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The longest "run" of \(S\) 's in the sequence SSFSSSSFFS has length 4 , corresponding to the \(S\) 's on the fourth, fifth, sixth, and seventh trials. Consider a binomial experiment with \(n=4\), and let \(y\) be the length (number of trials) in the longest run of \(S\) 's. a. When \(\pi=.5\), the 16 possible outcomes are equally likely. Determine the probability distribution of \(y\) in this case (first list all outcomes and the \(y\) value for each one). Then calculate \(\mu_{y}\) b. Repeat Part (a) for the case \(\pi=.6\). c. Let \(z\) denote the longest run of either \(S\) 's or \(F\) 's. Determine the probability distribution of \(z\) when \(\pi=.5\).

Short Answer

Expert verified
The solutions involves listing the total possible outcomes, identifying the longest runs, and calculating the probability distribution based on the frequency of each longest run. The mean is then calculated using these probabilities. Each of these steps are repeated adjusting for the given \(\pi\) values and the variable asked for ('S' or either 'S' or 'F').

Step by step solution

01

Determine all possible outcomes

For \(n=4\) trials of 'S' or 'F', list down the 16 (\(2^{4}\)) possibilities. They are: 'SSSS', 'SSSF', 'SSFS', 'SSFF', 'SFSS', 'SFSF', 'SFFS', 'SFFF', 'FSSS', 'FSSF', 'FSFS', 'FSFF', 'FFSS', 'FFSF', 'FFFS', 'FFFF'.
02

Identify the length for longest run of 'S' for each outcome

Now identify the longest run of 'S' in each outcome which is 'y'. For example, for the outcome 'SSSF', 'y=3'. Similarly calculate 'y' for all outcomes.
03

Calculate probability distribution for Part (a) where \(\pi=.5\)

The 16 outcomes listed in Step 1 are all equally likely when \(\pi=.5\) as the probability of 'S' or 'F' are the same. Thus, the probability of each outcome is \(1/16\). Count the frequency of each 'y' value observed in Step 2 and divide it by 16 to find the corresponding probability. This will be the probability distribution of 'y'.
04

Evaluate mean 'y' for \(\pi=.5\)

Calculate the mean \(\mu_{y}\) which will be \(\mu_{y} = \sum y \times P(y)\), where all distinct 'y' values and corresponding probabilities calculated in Step 3 is used.
05

Repeat Step 2,3 and 4 for \(\pi=.6\)

Repeat all previous steps for \(\pi=.6\) to determine the probability distribution of 'y' and \(\mu_{y}\) for this case. Note that while the longest runs of 'S' (identified in step 2) will remain the same, the probabilities will change as the outcomes are no longer equally likely.
06

Identify longest run of either 'S' or 'F' for part (c)

Here, recursively check 'z', the length of the longest 'run' of either 'S' or 'F' for all outcomes listed in Step 1.
07

Calculate probability distribution for 'z'

Count the frequency of each 'z' value from Step 6. Divide it by 16 to calculate their corresponding probability which will give the probability distribution of 'z' for \(\pi=.5\).
08

Evaluate average 'z'

Using the probability distribution of 'z' found in step 7, calculate the mean \(\mu_{z}\) using the formula \(\mu_{z} = \sum z \times P(z)\)

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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 statistical experiment that defines one of the foundational concepts of probability theory. It consists of running a trial, or an experiment, multiple times under identical and independent conditions while counting the number of successes and failures. This type of experiment has three main characteristics: there are a fixed number of trials (), each trial has only two possible outcomes (commonly referred to as 'success' and 'failure'), and the probability of success () is the same for each trial.

For example, flipping a fair coin four times is a binomial experiment where each flip is independent from the others, and the coin has an equal chance of landing heads (success) or tails (failure). The concept of a binomial experiment is crucial in calculating various probabilities, such as determining the probability distribution of the longest run of a specific outcome, like successive heads in coin flips.

When solving problems related to binomial experiments, one has to list all possible outcomes and assign probabilities to these outcomes. This enumeration of scenarios is vital because it forms the basis for computing the probability distribution of the experiment's results.
Probability
Probability is the measure of the likelihood that an event will occur. In the context of a binomial experiment, probability is quantified as a number between 0 and 1, inclusive, where 0 indicates an impossible event, and 1 represents a certain event. The probability () of success in a binomial experiment is a fixed value for each trial.

As seen in the problem, with a series of flips and the longest run of successes (), we calculate the probability of each outcome to construct a probability distribution. This distribution details the likelihood of observing each possible value of in our experiment. When is 0.5, the outcomes are equally likely, implying each sequence of coin flips is just as probable as the other. When changes, so does the probability of each outcome, leading to a different distribution.

Understanding probability is essential, as it allows us to make predictions about the likelihood of various outcomes in a binomial experiment. This application is particularly important in fields like finance, insurance, and any domain where risk assessment and statistical analysis are crucial.
Random Variable
A random variable is a numerical description of the outcome of a statistical experiment. In our context, the random variable represents the length of the longest run of 'S' in a series of coin flips. To each possible outcome of the experiment, a random variable assigns a numerical value. For a discrete random variable like , each value can be listed and is associated with a certain probability.

For example, let us consider each potential sequence of coin flips, ‘SSSF’, ‘SSF’, and so on - each sequence will have a longest run of successes, which is our random variable . After evaluating all such sequences, we compile the frequencies of , converging on a probability distribution that reflects the chances of each possible run length occurring.

A random variable is not just a standalone number but a function that relates each event with a numerical value, allowing us to quantify outcomes and work with them in mathematical expressions and calculations. Therefore, identifying the random variable in question is the first step toward applying probability theory to find expected values, variances, and other statistical measures.

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Most popular questions from this chapter

Suppose that \(5 \%\) of cereal boxes contain a prize and the other \(95 \%\) contain the message, "Sorry, try again." Consider the random variable \(x\), where \(x=\) number of boxes purchased until a prize is found. a. What is the probability that at most two boxes must be purchased? b. What is the probability that exactly four boxes must be purchased? c. What is the probability that more than four boxes must be purchased?

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Suppose that \(90 \%\) of all registered California voters favor banning the release of information from exit polls in presidential elections until after the polls in California close. A random sample of 25 California voters is to be selected. a. What is the probability that more than 20 voters favor the ban? b. What is the probability that at least 20 voters favor the ban? c. What are the mean value and standard deviation of the number of voters who favor the ban? d. If fewer than 20 voters in the sample favor the ban, is this at odds with the assertion that (at least) \(90 \%\) of the populace favors the ban? (Hint: Consider \(P(x<20)\) when \(\pi=.9 .)\)

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