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Chapter 4: Problem 23

Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability 0.5. In the long run, what proportion of trials are successes?

Short Answer

Expert verified
In the long run, the proportion of trials that are successes is approximately 43.0%. This is obtained by finding the steady-state probability of success using conditional probabilities and solving a quadratic equation. \(P(S) \approx 0.430\)

Step by step solution

01

Define the variables

Let: - \(P(S_k)\) be the probability that the \(k\)-th trial is a success - \(P(S_k | S_{k-1}, S_{k-2})\) be the conditional probability of a success on the \(k\)-th trial, given the last two trials were successes
02

Analyze the conditional probabilities

From the problem statement, we know that: - \(P(S_k | S_{k-1}, S_{k-2}) = 0.8\) - \(P(S_k | \overline{S_{k-1}S_{k-2}}) = 0.5\), where \(\overline{S_{k-1}S_{k-2}}\) means "not both of the last two trials were successes."
03

Determine the steady-state probabilities

We can write the overall probability of success on the \(k\)-th trial as: \(P(S_k) = P(S_k | S_{k-1}, S_{k-2})P(S_{k-1}, S_{k-2}) + P(S_k | \overline{S_{k-1}S_{k-2}})P(\overline{S_{k-1}S_{k-2}})\) In the steady state, we have \(P(S_k) = P(S_{k+1})\), \(P(S_{k-1}) = P(S_k)\), and \(P(S_{k-2}) = P(S_k)\). So, we can rewrite the equation as: \(P(S) = 0.8P(S)^2 + 0.5(1 - P(S)^2)\)
04

Solve for the steady-state probability of success

Now, let's solve for the steady-state probability of success, P(S): \(P(S) = 0.8P(S)^2 + 0.5 - 0.5P(S)^2\) \(P(S) = 0.3P(S)^2 + 0.5\) \(0.3P(S)^2 - P(S) + 0.5 = 0\) This is a quadratic equation. To solve, either factor it or use the quadratic formula. After analyzing the equation, it's clear that it cannot be factored easily, so we resort to using the quadratic formula: \(P(S) = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Here, a = 0.3, b = -1, and c = 0.5. \(P(S) = \dfrac{1 \pm \sqrt{(-1)^2 - 4(0.3)(0.5)}}{2(0.3)}\) \(P(S) = \dfrac{1 \pm \sqrt{1 - 0.6}}{0.6}\) \(P(S) = \dfrac{1 \pm \sqrt{0.4}}{0.6}\) We find that there are two possibilities for P(S): 1. \(P(S) = \dfrac{1 + \sqrt{0.4}}{0.6}\) (approximately 1.16) 2. \(P(S) = \dfrac{1 - \sqrt{0.4}}{0.6}\) (approximately 0.430) Since a probability must be between 0 and 1, the value 1.16 is not valid. Therefore, the valid solution is: \(P(S) \approx 0.430\)
05

Interpret the result

In the long run, the proportion of trials that are successes is approximately 43.0%.

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Key Concepts

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

Steady-State Probability
In the context of probability theory, steady-state probability is a concept from the realm of Markov processes and is used to refer to the probability that a system will be in a certain state after a large number of steps or over a long period of time. This probability becomes constant, or 'steady', hence the name. It's quite insightful in scenarios like the given exercise, where events occur sequentially and have probabilities that depend on the preceding outcomes.

In our problem, we are seeking the steady-state probability of success in a series of trials with conditional probabilities. It is found by setting up an equation where the probability of success, denoted by P(S), in the long run is equal to the probability of success on any given trial, since once the system reaches the steady state, these probabilities no longer change. This equation accommodates the potential outcomes from preceding trials and their respective probabilities. As seen in the provided step-by-step solution, by considering the steady-state condition鈥攚here the probability does not change from one trial to the next鈥攚e can derive an equation that represents the system in its long-run behavior.
Quadratic Formula in Probability
When confronted with problems in probability that involve squared terms, we may encounter a quadratic equation. Quadratic equations are polynomial equations of the second degree, usually in the form of ax^2 + bx + c = 0. The quadratic formula, which provides a method to solve for x, is given by \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. The formula is a powerful algebraic tool to find the values of x that make the equation true.

In the exercise, we apply the quadratic formula to solve for the steady-state probability. The step-by-step solution shows that no straightforward factoring is possible, prompting the use of the quadratic formula. In the scenario of the exercise, x corresponds to P(S), our probability of success. After simplification, the formula gives two possible solutions, and since probabilities are always between 0 and 1, we disregard the value above 1, identifying a valid probability of approximately 0.430, representing the long-term success rate in the sequence of trials.
Sequence of Trials
The concept of a sequence of trials refers to a series of experiments, games, or actions where each trial's outcome does not necessarily depend on the previous trials' outcomes. When dealing with a sequence of independent trials, we often observe consistent probabilities across each trial, such as flipping a fair coin; however, in some cases, such as this exercise, the probability of success in the current trial may depend on the outcomes of previous trials. This gives us a sequence of dependent trials.

The solution to the given exercise demonstrates how sequences of dependent trials can significantly complicate the process of calculating probabilities. We need to consider the conditional probabilities鈥攔epresented by P(S_k | S_{k-1}, S_{k-2})鈥攖o determine the likelihood of an event given that certain conditions are met. It's essential to understand how these contingencies factor into the broader calculation of successful trials, as this directly impacts our ultimate probability findings and determines the proportion of success within the defined system of trials.

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Most popular questions from this chapter

A Markov chain is said to be a tree process if (i) \(P_{i j}>0\) whenever \(P_{j i}>0\). (ii) for every pair of states \(i\) and \(j, i \neq j\), there is a unique sequence of distinct states \(i=i_{0}, i_{1}, \ldots, i_{n-1}, i_{n}=j\) such that $$ P_{i_{k}, i_{k+1}}>0, \quad k=0,1, \ldots, n-1 $$ In other words, a Markov chain is a tree process if for every pair of distinct states \(i\) and \(j\) there is a unique way for the process to go from \(i\) to \(j\) without reentering a state (and this path is the reverse of the unique path from \(j\) to \(i\) ). Argue that an ergodic tree process is time reversible.

Let \(\pi_{i}\) denote the long-run proportion of time a given Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i\). (b) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

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.

\(M\) balls are initially distributed among \(m\) urns. At each stage one of the balls is selected at random, taken from whichever urn it is in, and then placed, at random, in one of the other \(M-1\) urns. Consider the Markov chain whose state at any time is the vector \(\left(n_{1}, \ldots, n_{m}\right)\) where \(n_{i}\) denotes the number of balls in urn \(i\). Guess at the limiting probabilities for this Markov chain and then verify your guess and show at the same time that the Markov chain is time reversible.

Recall that state \(i\) is said to be positive recurrent if \(m_{i, i}<\infty\), where \(m_{i, 1}\) is the expected number of transitions until the Markov chain, starting in state \(i\), makes a transition back into that state. Because \(\pi_{i}\), the long run proportion of time the Markov chain, starting in state \(i\), spends in state \(i\), satisfies $$ \dot{\pi}_{i}=\frac{1}{m_{i, i}} $$ it follows that state \(i\) is positive recurrent if and only if \(\pi_{i}>0 .\) Suppose that state \(i\) is positive recurrent and that state \(i\) communicates with state \(j .\) Show that state \(j\) is also positive recurrent by arguing that there is an integer \(n\) such that $$ \pi_{j} \geqslant \pi_{i} P_{i, j}^{n}>0 $$
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