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Exam questions · Maths

Probability

  • 30 exam questions
  • 96 marks
  • 45 quick checks

Probability Basics and Relative Frequency

Just this lesson
  1. 1 Find [2 marks]

    The letters of the word MATHEMATICS are written on 11 cards, one letter on each card. One card is chosen at random. Find the probability that it is (a) the letter M, (b) a vowel. [2 marks]

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    Model answer

    (a) There are two Ms, so \(\dfrac{2}{11}\). (b) The vowels are A, E, A, I, so \(\dfrac{4}{11}\).

    Mark scheme

    • (a) \(\dfrac{2}{11}\) — B1
    • (b) \(\dfrac{4}{11}\) — B1
  2. 2 Calculate [3 marks]

    Hana spins a spinner 100 times. The results are red 38, blue 27 and green 35. (a) Write down the relative frequency of blue as a decimal. [1 mark] (b) Hana spins the spinner 500 times. Estimate the number of times it lands on blue. [2 marks]

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    Model answer

    (a) \(\dfrac{27}{100} = 0.27\). (b) \(0.27 \times 500 = 135\).

    Mark scheme

    • (a) \(0.27\) — B1
    • (b) \(0.27 \times 500\) — M1
    • (b) 135 — A1
  3. 3 Calculate [3 marks]

    The probability that a bus is on time is 0.72. (a) Calculate the probability that the bus is not on time. [1 mark] (b) Out of 50 days, estimate the number of days that the bus is on time. [2 marks]

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    Model answer

    (a) \(1 - 0.72 = 0.28\). (b) \(0.72 \times 50 = 36\).

    Mark scheme

    • (a) \(0.28\) — B1
    • (b) \(0.72 \times 50\) — M1
    • (b) 36 — A1
  4. 4 Calculate [3 marks]

    A bag contains only red, white and green counters. The probability of picking a red counter is \(x\). The probability of picking a white counter is \(2x\). The probability of picking a green counter is 0.4. (a) Calculate the value of \(x\). [2 marks] (b) Write down the probability of picking a white counter. [1 mark]

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    Model answer

    (a) \(x + 2x + 0.4 = 1\), so \(3x = 0.6\) and \(x = 0.2\). (b) \(2x = 0.4\).

    Mark scheme

    • (a) \(x + 2x + 0.4 = 1\) — M1
    • (a) \(0.2\) — A1
    • (b) \(0.4\) — B1
  5. 5 Explain [2 marks]

    Ben throws a normal dice. He says, “Either I get a six or I do not, so the probability of a six is \(\dfrac{1}{2}\).” Explain why Ben is wrong. [2 marks]

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    Model answer

    The two outcomes are not equally likely. There is only one way to get a six out of six equally likely outcomes, so the probability is \(\dfrac{1}{6}\).

    Mark scheme

    • States that the outcomes are not equally likely — B1
    • States that the probability of a six is \(\dfrac{1}{6}\) — B1
  6. 6 Calculate [3 marks]

    \(P(F) = 0.55\), \(P(T) = 0.35\) and the probability that neither \(F\) nor \(T\) happens is 0.2. Calculate \(P(F \text{ and } T)\). [3 marks]

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    Model answer

    \(P(F \text{ or } T) = 1 - 0.2 = 0.8\). Using \(P(F \text{ or } T) = P(F) + P(T) - P(F \text{ and } T)\) gives \(0.8 = 0.9 - P(F \text{ and } T)\), so \(P(F \text{ and } T) = 0.1\).

    Mark scheme

    • \(1 - 0.2 = 0.8\) — M1
    • \(0.55 + 0.35 - 0.8\) — M1
    • \(0.1\) — A1

Quick check

  1. 1

    What is the probability of an impossible event?

    1. A\(1\)
    2. B\(0\)
    3. C\(\dfrac{1}{2}\)
    4. D\(-1\)
    Show answerHide answer

    B: \(0\)

    An impossible event has probability 0.

  2. 2

    A bag has 3 red, 5 blue and 2 green counters. One is taken at random. What is the probability it is blue?

    1. A\(\dfrac{1}{2}\)
    2. B\(\dfrac{1}{5}\)
    3. C\(\dfrac{1}{3}\)
    4. D\(\dfrac{3}{5}\)
    Show answerHide answer

    A: \(\dfrac{1}{2}\)

    There are 10 counters and 5 are blue, so \(\dfrac{5}{10} = \dfrac{1}{2}\).

