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In an experiment, there are \(n\) independent trials. For each trial, there are three outcomes, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). For each trial, the probability of outcome \(\mathrm{A}\) is \(0.40\); the probability of outcome \(\mathrm{B}\) is \(0.50\); and the probability of outcome \(\mathrm{C}\) is \(0.10\). Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type \(\mathrm{B}\), and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type \(\mathrm{A}\) and six outcomes that are not of type A? Explain. What is the probability of success on each trial?

Short Answer

Expert verified
(a) No, use a multinomial model. (b) Yes, success probability is 0.40.

Step by step solution

01

Analyze the Conditions for Binomial Model

A binomial experiment consists of *n* independent trials, each with only two possible outcomes: success or failure. Since each trial in this problem has three possible outcomes (A, B, or C), it doesn't meet the criteria for a binomial model, which requires only two outcomes.
02

Assess Question (a)

For part (a), we cannot use the binomial model. Since we have three different outcomes in each trial, detailing specific numbers for each outcome (four A's, five B's, and one C) demands a multinomial model, which suits problems involving more than two outcomes per trial.
03

Determine if Question (b) Fits Binomial Model

Part (b) asks if we can use a binomial model for four outcomes of type A and six outcomes not type A. Since we are grouping non-A outcomes together, this sets up a two-category problem (A and not A), fitting a binomial model.
04

Calculate Probability of Success for Type A

In the context of a binomial model, let "success" be getting an outcome of type A. The probability of success (getting A) on each trial is given as 0.40.

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Key Concepts

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

Multinomial Distribution
When analyzing situations with more than two possible outcomes per trial, the multinomial distribution is the go-to tool. It's an extension of the binomial distribution model. While the binomial distribution deals with binary outcomes, the multinomial distribution accommodates scenarios with multiple outcomes.

In the given exercise, each trial presents three potential outcomes: A, B, and C. Hence, the multinomial distribution is apt because it handles three probabilities.

The essence of the multinomial distribution is in its ability to calculate the probability of any specific combination of outcomes over a series of trials. When you want to find the probability of specific instances like 4x outcome A, 5x outcome B, and 1x outcome C in 10 trials, apply the multinomial formula. This formula considers the factorial of the number of trials overall and each outcome frequency.
  • The formula involves each outcome's probability raised to the power of its occurrence number.
  • It then multiplies these results, ensuring every trial's result is incorporated.
  • The arrangement's complexity increases with each additional outcome.
Truly understanding this concept allows one to analyze more diverse scenarios in probability theory.
Probability Theory
Probability theory forms the foundation for understanding random phenomena. It's the mathematical framework used to quantify uncertainty. In our context, it helps determine the chance of particular outcomes in repeated trials, like in the exercise example.

Let's break down some main ideas in probability theory:
  • **Probability of an Event**: This is the likelihood that a certain event will occur. It ranges from 0 (impossible) to 1 (certain).
  • **Sample Space**: This encompasses all possible outcomes of an experiment. For the exercise, the sample space per trial is {A, B, C}.
  • **Independent Events**: Outcome of one event does not affect the others. In independent trials like in this exercise, knowing one trial's result gives zero information about others.
  • **Compound Events**: These events involve the combination of two or more individual events. For instance, calculating four A's over ten trials is a compound event in the exercise context.
Grasping probability theory empowers anyone to understand complex models, identify dependencies among events, and develop strategies for managing uncertainty in varied domains.
Independent Trials
In probability experiments, understanding whether trials are independent is crucial. Independent trials imply that the result of one trial has no effect on another. This hypothesis ensures that each trial's outcome remains unaffected by the preceding events.

In the given exercise, all trials are independent. That means the result of one trial (whether it's A, B, or C) doesn't change the trials that follow. This quality validates the use of probability models that assume independence, like the multinomial and binomial distribution models.

Some key points about independent trials include:
  • **Consistent Probability**: Each trial encounters the same probabilities, making predictions reliable when trials are numerous.
  • **Simplifies Calculations**: Independence allows use of the multiplication rule, simplifying the computation of probabilities for multiple trials.
  • **Real-World Relevance**: Many real-world scenarios model trials as independent, like flipping coins or quality testing batches of products.
Emphasizing independent trials in probability theory enhances foundational understanding, providing a clear path to model more complex statistical scenarios.

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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) Quota Problem 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?

Blood type A occurs in about \(41 \%\) of the population (Reference: Laboratory and Diagnostic Tests, F. Fischbach). A clinic needs 3 pints of type A blood. A donor usually gives a pint of blood. Let \(n\) be a random variable representing the number of donors needed to provide 3 pints of type \(\mathrm{A}\) blood. (a) Explain why a negative binomial distribution is appropriate for the random variable \(n\). Write out the formula for \(P(n)\) in the context of this application. Hint: See Problem \(26 .\) (b) Compute \(P(n=3), P(n=4), P(n=5)\), and \(P(n=6)\). (c) What is the probability that the clinic will need from three to six donors to obtain the needed 3 pints of type A blood? (d) What is the probability that the clinic will need more than six donors to obtain 3 pints of type A blood? (e) What are the expected value \(\mu\) and standard deviation \(\sigma\) of the random variable \(n\) ? Interpret these values in the context of this application.

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

: Syringes The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose the batch contains \(1 \%\) defective syringes. (a) Make a histogram showing the probabilities of \(r=0,1,2,3,4,5,6,7\), and 8 defective syringes in a random sample of eight syringes. (b) Find \(\mu .\) What is the expected number of defective syringes the inspector will find? (c) What is the probability that the batch will be accepted? (d) Find \(\sigma\).

Have you ever tried to get out of jury duty? About \(25 \%\) of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York). If 12 people are called for jury duty, (a) what is the probability that all 12 will be available to serve on the jury? (b) what is the probability that 6 or more will not be available to serve on the jury? (c) Find the expected number of those available to serve on the jury. What is the standard deviation? (d) Quota Problem How many people \(n\) must the jury commissioner contact to be \(95.9 \%\) sure of finding at least 12 people who are available to serve?

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