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Chapter 4: Problem 28

For a series of dependent trials the probability of success on any trial is \((k+1) /(k+2)\) where \(k\) is equal to the number of successes on the previous two trials. Compute \(\lim _{n \rightarrow \infty} P\) \\{success on the \(n\) th trial\\}.

Short Answer

Expert verified
The limit of the probability of success on the \(n\)th trial as \(n \rightarrow \infty\) does not exist in a precise way, as it will be between \(\frac{1}{2}\) and \(\frac{3}{4}\) depending on the number of successes on the previous two trials.

Step by step solution

01

Find the probability of success on the \(n\)th trial

Since the probability depends on the number of successes in the previous two trials, let's represent the probability of success on the \(n\)th trial as \(P_n\). Now, we want to find an expression for \(P_n\) in terms of \(n\). We know that \(k\) is equal to the number of successes on the previous two trials, so we can write the probability of success as \(P_n = \frac{k+1}{k+2}\).
02

Determine the possible values of \(k\)

Since \(k\) represents the number of successes on the previous two trials, there are only three possible values for \(k\): 0, 1, or 2. We need to find the probability of success on the \(n\)th trial for each possible value of \(k\). When \(k = 0\), there were no successes on the previous two trials, so the probability of success on the \(n\)th trial is \(P_n = \frac{0+1}{0+2} = \frac{1}{2}\). When \(k = 1\), there was one success on the previous two trials, so the probability of success on the \(n\)th trial is \(P_n = \frac{1+1}{1+2} = \frac{2}{3}\). And when \(k = 2\), there were two successes on the previous two trials, so the probability of success on the \(n\)th trial is \(P_n = \frac{2+1}{2+2} = \frac{3}{4}\).
03

Analyze the three possible values for the probability

We know that for any trial, the probability of success is either \(\frac{1}{2}\), \(\frac{2}{3}\), or \(\frac{3}{4}\). We also know that for some \(n\), \(P_n\) depends on the number of successes on the previous two trials (with \(k\) taking on the possible values of 0, 1, or 2). Let's analyze these cases: 1. If the probabilities of success in the two previous trials were both \(\frac{1}{2}\), the probability of success for the \(n\)th trial will be \(\frac{2}{3}\) (since \(k = 1\) in this case). 2. If the probabilities of success in the two previous trials were both \(\frac{3}{4}\), the probability of success for the \(n\)th trial will also be \(\frac{3}{4}\) (since \(k = 2\) in this case). 3. In any other case, the probability of success for the \(n\)th trial will be between \(\frac{1}{2}\) and \(\frac{3}{4}\), and it will either remain that way or start converging towards either \(\frac{2}{3}\) or \(\frac{3}{4}\). Taking this into account, we can see that the limit of the probability of success on the \(n\)th trial as \(n\) approaches infinity will be between \(\frac{1}{2}\) and \(\frac{3}{4}\) and it can never be exactly determined. Therefore, we conclude that the limit does not exist in a precise way.

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Key Concepts

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

Probability Limit
The probability limit in a series of trials refers to the value that the probability of success approaches as the number of trials goes to infinity. In mathematical terms, it is denoted by \( \lim_{n \rightarrow \infty} P \{\text{success on the } n\text{th trial}\} \). In this context, we deal with dependent trials, which means the probability of success on any trial depends on the outcomes of previous trials.
In our situation, the probability does not stabilize to a single number but instead remains within a range between \(\frac{1}{2}\) and \(\frac{3}{4}\). This lack of convergence to a specific number indicates that there is no exact probability limit that can be reached for the success probability as trials continue indefinitely.
Success Probability
The success probability is the chance of obtaining a favorable result in any given trial. In the problem at hand, the probability of success on the \( n \)th trial is given by the formula \( P_n = \frac{k+1}{k+2} \), where \(k\) represents the number of successes in the previous two trials.
The values of \(k\) and the resulting success probabilities are dynamically dependent on the outcomes of past trials.
  • If no successes occurred in the previous two trial (\(k = 0\)), the success probability is \(\frac{1}{2}\).
  • If there was one success in the previous two trial (\(k = 1\)), the probability increases to \(\frac{2}{3}\).
  • If both trials were successes (\(k = 2\)), then the probability is \(\frac{3}{4}\).
This dependence on past outcomes leads to fluctuations in success probability rather than a fixed value.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as it approaches a particular limit, often as a variable approaches infinity. In statistical trials, this involves examining what happens to the probability distributions as the number of trials increases.
For the given series of trials, even though we try to identify an asymptotic probability for successes, the probability distribution sits within a range between \( \frac{1}{2} \) and \( \frac{3}{4} \). This reflects an asymptotic behavior where the outcome does not converge to a single deterministic probability. Instead, the probability behavior continues to fall within a bounded region, revealing the influence of dependencies among past trials.
Series of Trials
A series of trials refers to repeated experiments or attempts to observe an outcome over multiple events. These could be independent, where each trial is unaffected by previous results, or dependent, like in our problem, where prior outcomes influence the current trial probability.

In dependent trials, such as this exercise, each trial's probability uses a unique condition determined by the previous two trials. This creates a sequence where outcome patterns can fluctuate and are influenced by their immediate predecessors. This dynamic makes series of trials a fascinating area of study, given its complex behavior and dependency dynamics compared to straightforward calculations in independent trials.

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Most popular questions from this chapter

Let \(\mathrm{P}\) be the transition probability matrix of a Markov chain. Argue that if for some positive integer \(r, \mathbf{P}^{r}\) has all positive entries, then so does \(\mathbf{P}^{n}\), for all integers \(n \geqslant r\).

On a chessboard compute the expected number of plays it takes a knight, starting in one of the four corners of the chessboard, to return to its initial position if we assume that at each play it is equally likely to choose any of its legal moves. (No other pieces are on the board.) Hint: Make use of Example 4.32.

A transition probability matrix \(\mathbf{P}\) is said to be doubly stochastic if the sum over each column equals one; that is, $$ \sum_{i} P_{i j}=1, \quad \text { for all } j $$ If such a chain is irreducible and aperiodic and consists of \(M+1\) states \(0,1, \ldots, M\), show that the limiting probabilities are given by $$ \pi_{j}=\frac{1}{M+1}, \quad j=0,1, \ldots, M $$

For a time reversible Markov chain, argue that the rate at which transitions from \(i\) to \(j\) to \(k\) occur must equal the rate at which transitions from \(k\) to \(j\) to \(i\) occur.

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(i, i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.
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