  3. 3

    A bag has 3 red, 5 blue and 2 green counters. What is the probability that a counter taken at random is not green?

    1. A\(\dfrac{1}{5}\)
    2. B\(\dfrac{2}{5}\)
    3. C\(\dfrac{3}{10}\)
    4. D\(\dfrac{4}{5}\)
    Show answerHide answer

    D: \(\dfrac{4}{5}\)

    \(1 - \dfrac{2}{10} = \dfrac{8}{10} = \dfrac{4}{5}\).

  4. 4

    A dice is thrown 120 times and a 6 comes up 30 times. What is the relative frequency of a 6?

    1. A\(\dfrac{1}{6}\)
    2. B\(\dfrac{1}{30}\)
    3. C\(\dfrac{1}{4}\)
    4. D\(4\)
    Show answerHide answer

    C: \(\dfrac{1}{4}\)

    \(\dfrac{30}{120} = \dfrac{1}{4}\).

  5. 5

    A fair dice is thrown 600 times. How many sixes are expected?

    1. A\(60\)
    2. B\(100\)
    3. C\(120\)
    4. D\(600\)
    Show answerHide answer

    B: \(100\)

    \(\dfrac{1}{6} \times 600 = 100\).

  6. 6

    Two fair dice are thrown. What is the probability that the total score is 7?

    1. A\(\dfrac{1}{6}\)
    2. B\(\dfrac{1}{12}\)
    3. C\(\dfrac{7}{36}\)
    4. D\(\dfrac{1}{36}\)
    Show answerHide answer

    A: \(\dfrac{1}{6}\)

    There are 6 ways to make 7 out of 36, so \(\dfrac{6}{36} = \dfrac{1}{6}\).

  7. 7

    A bag has 3 red, 5 blue and 2 green counters. What is the probability of taking a red or a green counter?

    1. A\(\dfrac{3}{5}\)
    2. B\(\dfrac{1}{5}\)
    3. C\(\dfrac{3}{10}\)
    4. D\(\dfrac{1}{2}\)
    Show answerHide answer

    D: \(\dfrac{1}{2}\)

    The events are mutually exclusive, so \(\dfrac{3}{10} + \dfrac{2}{10} = \dfrac{1}{2}\).

  8. 8

    A spinner is spun 200 times and lands on red 74 times. How many reds would be expected if the probability of red were \(\dfrac{1}{4}\)?

    1. A\(74\)
    2. B\(25\)
    3. C\(50\)
    4. D\(200\)
    Show answerHide answer

    C: \(50\)

    \(\dfrac{1}{4} \times 200 = 50\). The result of 74 suggests the spinner may be biased.

  9. 9

    A card is taken from a pack of 52. What is the probability that it is a heart or a king?

    1. A\(\dfrac{17}{52}\)
    2. B\(\dfrac{4}{13}\)
    3. C\(\dfrac{1}{13}\)
    4. D\(\dfrac{5}{26}\)
    Show answerHide answer

    B: \(\dfrac{4}{13}\)

    \(\dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}\).

Tree Diagrams

Just this lesson
  1. 1 Find [2 marks]

    A spinner lands on red with probability 0.4. It is spun twice. Find the probability that it lands on red both times. [2 marks]

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    Model answer

    \(0.4 \times 0.4 = 0.16\).

    Mark scheme

    • \(0.4 \times 0.4\) — M1
    • \(0.16\) — A1
  2. 2 Calculate [4 marks]

    Zara plays two games of chess against Ahmed. The probability that Zara wins a game is \(\dfrac{2}{3}\). The games are independent. The incomplete tree diagram shows the first game and the second game. (a) Complete the tree diagram. [2 marks] (b) Calculate the probability that Zara wins exactly one of the two games. [2 marks]

    A probability tree diagram for two games, with the probability of winning 2/3 on the first branch and the other probabilities missing.
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    Model answer

    (a) The probability of losing is \(1 - \dfrac{2}{3} = \dfrac{1}{3}\). Every second-game branch is \(\dfrac{2}{3}\) for win and \(\dfrac{1}{3}\) for lose. (b) \(\dfrac{2}{3} \times \dfrac{1}{3} + \dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{4}{9}\).

