Is My Venn Diagram Calculation for Tourist Destinations Correct?

In summary, the probability of a tourist being chosen at random having been to South Africa before is 1/2, the probability of them having been to Namibia before is 4/9, and the probability of them having been to neither country before is 1/3. By converting these fractions to percentages, adding them together, and multiplying by the sample size of 36, we can determine that out of the 36 tourists, 10 have been counted more than once. This means that 8 have been to South Africa but not Namibia, 6 have been to Namibia but not South Africa, 10 have been to both countries, and 12 have not been to either country. This aligns with the given
  • #1
Taylor_1989
402
14
I would just like someone to check maths for this venn diagram, just to make sure I have done the calculations correct. I have attached copy, with the numbering inside the diagram. I will show my working here.

Question: One of 36 tourists on holiday in Namibia and South Africa is chosen at random.
The probability that he has been to south africa before is 1/2
The probability that he has been to Namibia before is 4/9
The probability that he has been to neither country before is 1/3

So here is how I worked it out: I convert all the fractions to percentages and add them together:
1/2 = 0.50 = 50%
4/9 = 0.44444 = 44.4%
1/3 = 0.33333 = 33.3%

I the add the fraction or percentages up which gave me: 127.777777778.
I then did 36/100 = 0.36*127.777777778, which gave me 46.0000000001; I rounded this number to 46.
I then took 46-36 = 10, this gave me the intersection of the two circles.

Once I had done this, I took 1/2 of 36, 4/9 of 36, which respectively gave me 16 and 18.

So I then went on and did 16-10=6 so 8 people went to Namibia, 18-10 = 8 people went to south Africa. This gives me a grand total of 24. I then did 36-24 = 12 people which have not been to either. I did double check with the 1/3 of 36 which is correct.

I would appreciate if someone could take a look just to make sure this is correct method.
 

Attachments

  • venn.jpg
    venn.jpg
    9.4 KB · Views: 406
Mathematics news on Phys.org
  • #2
I really have no clue what you are doing. You give a "situation" but there is no question to be answered. What are you trying to determine? You change fractions to complicated decimals and then round off so they are only approximately correct. Why do that?

There are 36 tourists and half of them, 18, have been to South Africa before, 4/9 of them, 16, have been to Namibia before, and 1/3 of them, 12, have not been to either country before. Those add to 18+ 16+ 12= 46 which is greater than 36 so 46-36= 10 must have been counted more than once. Of course, it is impossible to have "not been to either country" and have been to either South Africa or Namibia before so the overlap must be in those who have been to South Africa or Nambia before.


Of the 36 tourists, 18- 10= 8 have been to South Africa before but NOT Namibia, 16- 10= 6 have been to Namibia before but NOT South Africa, 10 have been to both South Africa and Namibia before, and 12 have not been to either country before.

But I still don't know what question you are trying to answer.
You say
So I then went on and did 16-10=6 so 8 people went to Namibia, 18-10 = 8 people went to south Africa. This gives me a grand total of 24. I then did 36-24 = 12 people which have not been to either. I did double check with the 1/3 of 36 which is correct.
which is similar to what I did ("so 8 people went to Namibia" is a typo) but you do NOT say :only to Namibia and not to South Africa" and "only to South Africa and not Namibia" You do NOT say how many people went to both.
 
  • #3
Hey Taylor_1989.

You are not doing this quite right.

Your total probability will add to up 1. The probability of going to neither is 1/3 and its complement is the probability of going to either Namibia Or South Africa (or both) which means P(SA or Namibia) = 1 - 1/3 = 2/3.

Now from Kolmogorov Axioms, we have P(A OR B) = P(A) + P(B) - P(A and B). We know P(A) = 1/2 and P(B) = 4/9 so we get P(A AND B) = P(A) + P(B) - P(A OR B) = 1/2 + 4/9 - 2/3 = 5/18.

If you multiply all of these probabilities by the sample size (36) you will get everything in your Venn diagram the way it should be done probabilistically.
 
  • #4
Sorry I do apologise, I throught a add the question in.
The questions are: The set N represents those who have been to Namibia before.
The set S represents those who have been to south Africa before.

Copy the vent diagram, calculate and fill in the numbers of tourist in each part of the venn diagram.

What I have done in shown how I worked it out.
 
  • #5


I would suggest using a different approach to solve this problem. Instead of converting fractions to percentages and adding them together, it would be more accurate to use the probability formula: P(A or B) = P(A) + P(B) - P(A and B). This formula takes into account the overlap between the two events.

Using this formula, we can calculate the probability of a tourist having been to either South Africa or Namibia before as follows:

P(South Africa or Namibia) = P(South Africa) + P(Namibia) - P(South Africa and Namibia)
= 1/2 + 4/9 - 1/3
= 5/9

Therefore, the probability that a tourist has not been to either country before is:

P(neither country) = 1 - P(South Africa or Namibia)
= 1 - 5/9
= 4/9

Now, we can use this probability to calculate the number of tourists who have not been to either country before. Since there are 36 tourists in total, we can multiply 4/9 by 36 to get 16 tourists who have not been to either country before.

Similarly, we can use the probabilities for South Africa and Namibia to calculate the number of tourists who have been to each country before. This would give us 18 tourists who have been to South Africa before and 16 tourists who have been to Namibia before.

Overall, this approach takes into account the overlap between the two events and provides a more accurate result. I would recommend using this method for future calculations involving probabilities.
 

What is a Venn diagram?

A Venn diagram is a visual representation of the relationships between different sets of data. It consists of overlapping circles or other shapes that represent different groups or categories, and the areas where the circles overlap represent the similarities and differences between those groups.

How is a Venn diagram used in math?

In math, Venn diagrams are often used to represent the relationships between different sets of data, such as numbers, shapes, or objects. They can help visualize concepts such as unions, intersections, and complements, and can be used to solve problems involving logic and probability.

What is a "math check" in a Venn diagram?

A "math check" is a method of using a Venn diagram to verify the accuracy of a mathematical problem or statement. This involves visually representing the problem or statement in the form of a Venn diagram and checking if the result matches the expected outcome.

How do you create a Venn diagram for a math check?

To create a Venn diagram for a math check, you first need to identify the sets or categories involved in the problem or statement. Then, draw overlapping circles or other shapes to represent each set, and label them accordingly. Finally, use the diagram to visualize the relationships between the sets and solve the problem or verify the statement.

What are the benefits of using a Venn diagram for a math check?

Using a Venn diagram for a math check can help you better understand the relationships between different sets of data and visualize the problem or statement. It can also help you identify any errors or misunderstandings in your calculations and provide a more intuitive way to solve math problems.

Similar threads

  • General Math
Replies
4
Views
12K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
9
Views
2K
  • General Math
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • General Math
Replies
9
Views
2K
Back
Top