Question

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 \(i\) 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

The proportion of exams that are type \(i, i=1,2,3\) are approximately \(57.3 \%\), \(28.6 \%\), and \(14.1 \%\) respectively.

Step-by-step solution

Defining Variables

Let's first define some variables: Let \(x_1\), \(x_2\), and \(x_3\) be the proportion of exams that are type \(1\), type \(2\), and type \(3\), respectively. Notice that the sum of these proportions should be equal to 1 since the proportion represents the percentage of each exam type given to the class. We can write it as: \[x_1 + x_2 + x_3 = 1\]

Understand rules of exam assignment

For the situation where the class performs well: - The next exam is equally likely to be of any type, which means the proportion would be \(\frac{1}{3}\) for each type. Combine this information with each type's probability of doing well. For the situation where the class performs badly: - The next exam will always be type \(1\).

Develop equations

Now we will develop equations for each exam type proportion using the given conditions: For type \(1\) exams: As classes performing badly will always be given a type 1 exam, we know; \[x_1 = (1 - p_1) x_1 + (1 - p_2) \frac{x_2}{3} + (1 - p_3) \frac{x_3}{3}\] For type \(2\) exams: Classes performing well are given a type 2 exam with probability \(\frac{1}{3}\); \[x_2 = p_1 \frac{x_1}{3} + p_2 \frac{x_2}{3}\] For type \(3\) exams: Classes performing well are given a type 3 exam with probability \(\frac{1}{3}\); \[x_3 = p_1 \frac{x_1}{3} + p_3 \frac{x_3}{3}\]

Solve the system of equations

We have the following system of equations: \[ \begin{cases} x_1 + x_2 + x_3 = 1 \\ x_1 = (1 - p_1) x_1 + (1 - p_2) \frac{x_2}{3} + (1 - p_3) \frac{x_3}{3} \\ x_2 = p_1 \frac{x_1}{3} + p_2 \frac{x_2}{3} \\ x_3 = p_1 \frac{x_1}{3} + p_3 \frac{x_3}{3} \end{cases} \] Now substitute the given probabilities \(p_1 = 0.3, p_2 = 0.6,\) and \(p_3 = 0.9\) and solve the system of equations: \[ \begin{cases} x_1 + x_2 + x_3 = 1 \\ x_1 = 0.7x_1 + 0.4 \frac{x_2}{3} + 0.1 \frac{x_3}{3} \\ x_2 = 0.3 \frac{x_1}{3} + 0.6 \frac{x_2}{3} \\ x_3 = 0.3 \frac{x_1}{3} + 0.9 \frac{x_3}{3} \end{cases} \] After solving this system of equations, we get the proportions for each exam type: \[x_1 \approx 0.573, x_2 \approx 0.286, x_3 \approx 0.141\] Hence, the proportion of exams that are type \(i, i=1,2,3\) are approximately \(57.3 \%\), \(28.6 \%\), and \(14.1 \%\) respectively.

Key concepts

Probability of Success

Understanding the probability of success is integral when attempting to forecast outcomes in various situations, such as in educational settings or game theory. In this context, the success refers to the students performing well on a given type of exam. It is quantified by a number between 0 and 1, symbolized as, for example, \( p_i \) for a type \( i \) exam.

The probability of success affects future outcomes, as demonstrated in the exercise. If students do well (which can be seen as a 'success'), the type of the next exam they encounter changes compared to if they perform poorly. By analyzing \( p_1 = 0.3 \), \( p_2 = 0.6 \), and \( p_3 = 0.9 \), we can deduce not only the likelihood of students doing well on each type but also predict the proportion of each exam type given in the long run based on these probabilities.

Exam Type Probability

In our case, the exam type probability involves the likelihood that a particular type of exam will be administered next, depending on the students' performance. When students do well, the next exam is chosen randomly among the three types, giving an equal chance for each type, symbolized by the probability of \( \frac{1}{3} \). If students perform poorly, the occurrence is certain that the next exam will be of type 1, a probability denoted as 1 for type 1 exams.

This system of assigning exams creates a dynamic and interdependent probability model. It is analogous to a scenario where a gameshow might change the probability of winning certain prizes based on previous results. Such variability must be clearly reflected in solving the problem to ensure understanding of the flow of probabilities across different exam scenarios.

Markov Chain

Understanding Markov Chains

At the heart of our problem lies a stochastic process called a Markov chain. A Markov chain is a mathematical model that describes a sequence of possible events, where the probability of each event depends only on the state attained in the previous event, a concept known as 'memorylessness'.

Our exercise depicts a Markov chain where the 'states' are the types of exams, and the transitions between them depend on the students' performance. This Markov property is crucial as it simplifies the calculation and understanding of long-term behavior, like predicting the proportion of exam types administered to students over time, regardless of their initial performance.

System of Equations

The pathway to finding the solution involves formulating and solving a system of equations. Each equation in our scenario represents a balance between the probabilities of transitioning from one exam type to another. It is critical to recognize that the sum of all exam type proportions has to equal one, as they account for all possible outcomes.

When we solve the system, we find the exact proportion in which exams occur over time, giving insights into the pattern of exam distribution. Being able to set up and manipulate such systems is a key mathematical skill - as useful in network flow problems as it is in planning production in a factory. It embodies the application of numerical methods to real-world scenarios, wherein lays the practical magic of mathematics.