    Mark scheme

    • (a) \(\dfrac{1}{3}\) on the first lose branch — B1
    • (a) \(\dfrac{2}{3}\) and \(\dfrac{1}{3}\) on all second-game branches — B1
    • (b) \(\dfrac{2}{3} \times \dfrac{1}{3}\) and \(\dfrac{1}{3} \times \dfrac{2}{3}\) added — M1
    • (b) \(\dfrac{4}{9}\) — A1
  3. 3 Calculate [3 marks]

    A bag contains 4 lemon sweets and 3 mint sweets. Two sweets are taken at random without replacement. Calculate the probability that the two sweets have different flavours. [3 marks]

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    Model answer

    Lemon then mint: \(\dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{12}{42}\). Mint then lemon: \(\dfrac{3}{7} \times \dfrac{4}{6} = \dfrac{12}{42}\). The total is \(\dfrac{24}{42} = \dfrac{4}{7}\).

    Mark scheme

    • \(\dfrac{4}{7} \times \dfrac{3}{6}\) or \(\dfrac{3}{7} \times \dfrac{4}{6}\) — M1
    • Both products added — M1
    • \(\dfrac{4}{7}\) or \(\dfrac{24}{42}\) — A1
  4. 4 Calculate [3 marks]

    Two machines work independently. The probability that machine \(A\) works is 0.9 and the probability that machine \(B\) works is 0.8. Calculate the probability that at least one of the machines works. [3 marks]

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    Model answer

    The probability that neither works is \(0.1 \times 0.2 = 0.02\), so the probability that at least one works is \(1 - 0.02 = 0.98\).

    Mark scheme

    • \(0.1\) and \(0.2\) seen — M1
    • \(0.1 \times 0.2\) — M1
    • \(0.98\) — A1
  5. 5 Calculate [4 marks]

    A bag contains 5 blue counters and 3 green counters. Two counters are taken at random without replacement. (a) Show that the probability that both counters are blue is \(\dfrac{5}{14}\). [2 marks] (b) Calculate the probability that at least one counter is green. [2 marks]

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    Model answer

    (a) \(\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}\). (b) \(1 - \dfrac{5}{14} = \dfrac{9}{14}\).

    Mark scheme

    • (a) \(\dfrac{5}{8} \times \dfrac{4}{7}\) — M1
    • (a) \(\dfrac{20}{56} = \dfrac{5}{14}\) shown — A1
    • (b) \(1 - \dfrac{5}{14}\) — M1
    • (b) \(\dfrac{9}{14}\) — A1
  6. 6 Calculate [3 marks]

    Three cards are numbered 1, 2 and 3. Two cards are taken at random without replacement, and the two numbers are added. Calculate the probability that the total is even. [3 marks]

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    Model answer

    The possible pairs are 1 and 2, 1 and 3, and 2 and 3, which are equally likely. The totals are 3, 4 and 5, so only one total is even. The probability is \(\dfrac{1}{3}\).

    Mark scheme

    • Lists the three possible pairs or totals — M1
    • Totals 3, 4 and 5 — A1
    • \(\dfrac{1}{3}\) — A1

Quick check

  1. 1

    The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on both of two days?

    1. A\(0.6\)
    2. B\(0.3\)
    3. C\(0.09\)
    4. D\(0.9\)
    Show answerHide answer

    C: \(0.09\)

    \(0.3 \times 0.3 = 0.09\).

  2. 2

    The probability of rain on any day is 0.3, and days are independent. What is the probability of rain on at least one of two days?

    1. A\(0.09\)
    2. B\(0.51\)
    3. C\(0.42\)
    4. D\(0.49\)
    Show answerHide answer

    B: \(0.51\)

    \(1 - 0.7 \times 0.7 = 1 - 0.49 = 0.51\).

  3. 3

    Two fair coins are tossed. What is the probability of two heads?

    1. A\(\dfrac{1}{4}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{1}{3}\)
    4. D\(\dfrac{3}{4}\)
    Show answerHide answer

    A: \(\dfrac{1}{4}\)

    \(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).

  4. 4

    A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that both are red?

    1. A\(\dfrac{9}{25}\)
    2. B\(\dfrac{3}{5}\)
    3. C\(\dfrac{1}{10}\)
    4. D\(\dfrac{3}{10}\)
    Show answerHide answer

    D: \(\dfrac{3}{10}\)

    \(\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}\).

  5. 5

    A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are different colours?

