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Chapter 5: Problem 25

A professor continually gives exams to her students. She can give three possible types of exams, and her class is graded as either having done well or badly. Let \(p_{i}\) denote the probability that the class does well on a type 驴 exam, and suppose that \(p_{1}=0.3, p_{2}=0.6\), and \(p_{3}=0.9\). If the class does well on an exam, then the next exam is equally likely to be any of the three types. If the class does badly, then the next exam is always type 1. What proportion of exams are type \(i, i=1,2,3 ?\)

Short Answer

Expert verified
The proportions of exams in the long run are type 1: \(\frac{2}{3}\), type 2: \(\frac{1}{5}\), and type 3: \(\frac{1}{5}\).

Step by step solution

01

Define System of Recursive Equations

Let \(a_{i}\) denote the proportion of the exams that are type \(i\) with \(i=1, 2, 3\). We will use the given probabilities and the method for deciding the next exam to find the expressions for the recursive system: \[ \begin{cases} a_{1} = a_{1}(1-p_{1}) + \frac{1}{3}\sum\limits_{i=1}^3 a_{i}p_{i} \\ a_{2} = a_{1}p_{1} \\ a_{3} = a_{1}p_{1} \\ \end{cases} \] Since \(a_1 + a_2 + a_3 = 1\) (the total proportion of exams), we can then substitute the values of \(p_{1}=0.3, p_{2}=0.6\), and \(p_{3}=0.9\).
02

Solve Recursive Equations

After substituting the values of probabilities, we get the following system: \[ \begin{cases} a_{1} = a_{1}(0.7) + \frac{1}{3}(a_{1}(0.3) + a_{2}(0.6) + a_{3}(0.9)) \\ a_{2} = a_{1}(0.3) \\ a_{3} = a_{1}(0.3) \\ \end{cases} \] We can solve this system step by step. Let's find \(a_2\) and \(a_3\) in terms of \(a_1\). We have \(a_2=a_1\cdot0.3\) and \(a_3=a_1\cdot0.3\). Now, we will find the value for \(a_1\). Substituting \(a_2\) and \(a_3\) in the first equation, we have: \(a_1=0.7a_1+\frac{1}{3}(a_1\cdot0.3+0.3a_1\cdot0.6+0.3a_1\cdot0.9)\) Simplify and solve for \(a_1\): \(a_1=0.7a_1 + \frac{1}{3}(0.3a_1 + 0.18a_1 + 0.27a_1)\) \((0.7 - 1)a_1 = -0.3a_1\) \(a_1 = \frac{0.3}{0.45}\) \(a_1 = \frac{2}{3}\) Finally, substitute the found \(a_1\) value into the expressions for \(a_2\) and \(a_3\): \(a_2 = a_1\cdot0.3 = \frac{2}{3}\cdot0.3 = \frac{1}{5}\) \(a_3 = a_1\cdot0.3 = \frac{2}{3}\cdot0.3 = \frac{1}{5}\)
03

Interpret the Result

The proportions of exams are type 1: \(\frac{2}{3}\), type 2: \(\frac{1}{5}\), and type 3: \(\frac{1}{5}\). This means that 2/3 of the exams given in the long run will be type 1, whereas 1/5 of the exams will be type 2 or type 3.

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Key Concepts

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

Recursive Equations
Recursive equations are fundamental in the modeling of sequential events or processes that depend on previous outcomes. In educational settings, particularly regarding exam performance, these equations can model various scenarios, such as predicting future results based on past performances. The example reflective equation given is complex, involving probabilities that depend on the results of previous exams. To solve it, we apply a systematic approach where each variable is expressed in terms of the others.

Succeeding in understanding recursive equations in exam contexts, students must grasp the idea of the feedback loop, where the outcome of one exam influences the nature of the next. With strong problem-solving skills, students can employ recursive equations to predict patterns and proportions, which is particularly useful in fields like educational testing, economics, and computer science.
Exam Probabilities
The concept of exam probabilities plays a tremendous role in educational forecasting and planning. It hinges on mathematical probability, where the likelihood of certain outcomes 鈥� in this case, the performance in an exam 鈥� is quantitatively expressed. With the example provided, where three different exams types have respective probabilities of success, the analysis unfolds with real-world educational implications.

Students who understand exam probabilities can better prepare for their assessments by evaluating their strengths and weaknesses in relation to the probabilities of exam types. Moreover, educators may use these probabilities to tailor teaching methods to improve overall student success rates. It's crucial to approach exam probabilities with a mindset centered on informed predictions rather than definitive outcomes, as they are subject to variances and interpretations.
Proportional Analysis
Proportional analysis is a critical component in assorted academic disciplines, with notable emphasis in statistics and probability. In the classroom, this relates to the likelihood of certain events versus others, offering a figure of balance or disproportion among variables. In the given exercise, proportional analysis is used to determine the long-term proportion of exam types administered, based on previous results.

By understanding these principles, students can appreciate the predictions on the distribution of future exam types, which is indispensable for strategic study planning. When exams are tied to previous outcomes, as in the scenario provided, proportional analysis can reveal surprising trend data that affects both student readiness and instructor planning. It fosters an environment where the balance of curriculum and assessment types is more transparent and, arguably, fairer.

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Most popular questions from this chapter

\(M\) 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 \(1 .\) 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.

A particle moves among \(n+1\) vertices that are situated on a circle in the following manner: At cach step it moves one step either in the clockwise direction with probability \(p\) or the counterclockwise direetion with probability \(q=1-p\). Starting at a specified state, call it state 0 , let \(T\) be the time of the first return to state 0 . Find the probability that all states have been visited by time \(T\).

Let \(\left|X_{n}, n \geq 0\right|\) denote an ergodic Markov chain with limiting probabilities \(n\). Define the process \(\left[Y_{n}, n \geq 1\right\\}\) by \(Y_{n}=\left(X_{n-1}, X_{n}\right)\). That is, \(Y_{n}\) keeps track of the last two states of the original chain. Is \(\left\\{Y_{n}, n \geq 1 \mid\right.\) a Markov chain? If so, determine its transition probabilities and find $$ \lim _{n \rightarrow \infty} P\left(Y_{n}=(i, j)\right\\} $$

A fair coin is continually flipped. Compute the expected number of flips until the following patterns appear: (a) HHTTHT (b) HHTTHH (c) HHTHHT

Each morning an individual leaves his house and goes for a run. He is equally likely to leave either from his front or back door. Upon leaving the house, he chooses a pair of running shoes (or goes running barefoot if there are no shoes at the door from which he departed). On his return he is equally likely to enter, and leave his running shoes, either by the front or back door. If he owns a total of \(k\) pairs of running shoes, what proportion of the time does he run barefooted?
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