Question

A county containing a large number of rural homes is thought to have \(60 \%\) of those homes insured against fire. Four rural homeowners are chosen at random from the entire population, and \(x\) are found to be insured against fire. Find the probability distribution for \(x\). What is the probability that at least three of the four will be insured?

Short answer

Answer: The probability that at least three of the four randomly chosen homeowners will be insured is 0.4752.

Step-by-step solution

Identify the parameters and information given in the problem

The problem states that 60% of rural homes are insured against fire, meaning p(insured) = 0.6 and p(not_insured) = 0.4. We are given a random sample of four rural homeowners and need to find the probability distribution for x, where x is the number of insured homeowners.

Use the binomial probability formula to find the probability distribution for x = 0, 1, 2, 3, 4

The binomial probability formula is: \(P(X=k) = \binom{n}{k} (p)^k (1-p)^{(n-k)}\) where n is the number of trials (in our case, 4 homeowners), k is the number of successes (insured homeowners), p is the probability of success (0.6), and 1-p is the probability of failure (0.4). Let's calculate the probability for x = 0, 1, 2, 3, and 4. x = 0: \(P(X=0) = \binom{4}{0} (0.6)^0 (0.4)^{(4-0)} = 1(1)(0.4)^4 = 0.0256\) x = 1: \(P(X=1) = \binom{4}{1} (0.6)^1 (0.4)^{(4-1)} = 4(0.6)(0.4)^3 = 0.1536\) x = 2: \(P(X=2) = \binom{4}{2} (0.6)^2 (0.4)^{(4-2)} = 6(0.36)(0.4)^2 = 0.3456\) x = 3: \(P(X=3) = \binom{4}{3} (0.6)^3 (0.4)^{(4-3)} = 4(0.216)(0.4) = 0.3456\) x = 4: \(P(X=4) = \binom{4}{4} (0.6)^4 (0.4)^{(4-4)} = 1(0.1296)(1) = 0.1296\)

Calculate the probability that at least three of the four homeowners will be insured

We already found the probabilities for x = 3 and x = 4. To find the probability that at least three of the four homeowners will be insured, we simply add the probabilities for x = 3 and x = 4: \(P(X\ge3) = P(X=3) + P(X=4) = 0.3456 + 0.1296 = 0.4752\)

Summarize the probability distribution and the answer for the problem

The probability distribution for x, the number of insured homeowners, is: x | P(X) --|----- 0 | 0.0256 1 | 0.1536 2 | 0.3456 3 | 0.3456 4 | 0.1296 The probability that at least three of the four randomly chosen homeowners will be insured is 0.4752.

Key concepts

Probability Distribution

In the world of statistics, a probability distribution describes how the probabilities are distributed over the values of a random variable. This tells us how likely each outcome is in a certain scenario. For our exercise, the random variable is the number of insured rural homes, denoted as \(x\). Here's what you need to understand about the probability distribution of \(x\):

• Outcomes: The outcomes range from 0 to 4, representing the number of homes insured.
• Probabilities: Each outcome has an associated probability based on various factors, like how many homes can be insured simultaneously.

The probability distribution calculated in the exercise shows the likelihood of having 0, 1, 2, 3, or 4 homes insured. Knowing these individual probabilities helps predict how many homes might be insured in a random sample of four homeowners.

Binomial Probability Formula

The binomial probability formula is used when dealing with problems involving two outcomes, such as success or failure. In our exercise, a 'success' is an insured home, and 'failure' is an uninsured home. The key formula to remember is:
\[P(X=k) = \binom{n}{k} (p)^k (1-p)^{(n-k)}\]Here's a breakdown of the terms:

• \(P(X=k)\): the probability of \(k\) successes in \(n\) trials.
• \( \binom{n}{k} \): the number of ways to choose \(k\) successes out of \(n\) trials, also known as combinations.
• \(p\): probability of a single home being insured (in our exercise, it's 0.6).
• \((1-p)\): probability of a single home not being insured (here, it’s 0.4).

By plugging these values into the formula, you find the probability for each number of insured homes, \(x = 0, 1, 2, 3, 4\). This method gives us a clear mathematical approach to identifying the probabilities of these events.

Probability of Success

In any binomial setting, the probability of success plays a crucial role. It tells us the likelihood that a single trial results as hoped. In our exercise, the probability that one home is insured (a success) is 0.6. Here are a few important points to understand about the probability of success in such scenarios:

• Success Rate: If the success rate is more than 0.5, as it is here, successes are likely to occur often.
• Influence on Distribution: The higher the probability of success, the more we expect the random variable to attain higher values.

Understanding this concept is essential for calculating how likely it is to reach any given number of successes within the sample. This example shows that with a 60% probability of any home being insured, higher counts of insured homes are more probable.

Random Sampling

Random sampling is a key concept in probability and statistics, allowing us to gain insights from a smaller group that represents a larger population. In our case, the random sampling involves selecting four rural homes to analyze the likelihood of insurance. Here's why random sampling is vital:

• Unbiased Selection: Ensures that each home has an equal chance of being chosen, leading to fair results.
• Representative: A random sample often mirrors the diversity and characteristics of the entire population.

When applying random sampling in this binomial scenario, the calculated probabilities indicate whether the sample is consistent with the known distribution—in this case, 60% of homes being insured. Understanding random sampling helps validate the results as a true reflection of the population.