    1. A\(\dfrac{2}{5}\)
    2. B\(\dfrac{12}{25}\)
    3. C\(\dfrac{3}{5}\)
    4. D\(\dfrac{3}{10}\)
    Show answerHide answer

    C: \(\dfrac{3}{5}\)

    \(\dfrac{3}{5} \times \dfrac{2}{4} + \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{12}{20} = \dfrac{3}{5}\).

  6. 6

    A bag has 3 red and 2 blue counters. Two are taken without replacement. What is the probability that they are the same colour?

    1. A\(\dfrac{3}{5}\)
    2. B\(\dfrac{2}{5}\)
    3. C\(\dfrac{13}{25}\)
    4. D\(\dfrac{3}{10}\)
    Show answerHide answer

    B: \(\dfrac{2}{5}\)

    \(\dfrac{6}{20} + \dfrac{2}{20} = \dfrac{8}{20} = \dfrac{2}{5}\).

  7. 7

    On a tree diagram, one branch has probability 0.6. What is the other branch from the same point?

    1. A\(0.4\)
    2. B\(0.6\)
    3. C\(0.5\)
    4. D\(1.6\)
    Show answerHide answer

    A: \(0.4\)

    The branches from one point add up to 1.

  8. 8

    A bag has 4 red and 6 blue counters. Two are taken without replacement. What is the probability that both are blue?

    1. A\(\dfrac{9}{25}\)
    2. B\(\dfrac{2}{5}\)
    3. C\(\dfrac{3}{10}\)
    4. D\(\dfrac{1}{3}\)
    Show answerHide answer

    D: \(\dfrac{1}{3}\)

    \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).

  9. 9

    A bag has 4 red counters and some blue counters, \(n\) in all. Two are taken without replacement and \(P(\text{two reds}) = \dfrac{2}{15}\). How many counters are in the bag?

    1. A\(6\)
    2. B\(8\)
    3. C\(10\)
    4. D\(12\)
    Show answerHide answer

    C: \(10\)

    \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\), so \(n = 10\).

Venn Diagrams and Set Notation

Just this lesson
  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]

    Show answerHide answer

    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.
    Show answerHide answer

    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]

    Show answerHide answer

    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]

    Show answerHide answer

    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]

    Show answerHide answer

    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]

    Show answerHide answer

    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 \(B\)
    Show answerHide answer

    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 \(B\)
    Show answerHide answer

    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 \(A\)
    Show answerHide answer

    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\(\{\}\)
    Show answerHide answer

    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\(12\)
    Show answerHide answer

    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\(10\)
    Show answerHide answer

    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\(\dfrac{18}{30}\)
    Show answerHide answer

    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\(17\)
    Show answerHide answer

    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\(22\)
    Show answerHide answer

    D: \(22\)

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

Frequency Trees and Two-Way Tables

Just this lesson
  1. 1 Work out [3 marks]

    A club has 45 members. 27 of them are girls. 9 of the girls and 6 of the boys are left-handed. (a) Calculate the number of boys. [1 mark] (b) Calculate the number of left-handed members. [1 mark] (c) Calculate the number of girls who are right-handed. [1 mark]

    Show answerHide answer

    Model answer

    (a) \(45 - 27 = 18\). (b) \(9 + 6 = 15\). (c) \(27 - 9 = 18\).

    Mark scheme

    • (a) 18 — B1
    • (b) 15 — B1
    • (c) 18 — B1
  2. 2 Calculate [4 marks]

    The two-way table shows the drinks ordered by 130 customers in a cafe. Some of the numbers are missing. (a) Complete the two-way table. [3 marks] (b) A customer is chosen at random. Calculate the probability that the customer is a child who ordered tea. [1 mark]

    A two-way table of the drinks ordered by adults and children, with four missing values.
    Show answerHide answer

    Model answer

    (a) Adults who ordered juice: \(90 - 35 - 40 = 15\). Children who ordered tea: \(40 - 5 - 30 = 5\). Coffee total: \(40 + 5 = 45\). Overall total: \(90 + 40 = 130\). (b) \(\dfrac{5}{130} = \dfrac{1}{26}\).

