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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, it is appropriate to use a binomial distribution with \(n = 5\), \(r = 2\), and \(p = \frac{1}{6}\).

Step by step solution

01

Define a Binomial Experiment

A binomial experiment must satisfy four conditions: 1) A fixed number of trials, 2) Each trial is independent, 3) There are only two possible outcomes (success or failure), and 4) The probability of success is the same for each trial. We will check if these conditions are met by the carnival game scenario.
02

Check the Number of Trials

The number of trials, represented by \(n\), is the number of times the contestant plays the game. In this case, the contestant plays the game 5 times, so \(n = 5\).
03

Verify Independence

The outcome of each trial (whether the contestant wins or loses) is independent of the outcomes of other trials. As each game is played, the prize is repositioned randomly, ensuring independence between games.
04

Determine Success and Failure

Each trial (game) can result in either a success (winning the prize) or a failure (not winning the prize). There are only two possible outcomes for each trial.
05

Check Probability of Success

The probability of winning in each trial is \(\frac{1}{6}\) because there is one prize and six boxes. This probability remains constant for each game, satisfying the condition needed for a binomial distribution.
06

Identify the Binomial Parameters

Now that all conditions for a binomial experiment are met, identify the parameters: the number of trials \(n = 5\), the number of successes \(r = 2\) (winning exactly twice), and the probability of success \(p = \frac{1}{6}\).

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Key Concepts

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

Binomial Experiment Conditions
To determine if a scenario fits into a binomial experiment, we need to ensure four specific conditions are met. These conditions guide our understanding of the binomial probability distribution and its application. Let's explore these conditions and how they apply to the carnival game example:
  • Fixed Number of Trials: The number of attempts or trials is predefined. In the carnival game, it's the number of times the contestant plays the game.
  • Independent Trials: The outcome of one trial does not affect another. Each time the game is played, a new prize location is randomly selected, ensuring this independence.
  • Two Possible Outcomes: For each trial, there are two outcomes, typically labeled as "success" or "failure." In the context of the game, winning the prize is "success," and not winning is "failure."
  • Same Probability of Success: Each trial should have the same probability of success. In each round of the carnival game, the chance to win remains constant.
These conditions ensure the scenario is ripe for binomial probability analysis.
Probability of Success
The probability of success refers to the likelihood of achieving the desired outcome in a single trial. In our carnival game scenario, "success" means selecting the box with the prize. Since there are six identical boxes and only one contains a prize, the probability of choosing the correct box is \[\frac{1}{6}. \]This probability remains constant for each individual trial, aligning with one of the crucial requirements of a binomial experiment. Regardless of previous wins or losses, each game reset offers the same odds, which is essential for maintaining a fair and predictable binomial model.This consistent probability allows us to use binomial formulas to predict outcomes over multiple trials, such as winning precisely twice in five attempts.
Independent Trials
For a sequence of events to qualify as independent trials in a binomial experiment, the result of one event must not affect the results of another. In our carnival game, after each play, the prize is relocated randomly among the boxes. This random repositioning is a clear demonstration of independence between the game's rounds. Such independence is critical because without it, the probability of a "success" could change based on previous outcomes, which would violate the rules of a binomial distribution. No matter how many games are played, each decision to select a box remains an isolated event, unaffected by the contestant’s past choices or results. This independence provides a clean slate for every attempt, which is a key element in calculating binomial probabilities.
Fixed Number of Trials
Having a fixed number of trials is one of the fundamental conditions for a binomial probability distribution. This stipulation requires the exact number of attempts to be determined in advance. In the example of the carnival game, the contestant plans to play the game five times. Thus, the number of trials, denoted as \( n \), is equal to 5.This definite count is important because it limits the scenarios to a predetermined number of successes, which can be analyzed using binomial calculations. With \( n \) set, the binomial framework can accurately model the probability of various outcomes, such as winning exactly twice in our example.By establishing this limit, we ensure that all calculations are consistent and based on a known quantity, which is essential for using binomial distribution methods properly in predictive analyses.

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Most popular questions from this chapter

The probability that a single radar station will detect an enemy plane is \(0.65\). (a) Quota Problem How many such stations are required to be \(98 \%\) certain that an enemy plane flying over will be detected by at least one station? (b) If four stations are in use, what is the expected number of stations that will detect an enemy plane?

For a binomial experiment, what probability distribution is used to find the probability that the first success will occur on a specified trial?

What does the random variable for a binomial experiment of \(n\) trials measure?

USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 . (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) Quota Problem How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

Pyramid Lake is located in Nevada on the Paiute Indian Reservation. This lake is famous for large cutthroat trout. The mean number of trout (large and small) caught from a boat is \(0.667\) fish per hour (Reference: Creel Chronicle, Vol. 3, No. 2). Suppose you rent a boat and go fishing for 8 hours. Let \(r\) be a random variable that represents the number of fish you catch in the 8 -hour period. (a) Explain why a Poisson probability distribution is appropriate for \(r\). What is \(\lambda\) for the 8 -hour fishing trip? Round \(\lambda\) to the nearest tenth so that you can use Table 4 of Appendix II for Poisson probabilities. (b) If you have already caught three trout, what is the probability you will catch a total of seven or more trout? Compute \(P(r \geq 7 \mid r \geq 3)\). See Hint below. (c) If you have already caught four trout, what is the probability you will catch a total of less than nine trout? Compute \(P(r<9 \mid r \geq 4)\). See Hint below. (d) List at least three other areas besides fishing to which you think conditional Poisson probabilities can be applied. Hint for solution: Review item 6 , conditional probability, in the summary of basic probability rules at the end of Section \(4.2\). Note that $$ P(A \mid B)=\frac{P(A \text { and } B)}{P(B)} $$

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