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Wiris in the Classroom: Learn and Practice with Unlimited Exercises

Learning mathematics requires more than just understanding theories: the key to mastering complex concepts and honing skills lies in continuous practice. However, traditional methods often limit the learning process: once the student has solved a problem, the challenge disappears, and exams often repeat the same questions, thus losing their value as an assessment tool. Imagine having the possibility to generate unlimited exercises with random variables, adapted to the needs of each student. With Wiris, this is now possible.

In addition, it has an automatic correction functionality which means that while the exercise is being generated, the answer is calculated simultaneously. This allows instant feedback, making it easier to learn and correct mistakes immediately.
For teachers, this tool offers a key advantage: the possibility of generating an infinite number of exercises, with each one being unique. In this way, teachers can create varied and personalized learning experiences, ensuring that students face new challenges every time they practice.

For their part, students enjoy constant practice which allows them to reinforce their skills without the worry of repeating the same exercises. In addition, this ability to generate unlimited exercises gives them excellent preparation for exams.

Practical example: Solving a system of linear equations

To illustrate how this tool works and how it can be used in the classroom, we will focus on solving systems of linear equations. We chose these types of problems because they represent a common mathematical challenge faced by all students and offer the possibility to explore different problem-solving methods. In addition, systems of equations can have various characteristics, such as being compatible, incompatible, or indeterminate, allowing students to work with a wide range of situations.

Let us imagine that we have the following system of linear equations with three unknowns:

We are going to solve it using Gauss’s Method. The first step is to write the system in the form of an augmented matrix: 

Now, we will apply elementary operations to reduce the matrix to its row echelon form.

Therefore, we obtain: 

From equation 3 of the system, we find the variable z:

Unlimited generation of systems of linear equations with random variables

Now that we have solved this system, for further practice, we would like to be able to generate new systems of equations of the same form but with random values.

These systems would take the following form:

Below, we will show you a code example to implement this functionality and generate given random systems of equations. With Wiris, it is totally possible!

r() := random(-4,4)

sol = [r() with i in 1..3]

[a,b,c] = sol

repeat 

A = [[r() with i in 1..3] with j in 1..3]

until determinant (A) != 0 

b = A * sol

ec = {}

for i in 1..3 do

ecaux = A.i * [x,y,z]

ec = append(ec, ecaux = b.i)

end

{ec1, ec2, ec3} = ec

r() := random(-4,4): This function generates a random number between -4 and 4.

sol = [r() with i in 1..3]: Here, three random values are generated that will represent the solutions of the system of equations.

[a,b,c] = sol: The generated values are assigned to the unknowns of the system.

repeat…until determinant(A) != 0: This block ensures that the coefficient matrix A is invertible (i.e., its determinant is not zero), which guarantees the system has a unique solution.

b = A * sol: The result vector b is calculated by multiplying the coefficient matrix A by the solutions sol

for i in 1..3 do: In this cycle, the three equations are generated, using each of the rows of the matrix A and the solutions sol.

{ec1, ec2, ec3} = ec: Finally, the three generated equations are stored in the variables ec1, ec2, and ec3, ready to be used in new exercises.

Using WirisQuizzes and implementing this code as a question, we can obtain the following results:

Reloading another system with the symbol “=” generates a new set of equations, as shown below:

For more details, you can consult the complete product documentation here.

If you have found this article useful, we would love you to share it with other colleagues, teachers or students. We would also be delighted to hear your opinion and any suggestions you may have.

 

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What is the Birthday Paradox?

The Fascinating Probability of Sharing a Date

The birthday paradox is a mathematical concept that invites us to challenge our intuition. A paradox, in essence, is a statement or result that, although seemingly contradictory or contrary to common logic, proves to be true when analyzed closely. In the case of the birthday paradox, the scenario is simple but puzzling: how many people must be in a room for the probability of at least two sharing the same birthday to be greater than 50%?

The result will be surprisingly low, but do you dare to answer before you continue reading to reveal the answer?

The most intuitive way to approach the paradox would be to think that it takes at least 183 people, half the days of the year, for the probability of two people sharing the same birthday to be greater than 50%. However, the reality is very different. 

To understand how this probability is calculated, let us recall that the basic formula of any probability consists of dividing the favorable cases by the total cases. 

In this case, rather than calculating the probability of a coincidence directly, it is easier to first calculate the probability that no two people share a birthday. After determining this probability, subtracting it from 1 (that is, calculating the complement) gives the probability that at least two people share a birthday.

To calculate the probability that no two people share a birthday, we will proceed constructively, evaluating scenario by scenario.

  • If there is only one person in a room, there is no chance that they will share their birthday with someone else since no other people are present. Therefore, the probability that their birthday is unique is 100%, i.e., 365/365.
  • Now, if a second person comes in, the probability that their birthday will be different from the first person’s is ​364/365, since there are only 364 days remaining in which they can be born to avoid a coincidence.
  • If a third person is added, the probability that their birthday does not match anyone else’s is ​363/365 since there are now two busy days.

Therefore, for n people, all the above conditions must be met simultaneously: that the second person does not share a birthday with the first, that the third person does not share a birthday with the previous two, and so on. This implies that the probability that none of the n people share the same birthday is the product of the individual probabilities:

And finally, we obtain:

Analyzing probabilities for different values of n

Let us examine the results for different values of n:

  • For n=10 we obtain that P (at least one coincidence) = 11,61%

CalcMe representation of the equation

This calculation has been created using CalcMe
  • For n=15 we obtain that P (at least one coincidence) = 25,03%
  • For n=23 we obtain that P (at least one coincidence) = 50,05%
  • For n=50 we obtain that P (at least one coincidence) = 96,53%
  • For n=60 we obtain that P (at least one coincidence) = 99,22%

As we can see from the examples above, to achieve a probability of more than 50%, only 23 people in the same room are needed. This result may seem surprising, but it makes sense if we consider that with 23 people, 253 different pairs can be formed, and each of these pairs represents a chance for two people to share the same birthday. Even more surprising is that to have a probability greater than 99%, only 60 people are needed, demonstrating how the number of possible combinations grows rapidly with every new person added.

This is the graph of the probability distribution

Graph showing the probability of two people sharing a birthday, with the number of people on the X-axis and the probability on the Y-axis, expressed as a percentage.

Graph showing the probability of two people sharing a birthday, with the number of people on the X-axis and the probability on the Y-axis, expressed as a percentage.

In contrast, if instead of calculating the probability of coincidence between any pair of people we consider the probability that someone in a room of n people (excluding you) specifically shares your birthday, the calculation is different. This probability is given by: 

For n = 22, this probability is about 0.059 (5.9%), which is quite low. In fact, there would need to be at least 253 people in the room for this probability to exceed 50%. 

The birthday paradox is a fascinating reminder of how our intuition can fail when confronted with mathematical logic.

Now that you know the birthday paradox and its key concepts, can you calculate the probability that exactly k people within a group of n people share their birthday? We’d love to hear your results in the blog comments! If this article surprised you, share it with your friends so that they can learn more about this mathematical paradox. What’s more, if you’re interested in trying these calculations for yourself, have a look at CalcMe.

 

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