    Mark scheme

    • (a) At least two correct values — B1
    • (a) At least three correct values — B1
    • (a) 15, 5, 45 and 130 all correct — B1
    • (b) \(\dfrac{1}{26}\) or \(\dfrac{5}{130}\) — B1
  3. 3 Calculate [3 marks]

    There are 80 cars in a car park. 55% of the cars are petrol and the rest are diesel. \(\dfrac{1}{4}\) of the petrol cars and \(\dfrac{3}{4}\) of the diesel cars are red. Calculate the number of red cars. [3 marks]

    Show answerHide answer

    Model answer

    Petrol: \(0.55 \times 80 = 44\) and diesel: 36. Red cars: \(\dfrac{1}{4} \times 44 = 11\) and \(\dfrac{3}{4} \times 36 = 27\). The total is \(11 + 27 = 38\).

    Mark scheme

    • 44 petrol and 36 diesel — M1
    • 11 and 27 seen — M1
    • 38 — A1
  4. 4 Calculate [3 marks]

    There are 28 pupils in a class. 16 of them are girls. 9 of the girls and 5 of the boys have a pet. A pupil is chosen at random. (a) Calculate the number of pupils who do not have a pet. [1 mark] (b) Calculate the probability that the pupil has a pet. [2 marks]

    Show answerHide answer

    Model answer

    (a) \(9 + 5 = 14\) have a pet, so \(28 - 14 = 14\) do not. (b) \(\dfrac{14}{28} = \dfrac{1}{2}\).

    Mark scheme

    • (a) 14 — B1
    • (b) \(\dfrac{14}{28}\) — M1
    • (b) \(\dfrac{1}{2}\) — A1
  5. 5 Calculate [3 marks]

    In a school, 40% of the pupils are in the lower school. 15% of the lower school and 35% of the upper school walk home. A pupil is chosen at random. Calculate the probability that the pupil walks home. [3 marks]

    Show answerHide answer

    Model answer

    Lower school: \(0.4 \times 0.15 = 0.06\). Upper school: \(0.6 \times 0.35 = 0.21\). The total is \(0.06 + 0.21 = 0.27\).

    Mark scheme

    • \(0.4 \times 0.15 = 0.06\) — M1
    • \(0.6 \times 0.35 = 0.21\) — M1
    • \(0.27\) — A1
  6. 6 Calculate [4 marks]

    There are 120 people at a party. The ratio of men to women is \(2 : 3\). \(\dfrac{3}{4}\) of the men and \(\dfrac{1}{2}\) of the women wear a hat. A person is chosen at random. Calculate the probability that the person wears a hat. [4 marks]

    Show answerHide answer

    Model answer

    Men: \(120 \div 5 \times 2 = 48\) and women: 72. Hats: \(\dfrac{3}{4} \times 48 = 36\) and \(\dfrac{1}{2} \times 72 = 36\), a total of 72. The probability is \(\dfrac{72}{120} = \dfrac{3}{5}\).

    Mark scheme

    • 48 men and 72 women — M1
    • 36 and 36 seen — M1
    • \(\dfrac{72}{120}\) — A1
    • \(\dfrac{3}{5}\) — A1

Quick check

  1. 1

    A frequency tree starts with 100 pupils, and 60 are girls. How many are boys?

    1. A\(40\)
    2. B\(60\)
    3. C\(160\)
    4. D\(100\)
    Show answerHide answer

    A: \(40\)

    \(100 - 60 = 40\).

  2. 2

    40% of 60 girls walk to school. How many girls walk?

    1. A\(36\)
    2. B\(40\)
    3. C\(20\)
    4. D\(24\)
    Show answerHide answer

    D: \(24\)

    \(0.4 \times 60 = 24\).

  3. 3

    In a two-way table the Year 10 row total is 50. 24 take the bus and 18 walk. How many cycle?

    1. A\(42\)
    2. B\(6\)
    3. C\(8\)
    4. D\(18\)
    Show answerHide answer

    C: \(8\)

    \(50 - 24 - 18 = 8\).

  4. 4

    In a survey of 100 students, 24 are in Year 10 and take the bus. What is the probability that a student is in Year 10 and takes the bus?

    1. A\(\dfrac{1}{24}\)
    2. B\(\dfrac{6}{25}\)
    3. C\(\dfrac{24}{50}\)
    4. D\(\dfrac{1}{4}\)
    Show answerHide answer

    B: \(\dfrac{6}{25}\)

    \(\dfrac{24}{100} = \dfrac{6}{25}\).

