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Critical Thinking 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 \(B\), 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 \(\mathrm{A}\) ? Explain. What is the probability of success on each trial?

Short Answer

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

Step by step solution

01

Understanding Problem (a)

In part (a), we are asked whether we can use the binomial model for an experiment with three possible outcomes (A, B, and C). A binomial experiment is characterized by having exactly two possible outcomes in each trial, which essentially makes it incompatible with this scenario where there are three distinct outcomes. Therefore, the binomial model is not suitable for this situation.
02

Concluding Part (a)

Since a binomial experiment is defined by two possible outcomes, and our experiment has three outcomes, we cannot use the binomial model for determining the probabilities of specific counts of outcomes A, B, and C. Instead, a multinomial distribution is appropriate here.
03

Analyzing Problem (b)

In part (b), we are asked about the scenario with outcomes of type A versus outcomes that are not type A. This situation boils down to simplifying the three outcomes into two: 'type A' and 'not type A'. This is suitable for a binomial distribution as there are now only two possible outcomes in each trial.
04

Determining Probability of Success in Trials

To use the binomial experiment model, we need to define 'success'. Here, 'success' would be getting type A. Therefore, the probability of success in each trial is the probability of getting A, which is 0.40.

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Key Concepts

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

Binomial Experiment
A binomial experiment is a fundamental concept in probability theory. It refers to an experiment or process that satisfies specific criteria, allowing us to model the situation with a binomial distribution.
  • Each trial has only two possible outcomes, commonly labeled as "success" and "failure".
  • The number of trials, usually denoted by \( n \), is fixed.
  • The probability of success, denoted by \( p \), remains the same for each trial.
Important aspects of a binomial experiment include the requirement of just two outcomes, making it unique from experiments with more than two possible results.
In practice, binomial experiments are frequently used to model situations such as flipping a coin, where each outcome is binary: heads or tails. This concept allows us to calculate probabilities using the binomial formula:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is a binomial coefficient representing the number of ways \( k \) successes can occur in \( n \) trials, \( p \) is the probability of success on each trial, and \( 1-p \) is the probability of failure.
Independent Trials
Independent trials are crucial in understanding probability scenarios, especially those involving multiple trials like binomial or multinomial experiments. Each trial's outcome does not affect the outcome of another.
Independence of trials guarantees that the probability of outcomes remains constant across all trials. For example, in a situation where a die is rolled multiple times, each roll is an independent trial as the result of one roll does not influence another.
This independence is essential because it maintains the consistency and reliability of probability calculations across experiments. Without independent trials, predicting probabilities becomes far more complex and less reliable, as outcomes become contingent on previous results.
In the context of experiments, ensuring trials are independent aids in simplifying analysis, allowing for straightforward applications of probability formulas, such as those in the binomial setup.
Probability Theory
Probability theory is the branch of mathematics that deals with the likelihood of different outcomes in an event or experiment. It provides the framework needed to calculate how likely something is to occur.
Fundamental concepts include:
  • Sample Space: The set of all possible outcomes of an experiment.
  • Events: A subset of the sample space. Events can be single outcomes or more complex combinations.
  • Probability: A measure that quantifies uncertainty, often ranging from 0 to 1. A probability of 0 indicates an impossible event, while 1 denotes certainty.
Probability theory underpins everything from predicting weather patterns to ensuring fairness in games of chance.
In experiments, probability helps in determining the expected frequency of outcomes based on given success probabilities. For example, in the multinomial distribution (an extension of the binomial distribution for three or more outcomes), probability theory is used to calculate the likelihood of each possible mix of outcomes.
By applying probability theory principles, we can make informed predictions and decisions in uncertain situations, bringing clarity and precision to fields like finance, science, and engineering.

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Most popular questions from this chapter

Statistical Literacy Consider the probability distribution of a random variable \(x .\) Is the expected value of the distribution necessarily one of the possible values of \(x\) ? Explain or give an example.

Statistical Literacy 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?

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

Combination of Random Variables: Insurance Risk Insurance companies know the risk of insurance is greatly reduced if the company insures not just one person, but many people. How does this work? Let \(x\) be a random variable representing the expectation of life in years for a 25 -year-old male (i.e., number of years until death). Then the mean and standard deviation of \(x\) are \(\mu=50.2\) years and \(\sigma=11.5\) years (Vital Statistics Section of the Statistical Abstract of the United States, 116 th Edition). Suppose Big Rock Insurance Company has sold life insurance policies to Joel and David. Both are 25 years old, unrelated, live in different states, and have about the same health record. Let \(x_{1}\) and \(x_{2}\) be random variables representing Joel's and David's life expectancies. It is reasonable to assume \(x_{1}\) and \(x_{2}\) are independent. Joel, \(x_{1}: \mu_{1}=50.2 ; \sigma_{1}=11.5\) David, \(x_{2}: \mu_{2}=50.2 ; \sigma_{2}=11.5\) If life expectancy can be predicted with more accuracy, Big Rock will have less risk in its insurance business. Risk in this case is measured by \(\sigma\) (larger \(\sigma\) means more risk). (a) The average life expectancy for Joel and David is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation of \(W\). (b) Compare the mean life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W)\) (c) Compare the standard deviation of the life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W)\). (d) The mean life expectancy is the same for a single policy \(\left(x_{1}\right)\) as it is for two policies (W), but the standard deviation is smaller for two policies. What happens to the mean life expectancy and the standard deviation when we include more policies issued to people whose life expectancies have the same mean and standard deviation (i.e., 25 -year-old males)? For instance, for three policies, \(W=(\mu+\mu+\mu) / 3=\mu\) and \(\sigma_{W}^{2}=(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}=\) \((1 / 3)^{2}\left(3 \sigma^{2}\right)=(1 / 3) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{3}} \sigma .\) Likewise, for \(n\) such policies, \(W=\mu\) and \(\sigma_{W}^{2}=(1 / n) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{n}} \sigma .\) Looking at the general result, is it appropriate to say that when we increase the number of policies to \(n\), the risk decreases by a factor of \(\sigma_{w}=\frac{1}{\sqrt{n}} ?\)

Law Enforcement: Shoplifting The Denver Post reported that, on average, a large shopping center has had an incident of shoplifting caught by security once every three hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Let \(r\) be the number of shoplifting incidents caught by security in the 11-hour period during which the center is open. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r\). What is \(\lambda\) ? (b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security?

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