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We use n ( S ) to refer to the number of elements in a set S , n ( X ) for the number of elements in X , etc.

In a box there are pieces of paper with the numbers from 1 to 9 written on them. A piece of paper is drawn from the box andthe number on it is noted. Let S denote the sample space, let P denote the event 'drawing a prime number', and let E denote the event 'drawing an even number'. Using appropriate notation, in how many ways is it possible to draw: i)any number? ii) a prime number? iii) an even number? iv) a number that is either prime or even? v) a number that is both prime and even?

    • Drawing a prime number: P = { 2 ; 3 ; 5 ; 7 }
    • Drawing an even number: E = { 2 ; 4 ; 6 ; 8 }
  1. The union of P and E is the set of all elements in P or in E (or in both). P E = 2 , 3 , 4 , 5 , 6 , 7 , 8 .

  2. The intersection of P and E is the set of all elements in both P and E . P E = 2 .

  3. n ( S ) = 9 n ( P ) = 4 n ( E ) = 4 n ( P E ) = 7 n ( P E ) = 2
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100 people were surveyed to find out which fast food chain (Nandos, Debonairs or Steers) they preferred. The following results were obtained:

  • 50 liked Nandos
  • 66 liked Debonairs
  • 40 liked Steers
  • 27 liked Nandos and Debonairs but not Steers
  • 13 liked Debonairs and Steers but not Nandos
  • 4 liked all three
  • 94 liked at least one
  1. How many people did not like any of the fast food chains?
  2. How many people liked Nandos and Steers, but not Debonairs?

  1. The number of people who liked Nandos and Debonairs is 27, so this is the intersection of these two events. The number of people who liked Debonairs and Steers is 13, so the intersection of Debonairs and Steers is 13. We are told that 4 people like all three options, and so this means that there are 4 people in the intersection of all three options. So we can work out that the number of people who like just Debonairs is 66 4 27 - 13 = 22 (This is simply the total number who like Debonairs minus the number of people who like Debonairs and Steers, or Debonairs and Nandos or all three). We draw the following diagram to represent the data:
  2. We are told that there were 100 people and that 94 liked at least one. So the number of people that liked none is: 100 94 = 6 . This is the answer to a).
  3. We can redraw the part of the Venn diagram that is of interest:
    Total people who like Nandos: 50
    Of these 27 like both Nandos and Debonairs, and 4 people like all three options. So we find that the total number of people who just like Nandos is: 50 27 4 = 19
    Total people who like Steers: 40
    Of these 13 like both Steers and Debonairs, and 4 like all three options. So we find that the total number of people who like just Steers is: 40 13 4 = 23
    Now use the identity n(N or S) = n(N) + n(S) n(N and S) to find the number of people who like Nandos and Steers, but not Debonairs.
    n(N or S) = n(N) + n(S) n(N and S) 28 = 23 + 19 n(N and S) n(N and S) = 14
    The Venn diagram with that represents all this information is given:
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Activity: venn diagrams

Which cellphone networks have you used or are you signed up for (e.g. Vodacom, Mtn or CellC)? Collect this information from your classmates as well. Then use the information to draw a Venn diagram (if you have more than three networks, then choose only the three most popular or draw Venn diagrams for all the combinations). Try to see if you can work out the number of people who use just one network, or the number of people who use all the networks.

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Source:  OpenStax, Siyavula textbooks: grade 10 maths [caps]. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11306/1.4
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