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Some parts of Califomia are particularly earthquake-prone. Suppose that in one such area, \(30 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. One possible outcome is SFSS, with probability \((.3)(.7)(.3)(.3)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

Short Answer

Expert verified
a) Binomial distribution with n=4, p=0.3. b) Histogram shows highest bar at X=1. c) Most likely value is X=1. d) Probability that at least two are insured: 0.3483.

Step by step solution

01

Understand the Probability Distribution

This problem follows a Binomial distribution because each homeowner can either have insurance or not, and there are a fixed number of trials (4 homeowners). The random variable \(X\) represents the number of successes (homeowners with insurance) out of 4, with probability of success \(p = 0.3\) for each homeowner.
02

Calculate Probability for Each \(X\) Value

For a Binomial distribution, the probability of \(X = k\) is given by the formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) where \(n = 4\), \(p = 0.3\), and \(k\) varies from \(0\) to \(4\). Calculate each outcome:- \(P(X = 0) = \binom{4}{0} (0.3)^0 (0.7)^4 = 0.2401\)- \(P(X = 1) = \binom{4}{1} (0.3)^1 (0.7)^3 = 0.4116\)- \(P(X = 2) = \binom{4}{2} (0.3)^2 (0.7)^2 = 0.2646\)- \(P(X = 3) = \binom{4}{3} (0.3)^3 (0.7)^1 = 0.0756\)- \(P(X = 4) = \binom{4}{4} (0.3)^4 (0.7)^0 = 0.0081\).
03

Draw the Probability Histogram

Plot a histogram with \(X\) values on the x-axis (0, 1, 2, 3, 4) and their corresponding probabilities on the y-axis calculated in Step 2. The height of each bar represents the probability value of \(X\). The bar for \(X = 1\) will be the tallest.
04

Determine the Most Likely Value for \(X\)

The most likely value for \(X\) is the one with the highest probability. From Step 2, \(X = 1\) has the highest probability of 0.4116.
05

Calculate Probability of At Least Two Insured

We need \(P(X \geq 2)\), which is \(P(X = 2) + P(X = 3) + P(X = 4)\). From Step 2, this is: - \(P(X = 2) = 0.2646\)- \(P(X = 3) = 0.0756\)- \(P(X = 4) = 0.0081\)Thus, \( P(X \geq 2) = 0.2646 + 0.0756 + 0.0081 = 0.3483.\)

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Key Concepts

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

Probability Distribution
In the study of probability, a probability distribution provides a mathematical framework to describe the likelihood of different outcomes for a random variable. When we work with probability distributions, we focus on two key aspects:
  • The set of possible values a random variable can take.
  • The probability associated with each value.
In this exercise, we view the situation through the lens of a binomial distribution, which captures scenarios involving a fixed number of independent trials (in this case, four homeowners) with two possible outcomes: either having earthquake insurance or not.
Here's how it applies: each homeowner being insured can be seen as a 'success', and the probability of a homeowner having insurance is 0.3. The challenge lies in finding the probability for each possible number of successes (insured homeowners) among the four selected homeowners.
The probability distribution for our random variable "X," represents this data by listing out every outcome – from none being insured (X=0) to all four being insured (X=4) – along with their respective probabilities.
Probability Histogram
A probability histogram provides a visual representation of the probability distribution of a random variable. In this problem, plotting the histogram means portraying the probabilities calculated for each possible value of the variable X - the number of insured homeowners selected out of four.
Each bar in the histogram corresponds to one of these outcomes (0, 1, 2, 3, or 4 insured homeowners) with the height of the bar indicating the probability derived from the binomial formula.
This graphically illustrates the distribution pattern – notably that "X=1" has the tallest bar, meaning it reflects the most likely scenario, having the highest probability of occurrence at 0.4116. Probability histograms help make sense of which outcomes are more common and which are more rare, conveying this information quickly at a glance.
Random Variable
A random variable is a critical concept in probability theory, representing a variable whose value results from the randomness inherent in probability-based situations. We denote this variable often as "X" in mathematical formulations.
In our exercise, the random variable X represents the number of homeowners out of four who have insurance. Each outcome of the random variable is influenced by the chance of a homeowner being insured (p=0.3) and occurs across a set number of trials (n=4).
Random variables with a binomial distribution, like X in this scenario, are subject to specific rules—a finite number of trials, two distinct outcomes per trial, and the probability of success is constant for each trial.
Understanding random variables helps us quantify and manage uncertainty in statistical models, allowing us to make predictions and informed decisions based on probabilistic events.
Probability of Success
Central to the concept of the binomial distribution is the "probability of success," denoted by "p". In this context, it refers to the likelihood of a single trial resulting in a successful outcome, which for us is a homeowner having earthquake insurance.
The given probability of success is 0.3. This means that each homeowner has a 30% chance of being insured. The binomial probability distribution leverages this probability to calculate the likelihood of various combinations of successes across multiple trials.
By understanding the probability of success, we frame our expectations and prepare to compute the chances for specific numbers of successes (like 0, 1, 2, etc.) within a set of trials. It is a fundamental concept when dealing with any form of binomially distributed probabilities, impacting decisions, risk assessments, and planning for individuals and organizations alike.

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