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Exam questions · Maths · Probability

Venn Diagrams and Set Notation

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  1. 1 Write down [2 marks]

    \(\xi = \{1, 2, 3, \ldots, 12\}\). \(A\) is the set of factors of 12 and \(B\) is the set of odd numbers. (a) Write down \(A \cap B\). [1 mark] (b) Work out \(n(A' \cap B')\). [1 mark]

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    Model answer

    (a) \(A = \{1, 2, 3, 4, 6, 12\}\) and \(B = \{1, 3, 5, 7, 9, 11\}\), so \(A \cap B = \{1, 3\}\). (b) Neither set: \(\{8, 10\}\), so \(n(A' \cap B') = 2\).

    Mark scheme

    • (a) \(\{1, 3\}\) — B1
    • (b) 2 — B1
  2. 2 Calculate [4 marks]

    There are 40 students in a class. 19 have a cat and 15 have a dog. The incomplete Venn diagram shows 4 students with both and 10 with neither. (a) Complete the Venn diagram. [2 marks] (b) A student is chosen at random. Calculate the probability that the student has a dog but not a cat. [2 marks]

    A Venn diagram of cats and dogs with 4 in the overlap, 10 outside both circles and the other two regions empty.
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    Model answer

    (a) Cats only is \(19 - 4 = 15\) and dogs only is \(15 - 4 = 11\). Check: \(15 + 4 + 11 + 10 = 40\). (b) \(\dfrac{11}{40}\).

    Mark scheme

    • (a) 15 — B1
    • (a) 11 — B1
    • (b) \(\dfrac{11}{40}\) with the correct denominator 40 — M1
    • (b) \(\dfrac{11}{40}\) — A1
  3. 3 Calculate [3 marks]

    In a survey of 80 people, 45 read newspaper \(N\), 30 read magazine \(M\) and 15 read both. Calculate the number of people who read neither. [3 marks]

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    Model answer

    At least one: \(45 + 30 - 15 = 60\). Neither: \(80 - 60 = 20\).

    Mark scheme

    • \(45 + 30 - 15\) — M1
    • 60 — A1
    • 20 — A1
  4. 4 Calculate [3 marks]

    \(A\) and \(B\) are mutually exclusive events. \(P(A) = 0.5\) and \(P(B) = 0.35\). (a) Calculate \(P(A \cup B)\). [1 mark] (b) Calculate the probability that neither event happens. [1 mark] (c) Write down \(P(A \cap B)\). [1 mark]

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    Model answer

    (a) \(0.5 + 0.35 = 0.85\). (b) \(1 - 0.85 = 0.15\). (c) The events cannot both happen, so \(P(A \cap B) = 0\).

    Mark scheme

    • (a) \(0.85\) — B1
    • (b) \(0.15\) — B1
    • (c) 0 — B1
  5. 5 Calculate [4 marks]

    In a Venn diagram of 68 people, \(x + 5\) are in set \(A\) only, \(2x\) are in both sets, \(x\) are in set \(B\) only and \(3x\) are in neither set. (a) Calculate the value of \(x\). [2 marks] (b) A person is chosen at random. Calculate the probability that the person is in both sets. [2 marks]

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    Model answer

    (a) \(x + 5 + 2x + x + 3x = 68\), so \(7x + 5 = 68\) and \(x = 9\). (b) Both sets: \(2x = 18\), so \(\dfrac{18}{68} = \dfrac{9}{34}\).

    Mark scheme

    • (a) \(7x + 5 = 68\) — M1
    • (a) \(x = 9\) — A1
    • (b) \(\dfrac{18}{68}\) — M1
    • (b) \(\dfrac{9}{34}\) — A1
  6. 6 Calculate [3 marks]

    \(\xi = \{1, 2, 3, \ldots, 20\}\). \(A\) is the set of multiples of 4 and \(B\) is the set of square numbers. (a) List the members of \(A \cap B\). [1 mark] (b) Calculate \(n(A \cup B)\). [2 marks]

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    Model answer

    (a) \(A = \{4, 8, 12, 16, 20\}\) and \(B = \{1, 4, 9, 16\}\), so \(A \cap B = \{4, 16\}\). (b) \(n(A) + n(B) - n(A \cap B) = 5 + 4 - 2 = 7\).

    Mark scheme

    • (a) \(\{4, 16\}\) — B1
    • (b) \(5 + 4 - 2\) or a list of the union — M1
    • (b) 7 — A1

Quick check

  1. 1

    What does \(A \cap B\) mean?

    1. AThe items in \(A\) or \(B\)
    2. BThe items not in \(A\)
    3. CThe items only in \(A\)
    4. DThe items in both \(A\) and \(B\)
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    D: The items in both \(A\) and \(B\)

    \(\cap\) is the intersection, the overlap.

  2. 2

    What does \(A \cup B\) mean?

    1. AThe items in both \(A\) and \(B\)
    2. BThe items not in \(B\)
    3. CThe items in \(A\) or \(B\) or both
    4. DThe items only in \(B\)
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    C: The items in \(A\) or \(B\) or both

    \(\cup\) is the union, everything in either circle.

  3. 3

    What does \(A'\) mean?

    1. AThe items in \(A\) only
    2. BThe items not in \(A\)
    3. CThe items in both sets
    4. DThe number of items in \(A\)
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    B: The items not in \(A\)

    \(A'\) is the complement of \(A\).

  4. 4

    \(A = \{2, 4, 6, 8, 10\}\) and \(B = \{3, 6, 9\}\). What is \(A \cap B\)?

    1. A\(\{6\}\)
    2. B\(\{3, 6, 9\}\)
    3. C\(\{2, 3, 4, 6, 8, 9, 10\}\)
    4. D\(\{\}\)
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    A: \(\{6\}\)

    Only 6 is in both sets.

  5. 5

    18 students play football and 6 of them also play tennis. How many play football only?

    1. A\(18\)
    2. B\(24\)
    3. C\(6\)
    4. D\(12\)
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    D: \(12\)

    \(18 - 6 = 12\).

  6. 6

    30 students: 18 play football, 14 play tennis and 6 play both. How many play neither?

    1. A\(2\)
    2. B\(8\)
    3. C\(4\)
    4. D\(10\)
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    C: \(4\)

    \(12 + 6 + 8 = 26\) play at least one, so \(30 - 26 = 4\).

  7. 7

    In the same survey, what is the probability that a student chosen at random plays football only?

    1. A\(\dfrac{3}{5}\)
    2. B\(\dfrac{2}{5}\)
    3. C\(\dfrac{1}{5}\)
    4. D\(\dfrac{18}{30}\)
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    B: \(\dfrac{2}{5}\)

    \(\dfrac{12}{30} = \dfrac{2}{5}\). The fraction \(\dfrac{18}{30}\) would include those who play both.

  8. 8

    In a class of 40, 22 play football, 19 play tennis and 5 play neither. How many play both?

    1. A\(6\)
    2. B\(3\)
    3. C\(9\)
    4. D\(17\)
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    A: \(6\)

    \(40 - 5 = 35\) play at least one, and \(22 + 19 - 35 = 6\).

  9. 9

    \(n(A) = 15\), \(n(B) = 12\) and \(n(A \cap B) = 5\). What is \(n(A \cup B)\)?

    1. A\(27\)
    2. B\(32\)
    3. C\(17\)
    4. D\(22\)
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    D: \(22\)

    \(15 + 12 - 5 = 22\).