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In a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate to use the binomial probability distribution to find the probability that a contestant who plays the game five times wins exactly twice? Check each of the requirements of a binomial experiment and give the values of \(n, r,\) and \(p\)

Short Answer

Expert verified
Yes, the binomial distribution is appropriate with \(n = 5\), \(r = 2\), \(p = \frac{1}{6}\).

Step by step solution

01

Identify Trials

A binomial experiment consists of a fixed number of independent trials. Since the contestant plays the game five times, we have a fixed number of trials: \( n = 5 \).
02

Define Outcomes

Each trial results in one of two outcomes: winning (selecting the prize box) or losing (not selecting the prize box). This satisfies the condition of having only two possible outcomes for binomial distribution.
03

Check Independence

Each playing of the game is independent. The outcome of one game does not affect the others because the placement of the prize in a random box is independent for each game.
04

Constant Probability

The probability of winning (selecting the prize box in one of the six boxes) is constant for each trial. This probability is \( p = \frac{1}{6} \).
05

Evaluating Conditions

Since all conditions (fixed number of trials, two outcomes per trial, independent trials, and constant probability of success) are met, we conclude that this scenario can be modeled using a binomial distribution.
06

Assign Variables

For this binomial experiment, we have \( n = 5 \) (trials), \( r = 2 \) (desired number of wins), and \( p = \frac{1}{6} \) (probability of success on a single trial).

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Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In the case of a binomial distribution, which is a type of probability distribution, we deal specifically with experiments where there are only two possible outcomes. These outcomes are often referred to as "success" and "failure."
Think of it like flipping a coin, where you only get heads or tails. In our carnival game, the outcomes are winning (success) or losing (failure).
  • Binomial distribution is key when each trial has a fixed probability of success.
  • It helps calculate the probability of having a specific number of successes over a set number of trials.
In mathematical terms, the probability of obtaining exactly \( r \) successes in \( n \) independent trials is given by the formula:\[P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\]where \( \binom{n}{r} \) is the binomial coefficient, \( p \) is the probability of success, and \( (1-p) \) is the probability of failure. This formula allows one to calculate probabilities in scenarios of repeated independent trials such as our box-picking game.
Independent Trials
Independent trials are a crucial concept in understanding binomial distribution. Independence means that the result of one trial does not affect the results of other trials.
For instance, in our carnival game, the outcome of one box selection does not influence the next. Each time a contestant plays, the prize is randomly placed back, ensuring that each trial's outcome has no impact on subsequent ones.
  • Independence keeps the probability consistent from one trial to another.
  • This property simplifies analysis as you're able to apply the same probability computation to each trial.
Recognizing independent trials allows us to correctly use the binomial model, as it applies to experiments where each trial has an independent and identical probability of success.
Fixed Number of Trials
The concept of having a fixed number of trials in a binomial distribution is essential. This refers to conducting a predetermined number of trials or repetitions of an experiment. In the context of our game, the contestant attempts the experiment five times, making \( n = 5 \).
The fixed nature ensures clarity in computation; we already know the number of times the game will occur, which assists in setting up the binomial distribution formula correctly.
  • This condition helps formulate precise calculations of probability across trials.
  • It provides a basis for determining the specific probable outcomes over these trials.
Identifying and setting this fixed number allows for structured probability calculations, such as determining the number of successes (winning) within these trials.
Probability of Success
Understanding the probability of success is a fundamental component of the binomial distribution. It is the chance of achieving the "success" outcome on any given trial. In our carnival game scenario, the probability of correctly guessing the prize box is \( p = \frac{1}{6} \), since there is one prize among six boxes.
This consistent probability is a requirement for using the binomial model, ensuring uniformity in calculations across trials.
  • Maintains consistent odds for each experiment.
  • Ensures that the same probability applies to every independent trial.
The probability of success must remain constant across each trial to apply the binomial distribution effectively, allowing us to predict outcomes accurately over multiple trials.

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Most popular questions from this chapter

Innocent until proven guilty? In Japanese criminal trials, about \(95 \%\) of the defendants are found guilty. In the United States, about \(60 \%\) of the defendants are found guilty in criminal trials (Source: The Book of Risks, by Larry Laudan, John Wiley and Sons). Suppose you are a news reporter following seven criminal trials. (a) If the trials were in Japan, what is the probability that all the defendants would be found guilty? What is this probability if the trials were in the United States? (b) Of the seven trials, what is the expected number of guilty verdicts in Japan? What is the expected number in the United States? What is the standard deviation in each case? (c) As a U.S. news reporter, how many trials \(n\) would you need to cover to be at least \(99 \%\) sure of two or more convictions? How many trials \(n\) would you need if you covered trials in Japan?

What does it mean to say that the trials of an experiment are independent?

Repair Service A computer repair shop has two work centers. The first center examines the computer to see what is wrong, and the second center repairs the computer. Let \(x_{1}\) and \(x_{2}\) be random variables representing the lengths of time in minutes to examine a computer \(\left(x_{1}\right)\) and to repair a computer \(\left(x_{2}\right) .\) Assume \(x_{1}\) and \(x_{2}\) are independent random variables. Long-term history has shown the following times: Examine computer, \(x_{1}: \mu_{1}=28.1\) minutes; \(\sigma_{1}=8.2\) minutes Repair computer, \(x_{2}: \mu_{2}=90.5\) minutes; \(\sigma_{2}=15.2\) minutes (a) Let \(W=x_{1}+x_{2}\) be a random variable representing the total time to examine and repair the computer. Compute the mean, variance, and standard deviation of \(W\). (b) Suppose it costs 1.50 dollar per minute to examine the computer and 2.75 dollar per minute to repair the computer. Then \(W=1.50 x_{1}+2.75 x_{2}\) is a random variable representing the service charges (without parts). Compute the mean, variance, and standard deviation of \(W .\) (c) The shop charges a flat rate of 1.50 dollar per minute to examine the computer, and if no repairs are ordered, there is also an additional 50 dollar service charge. Let \(L=1.5 x_{1}+50 .\) Compute the mean, variance, and standard deviation of \(L.\)

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, \(37 \%\) of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that (a) out of five adults, none is concerned that employers are monitoring phone calls. (b) out of five adults, all are concerned that employers are monitoring phone calls. (c) out of five adults, exactly three are concerned that employers are monitoring phone calls.

The Honolulu Advertiser stated that in Honolulu there was an average of 661 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 316 homes. Let \(r=\) number of these homes that will be burglarized in a year. (a) Explain why the Poisson approximation to the binomial would be a good choice for the random variable \(r\). What is \(n\) ? What is \(p ?\) What is \(\lambda\) to the nearest tenth? (b) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood? (c) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood? (d) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood?

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