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Chapter 6: Problem 13

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as 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 \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

Short Answer

Expert verified
The probability distribution of x is: \(P(x = 0) = 0.4096, P(x = 1) = 0.4096, P(x = 2) = 0.1536, P(x = 3) = 0.0256, P(x = 4) = 0.0016\). The most likely value of x is 0 or 1, as they both have the highest probability (0.4096). The probability that at least two of the four selected homeowners have earthquake insurance is 0.1808.

Step by step solution

01

1a. Calculate Probability: x = 0

\( P(x = 0) = \dbinom{4}{0}(0.2)^0(1 - 0.2)^{4 - 0} = \dbinom{4}{0}(1)(0.8)^4 = 1 \cdot 0.4096 = 0.4096 \)
02

1b. Calculate Probability: x = 1

\( P(x = 1) = \dbinom{4}{1}(0.2)^1(1 - 0.2)^{4 - 1} = \dbinom{4}{1}(0.2)(0.8)^3 = 4 \cdot 0.1024 = 0.4096 \)
03

1c. Calculate Probability: x = 2

\( P(x = 2) = \dbinom{4}{2}(0.2)^2(1 - 0.2)^{4 - 2} = \dbinom{4}{2}(0.2)^2(0.8)^2 = 6 \cdot 0.0256 = 0.1536 \)
04

1d. Calculate Probability: x = 3

\( P(x = 3) = \dbinom{4}{3}(0.2)^3(1 - 0.2)^{4 - 3} = \dbinom{4}{3}(0.2)^3(0.8) = 4 \cdot 0.0064 = 0.0256 \)
05

1e. Calculate Probability: x = 4

\( P(x = 4) = \dbinom{4}{4}(0.2)^4(1 - 0.2)^{4 - 4} = \dbinom{4}{4}(0.2)^4(1) = 1 \cdot 0.0016 = 0.0016 \) The probability distribution of x is: \(P(x = 0) = 0.4096, P(x = 1) = 0.4096, P(x = 2) = 0.1536, P(x = 3) = 0.0256, P(x = 4) = 0.0016\). #Step 2: Find the most likely value of x# To find the most likely value of x, we simply compare the probabilities for each value of x and choose the one with the highest probability.
06

2. Most likely value of x

In this case, the most likely value of x is 0 or 1, as they both have the highest probability (0.4096). #Step 3: Calculate the probability that at least two of the four selected homeowners have earthquake insurance# To find the probability that at least two of the four selected homeowners have earthquake insurance, we can sum the probabilities for x = 2, x = 3, and x = 4.
07

3. Calculate probability of at least two homeowners with insurance

\(P(x \geq 2) = P(x = 2) + P(x = 3) + P(x = 4) = 0.1536 + 0.0256 + 0.0016 = 0.1808\) So the probability that at least two of the four selected homeowners have earthquake insurance is 0.1808.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a key concept in probability theory, often used to model situations where there are exactly two possible outcomes. These outcomes are commonly referred to as "success" and "failure," and they can be applied to various real-world scenarios. In the case of our exercise, being insured can be considered a "success," while not being insured would be the "failure."

For a scenario to be modeled by a binomial distribution, it must meet certain criteria:
  • There is a fixed number of trials, denoted by \(n\). In our case, \(n = 4\) because four homeowners are being considered.
  • Each trial has two possible outcomes.
  • The probability of "success," noted as \(p\), remains constant across trials. Here, \(p = 0.2\) as 20% of the homeowners are insured.
  • The trials are independent, meaning that the outcome of one trial does not impact the others.
The formula to find the probability of exactly \(k\) successes in \(n\) trials is given by:\[P(x=k) = \binom{n}{k}p^k(1-p)^{n-k}\]where \(\binom{n}{k}\) is a binomial coefficient, representing the number of ways to pick \(k\) successes out of \(n\) trials.
Random Variable
A random variable is a concept in probability and statistics used to represent outcomes of a random event in numerical form. For our homework problem, the random variable \(x\) expresses the number of insured homeowners among the four selected.

It's essential to understand that a random variable can take on various values, each with a specific probability. In this case, \(x\) could be 0, 1, 2, 3, or 4, depending on how many of the homeowners have insurance.

To make sense of this, we calculate probabilities for each possible value of \(x\) using the binomial distribution. This provides a full probability distribution of \(x\), outlining the likelihood of each value occurring. This distribution serves as a guide to predict and analyze real-world random events in structured terms.
Combinatorics
Combinatorics is the branch of mathematics dealing with counting and arrangement possibilities. It plays a crucial role in calculating probabilities within the binomial distribution.

In the context of our exercise, combinatorics helps determine the number of different ways homeowners can be insured or not. When calculating \(P(x=k)\), the term \(\binom{n}{k}\), called a binomial coefficient, arises. It represents the number of ways to choose \(k\) successes (insured homeowners) from \(n\) trials (total homeowners).

The binomial coefficient is calculated using the formula:
  • \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(!\) denotes factorial, meaning the product of all positive integers up to that number.
Using this tool, one can systematically explore all possible outcomes in a scenario and, thus, ensure a comprehensive probability distribution. By understanding and applying combinatorics, we can solve complex probability problems with precision.

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Most popular questions from this chapter

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