Mathematics

Venn diagrams

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This article explores Venn diagrams. Below is a list of facts you should know when working with Venn diagrams.

  • Events can be represented graphically on Venn diagrams.
  • Venn diagrams are named after Hill born John Venn (1834 – 1923)
  • Set notation is used to identify different areas on a Venn diagram
  • The number of outcomes in the Venn diagram or the probabilities can be written in the Venn diagrams to help solve the probability problems.
  • Venn diagrams can be used to solve probability problems of more than three events.

Properties of Venn diagrams

A rectangle of a Venn diagram represents the sample space. Closed curves are drawn in the rectangle to represent events.

The Venn diagram below shows events A and B in a sample space S; In this diagram the rectangle S represents the sample space. The sample space includes all possible outcomes of an experiment therefore the probability of the entire sample space is 1 i.e P(S) = 1

The closed curve labelled A represents all outcomes of event A and the closed curve labelled B represents outcomes of event B.

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Event A n B

The diagram below shows event A n B “A intersection B”
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The overlap between event A and event B is written as A n B. This diagram represents ‘A intersection B’ where both events A and B occur.

Event A u B

The diagram below represents the union of event A and event B (A u B)

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The diagram represents the event that either A or B occurs or both occur.

Event A’

The diagram below shows A’ (the complement of A) which is the event that ‘A does not occur’

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Remember that the sample space has a probability of 1. If you know the probability of A P(A) that must mean that P(A’) = 1 – P(A)

Example
A card is selected at random from a pack of 53 playing cards. Let A be the event that the card is an ace and D the event that the card is a diamond. Find the following;

  1. P(A n D)
  2. P(A u D)
  3. P(A’)
  4. P(A’ n D)
Answer
We shall first draw a Venn diagram; There are 52 cards and each card is selected randomly that must mean that there are 52 equally likely outcomes. It is easier to start at the intersection of the curves and work outwards by subtracting.
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Note that the total of all the numbers in the Venn diagram add up to 52.

There are 13 diamonds and 4 aces. There is only one card that is both a diamond and an ace. That must mean that there is only one outcome in the intersection which is the ace of diamonds.

Since there is 13 diamonds, 4 aces and 1 ace of diamonds. That means there are 12 cards that are diamonds but not aces and 3 cards that are aces but not diamonds. There are 52 cards of which 16 are counted as aces and diamonds. So that leaves 52-16 = 36 cards that make up the rest of the sample space S

  1. A n D is the event that ‘a card chosen is an ace of diamonds’. We can read directly from the Venn diagram P(A n D). In the diagram there is 1 at the overlap of A and D and 52 outcomes in S that must mean that;
    P(A n D) = 1/52
  2. A u B is the event that ‘a card chosen is an ace or a diamond or the ace of diamonds’

    There are (3 + 1 + 12) = 16 outcomes in A u D and 52 outcomes in S. That must mean that;

    P(A U D) = 16/52 = 4/13

  3. Using the Venn diagram we must add up all values outside A;

    12 + 36 = 48

    A’ is the event that ‘the card chosen is not ace’

    P(A’) = 48/52 = 12/13
  4. A’ n D is the event that ‘the card chosen is not an ace and is diamond’ The event that we’re after has been shaded on the diagram
    [IMAGE]

    P(A’ n D) = 12/52 = 3/13

Example

In a class of 40 students, 20 are in the choir , 16 are in the school band and 4 are in the choir and the band. A student is chosen at random from the class.

  1. Draw a Venn diagram to represent this information.
  2. Find the probability that;
    • the student is not in the band
    • the student is not in the choir nor in the band.
Answer

  1. When drawing Venn diagrams it is easier to start with the intersection (B n C). Note there are 40 students in the sample space so all numbers must add up to 40

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    We have started with the intersection B n C which stands for the 4 students in the choir and band at the same time. There are 20 students that are in the band so we subtract 20-4 to find the students in the school band which results in a 16. And subtract 1604 to get 12 students in the choir alone. To find the rest we add and subtract; 40 – (16+4+12) = 8.

  2. We must find the probability that the student is not in the band. The probability a student is not in the band is given by P(B’)
    P(B’) = 1 – P(B)

    We can look in the diagram to find the probability P(B);

    P(B) = 16+4/40 = 1/2 = 0.5
  3. We must find;
    P(the student is not in the choir or the band)

    We can use the Venn diagram by adding up all the numbers in the curves;

    16 + 4 + 12 = 32

    There are 32 students in total that are in the choir or the band, so that would leave (40 – 32) = 8 students not in the band or choir.

    P(the student is not in the choir or the band)

    = 8/40 = 1/5 = 0.2
    When there is more than one decimal place in a probability it is better to leave it as a fraction.

Example

A vet decided to survey her clients to find out what kind of pet each client had. The results are shown that;

25 own dogs, 15 own dogs and cats, 11 own dogs and tropical fish, 53 own cats, 10 own cats and tropical fish, 7 own dogs, cats and tropical fish and 40 own tropical fish.

A client is chosen at random.

  1. Draw a Venn diagram to represent this information
  2. Find the probability that the client;
    • owns dogs only
    • does not own tropical fish
    • does not own dogs, cats or tropical fish.
Answer

There are three types of events; client owning a cat, dog or fish, therefore the Venn diagram is going to contain three closed curves.

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First you’ll need to draw a Venn diagram such as that shown above. We must start with the intersection of all events. This has a probability of 7 out of 100. P(DnCnF) = 7/100

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We continue working outwards to find other intersections;

15/100 – 7/1000 = 8/100
10/100 – 7/100 = 3/100
11/100 – 7/100 = 4/100

We can update these in the Venn diagram;
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We can subtract further to find the probabilities for ‘dogs only’, ‘cats only’ and ‘tropical fish only’

P(dogs only) = 25/100 – 8+7+4/100 = 6/100
P(fish only) = 40/100 – (4+7+3)/100 = 26/100
P(cats only) = 53/100 – (8+7+3)/100 = 35/100

We can update these in the Venn diagram;

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  1. own dogs only

    We have to find the probability that a client owns a dog only. P(client owns dogs only)

    = 25-(8+7+4)/100 – 6/100 = 3/50

  2. does not own tropical fish

    P(client does not own tropical fish)

    = 100-40/100 = 60/100 = 3/5

  3. does not own dogs, cats or tropical fish

    The probability that a client does not own dogs, cats or tropical fish is the value outside the curve D, C, and F on the Venn diagram;

    P(client does not own any of the pets) = 11/100

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