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Chapter 6: Problem 21

In a recent study 90 percent of the homes in the United States were found to have largescreen TVs. In a sample of nine homes, what is the probability that: a. All nine have large-screen TVs? b. Less than five have large-screen TVs? c. More than five have large-screen TVs? d. At least seven homes have large-screen TVs?

Short Answer

Expert verified
a. 0.3874, b. Add the probabilities for X=0 to 4, c. 1 - Sum for X=0 to 5, d. Sum probabilities for X=7 to 9.

Step by step solution

01

Determine the Type of Probability Distribution

This problem involves a fixed number of independent trials (homes), each with the same probability of success (having a large-screen TV). Therefore, the binomial distribution is appropriate. Here, the probability of success (p) is 0.9, and the number of trials (n) is 9.
02

Calculate the Probability for All Nine Homes

To find the probability that all nine homes have large-screen TVs, use the formula:\[P(X = 9) = \binom{9}{9} (0.9)^9 (0.1)^0\]Since \(\binom{9}{9} = 1\) and \((0.1)^0 = 1\), the expression simplifies to \[(0.9)^9 \approx 0.3874\]
03

Calculate the Complement Probability for Less Than Five Homes

To find the probability that less than five homes have large-screen TVs, calculate the sum of probabilities from 0 to 4. Each probability is given by the binomial probability formula:\[P(X = k) = \binom{9}{k} (0.9)^k (0.1)^{9-k}\]Calculate for \(k = 0, 1, 2, 3, 4\), and then find the sum.
04

Use the Complement Rule for More Than Five Homes

To find the probability that more than five homes have large-screen TVs, we use the complement rule:\[P(X > 5) = 1 - P(X \leq 5)\]Calculate \(P(X \leq 5)\) by summing the probabilities from 0 to 5 using the binomial formula, then subtract from 1.
05

Calculate the Probability for At Least Seven Homes

To find the probability that at least seven homes have large-screen TVs, sum the probabilities for seven, eight, and nine homes:\[P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)\]Use the binomial formula for each case and add the results.

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Key Concepts

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

Probability Calculation
Probability Calculation is a fundamental concept in statistics that involves determining the likelihood of a particular outcome. When thinking about probability, in any event, the basic rule is the ratio of favorable outcomes to possible outcomes. This can be easily expressed by the formula:
  • Probability of Event = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)
In the context of the problem, we are considering a binomial scenario, which means we are interested in the probability of a certain number of successes across several trials.
The probability of each outcome will depend on the fixed probability of success in each trial and the number of trials. Calculating these probabilities requires careful consideration of each possible outcome and is ideally done using formulas like the Binomial Probability Formula.
Independent Trials
In probability theory, independent trials are experiments in which the outcome of one trial does not affect the outcomes of others. This independence is crucial when dealing with binomial distributions because it assumes that each event is separate and unaffected by the others.
In our example of homes with large-screen TVs, the presence of a TV in one home doesn't impact the presence in another. This assumption of independence allows us to treat each home as a separate trial.
  • Each trial (home) has the same probability of success (having a large-screen TV), denoted as 0.9 in this case.
  • The independence ensures that the binomial distribution is applicable.
Recognizing independent trials is essential for applying the binomial probability formula accurately.
Complement Rule
The Complement Rule is an invaluable tool in probability that helps in finding the probability of the complement of an event. The complement of an event is simply all outcomes not in the event.
The rule can be stated as:
  • If an event has a probability of occurring, then the probability of it not occurring is 1 minus the probability of the event.
For example, to find out the probability that more than five homes have large-screen TVs, we can use the complement rule:
  • Calculate the probability of five or fewer having TVs (i.e., the complement).
  • Subtract this from 1 to find the probability of more than five homes.
This means you're looking at the probability of the opposite scenario to find the needed probability.
Binomial Probability Formula
The Binomial Probability Formula is used to calculate the probability of achieving exactly k successes over n independent trials with the same probability of success p in each trial.
The formula looks like:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Here鈥檚 a breakdown of the components:
  • \( \binom{n}{k} \) is the number of ways to choose k successes in n trials (also known as 鈥渘 choose k鈥�).
  • \( p^k \) is the probability of success raised to the power of k.
  • \( (1-p)^{n-k} \) is the probability of failure raised to the power of (n-k).
Applying this formula allows for calculating the probability of any specific number of "success" instances, like all nine homes having large-screen TVs, by plugging the correct values into the formula.

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Most popular questions from this chapter

Dan Woodward is the owner and manager of Dan's Truck Stop. Dan offers free refills on all coffee orders. He gathered the following information on coffee refills. Compute the mean, variance, and standard deviation for the distribution of number of refills. $$\begin{array}{|cc|}\hline \text { Refills } & \text { Percent } \\\\\hline 0 & 30 \\\1 & 40 \\\2 & 20 \\\3 & 10 \\\\\hline\end{array}$$

The vice president of human resources. at Lowes is studying the number of on- the-job accidents over a period of one month. He developed the following probability distribution. Compute the mean, variance, and standard deviation of the number of accidents in a month. $$\begin{array}{|cc|}\hline \text { Number of } & \\\\\text { Accidents } & \text { Probability } \\\\\hline 0 & .40 \\\1 & .20 \\\2 & .20 \\\3 & .10 \\\4 & .10 \\\\\hline\end{array}$$

Recent crime reports indicate that 3.1 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution. a. Calculate the probability exactly four thefts occur in a minute. b. What is the probability there are no thefts in a minute? c. What is the probability there is at least one theft in a minute?

Which of these variables are discrete and which are continuous random variables? a. The number of new accounts established by a salesperson in a year. b. The time between customer arrivals to a bank ATM. c. The number of customers in Big Nick's barber shop. d. The amount of fuel in your car's gas tank last week. e. The number of minorities on a jury. f. The outside temperature today.

The industry standards suggest that 10 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday. a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service? c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.
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