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In a list of 15 households, 9 own homes and 6 do not own homes. Four households are randomly selected from these 15 households. Find the probability that the number of households in these 4 who own homes is a. exactly 3 b. at most \(\overline{1}\) c. exactly 4

Short Answer

Expert verified
The probability that exactly 3 households own homes is calculated using the binomial probability formula. The same formula is used to calculate the probabilities for 'at least 1' and 'exactly 4' homeowners, but keep in mind that 'at least 1' means that you should sum up the probabilities of 0 and 1 homeowners. Once the calculations are done, the probabilities are obtained.

Step by step solution

01

Understand and Analyse The Binomial Random Variable

Here, we have a binomial random variable with n=4 trials (since 4 households are randomly selected). We also have the success probability \(p = \frac{9}{15}\) (probability that a household owns a home) and it follows, failure probability \(q = 1-p = \frac{6}{15}\) (probability that a household does not own a home). Our task is to find the probability PMF (Probability Mass Function) of different values of random variable X - number of households owning homes among the 4 selected.
02

Calculate the Probability When Exactly 3 Households Own Homes

To find the probability that exactly 3 households own homes, we use the formula for binomial probability: \(Pr(X=k) = C(n, k) * (p^k) * (q^{n - k})\), where \(C(n, k)\) is the binomial coefficient. So, in this exercise, k=3, n=4, p=\(\frac{9}{15}\), and q=\(\frac{6}{15}\). Substituting these values in the formula, we get the desired probability.
03

Calculate the Probability When At Most 1 Home is Owned

'At most 1' means that k can be 0 or 1. So, we need to calculate two probabilities and sum them up. We use the same binomial formula, first for k=0 and then k=1, and sum up the obtained probabilities.
04

Calculate the Probability When Exactly 4 Households Own Homes

We again use the binomial formula. Here, k = 4, n = 4, and the success and failure probabilities remain the same. This gives us the probability that all four households selected own homes.

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Key Concepts

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

Probability Mass Function
The Probability Mass Function, often abbreviated as PMF, is a crucial concept when dealing with discrete random variables such as those found in a binomial distribution. PMF helps us understand how the probabilities are distributed over different possible values of a random variable. For a binomial distribution, the PMF gives you the probability of achieving a specific number of successes in a fixed number of trials. Consider n as the number of trials, k as the required number of successes, and p as the success probability for each trial. The binomial PMF is given by the formula: \[ Pr(X=k) = C(n, k) * p^k * (q^{n-k}) \]Where \(C(n, k)\) is the binomial coefficient, and q is the failure probability (1-p). This formula helps calculate the probability of getting k successes in n trials, which is particularly useful in scenarios like the exercise provided. It shows the probability of specific outcomes when selecting households that own homes from a given set.
Binomial Coefficient
The binomial coefficient is a mathematical concept that determines the number of ways to choose k successes in n trials, and it is denoted by \(C(n, k)\) or sometimes by \(\binom{n}{k} \). It forms a fundamental part of the binomial probability formula. Essentially, it calculates the number of combinations without regard to order. The binomial coefficient can be calculated using the formula:\[ C(n, k) = \frac{n!}{k!(n-k)!} \]Where \(!\) represents the factorial operation. For instance, if you want to find the number of ways to choose 3 households that own homes out of 4 selected ones, you would calculate \(C(4, 3)\). This part of the binomial formula shows how combinations affect the probability outcomes in the overall calculation of the PMF. By understanding how to use the binomial coefficient, students can better understand how different combinations of successes and failures affect overall probability in binomial distributions.
Random Variable
In probability, a random variable is a variable that represents the outcomes of a statistical experiment. Specifically, in the case of the binomial distribution, a random variable can take on a fixed number of possible outcomes, each associated with a particular probability. In the exercise example, the random variable X indicates the number of households that own homes among the 4 selected. This is a discrete random variable because it can only take whole number values (0, 1, 2, 3, or 4). Random variables are essential in connecting real-world scenarios to mathematical probability models. By mapping possible outcomes to numerical values, we can apply probability and statistics to analyze and interpret data efficiently. When dealing with random variables, it becomes easier to compute expected values and variances, which, in turn, aids in making informed decisions based on probabilistic models.
Success Probability
Success probability, often denoted by p, is a fundamental element of binomial experiments. It represents the likelihood of a single trial resulting in the desired success outcome. In a binomial setting, each trial is independent, which means the success probability remains constant throughout each trial. In the given exercise, the success probability is \(p = \frac{9}{15} \), which means, of the total households considered, 9 out of 15 are home-owning. This probability is key to calculating various outcomes using the binomial formula. Each trial's outcome depends on whether it results in a success, contributing to the total number of successes. This persistent probability is crucial for determining probabilities such as exactly 3 households owning homes, as calculated in the exercise. By understanding success probability, one can better analyze situations involving repeated trials, helping in evaluating outcomes systematically.

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