binomial Distribution
The binomial distribution is a specific probability distribution that applies to scenarios where there are exactly two potential outcomes. It is a special case of the multinomial distribution when there are only two categories to consider.
In our residential fire problem, however, we're dealing with three categories, thus requiring the multinomial instead of the binomial distribution. But understanding the binomial is foundational:
  • Binary Outcomes: Unlike a multinomial, the binomial focuses on success/failure (or yes/no) outcomes.

  • Fixed Number of Trials: You have a set number of trials, like tossing a coin multiple times.

  • Formula Applicability: For situations needing the binomial formula, we'd use this to compute the probabilities of a particular number of successes out of a fixed number of trials.
Although our problem requires more than two outcomes, the principles of a fixed number of trials and independent probabilities seen in both distributions are fundamentally important.
Statistical Analysis
Statistical analysis is key to interpreting data and drawing meaningful conclusions. It encompasses a range of techniques, including probability distributions like the multinomial, that help us understand the patterns in data.
In the scenario with fire incidents, statistical analysis involves interpreting the distribution of reported fire events:
  • Descriptive Statistics: Initially, statistics help describe the basic features of the data, such as percentages of fires in different residences.

  • Inferential Statistics: They involve methods like the multinomial distribution to infer the probability of different configurations happening, based on the data provided.

  • Real-world Application: Statistical analysis of such events allows authorities and planners to allocate resources effectively and develop safety measures.
Through using statistical analysis, we can make informed predictions and decisions, even when dealing with complex data sets and multiple outcome paths. It provides a structured approach to understanding randomness and uncertainty in real-world events.

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Most popular questions from this chapter

In Exercise 5.9 , we determined that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find a. the marginal density functions for \(Y_{1}\) and \(Y_{2}\) b. \(P\left(Y_{2} \leq 1 / 2 | Y_{1} \leq 3 / 4\right)\) 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. \(P\left(Y_{2} \geq 3 / 4 | Y_{1}=1 / 2\right)\)

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