  5. 5

    13 out of 100 students cycle to school. How many of 500 students would you expect to cycle?

    1. A\(65\)
    2. B\(13\)
    3. C\(130\)
    4. D\(6.5\)
    Show answerHide answer

    A: \(65\)

    \(\dfrac{13}{100} \times 500 = 65\).

  6. 6

    What is 25% of 120?

    1. A\(25\)
    2. B\(48\)
    3. C\(95\)
    4. D\(30\)
    Show answerHide answer

    D: \(30\)

    \(0.25 \times 120 = 30\).

  7. 7

    200 people go to a gym. 60% are women. 25% of the women and 40% of the men go in the morning. How many people go in the morning?

    1. A\(54\)
    2. B\(60\)
    3. C\(62\)
    4. D\(70\)
    Show answerHide answer

    C: \(62\)

    120 women and 80 men. \(30 + 32 = 62\).

  8. 8

    50 people were asked about pets. 28 have a cat, 20 have a dog and 6 have both. How many have neither?

    1. A\(6\)
    2. B\(8\)
    3. C\(14\)
    4. D\(22\)
    Show answerHide answer

    B: \(8\)

    At least one: \(22 + 6 + 14 = 42\). \(50 - 42 = 8\).

  9. 9

    In a frequency tree, the branches after “60 girls” show 24 who walk and some who take the bus. How many take the bus?

    1. A\(36\)
    2. B\(24\)
    3. C\(84\)
    4. D\(60\)
    Show answerHide answer

    A: \(36\)

    The branches add up to the number they came from: \(60 - 24 = 36\).

Conditional Probability

Just this lesson
  1. 1 Calculate [2 marks]

    A bag contains 4 blue counters and 3 red counters. One counter is taken at random and not replaced. Another counter is then taken. Given that the first counter is blue, calculate the probability that the second counter is blue. [2 marks]

    Show answerHide answer

    Model answer

    After a blue counter is taken there are 6 counters left and 3 are blue, so \(\dfrac{3}{6} = \dfrac{1}{2}\).

    Mark scheme

    • 3 blue out of 6 remaining — M1
    • \(\dfrac{1}{2}\) — A1
  2. 2 Calculate [5 marks]

    The table shows the results of a test for 80 students. (a) A student is chosen at random. Calculate the probability that the student passed. [1 mark] (b) A student who revised is chosen at random. Calculate the probability that the student passed. [2 marks] (c) A student who failed is chosen at random. Calculate the probability that the student did not revise. [2 marks]

    A two-way table of test results for 50 students who revised and 30 who did not.
    Show answerHide answer

    Model answer

    (a) \(\dfrac{54}{80} = \dfrac{27}{40}\). (b) \(\dfrac{42}{50} = \dfrac{21}{25}\). (c) 26 students failed and 18 of them did not revise, so \(\dfrac{18}{26} = \dfrac{9}{13}\).

    Mark scheme

    • (a) \(\dfrac{27}{40}\) or \(\dfrac{54}{80}\) — B1
    • (b) \(\dfrac{42}{50}\) — M1
    • (b) \(\dfrac{21}{25}\) — A1
    • (c) \(\dfrac{18}{26}\) — M1
    • (c) \(\dfrac{9}{13}\) — A1
  3. 3 Calculate [3 marks]

    \(P(A \text{ and } B) = 0.12\) and \(P(B) = 0.4\). (a) Calculate \(P(A \text{ given } B)\). [2 marks] Given also that \(P(A) = 0.3\), (b) state whether \(A\) and \(B\) are independent, giving a reason. [1 mark]

    Show answerHide answer

    Model answer

    (a) \(\dfrac{0.12}{0.4} = 0.3\). (b) Yes, because \(P(A \text{ given } B) = 0.3 = P(A)\).

    Mark scheme

    • (a) \(\dfrac{0.12}{0.4}\) — M1
    • (a) \(0.3\) — A1
    • (b) Independent, because \(P(A \text{ given } B) = P(A)\) — B1
  4. 4 Calculate [4 marks]

    In a group of 60 people, 28 cycle (\(C\)), 22 swim (\(S\)) and 8 do both. (a) Calculate \(P(S \text{ given } C)\). [2 marks] (b) Calculate the probability that a person who does not swim cycles. [2 marks]

    Show answerHide answer

    Model answer

    (a) \(\dfrac{8}{28} = \dfrac{2}{7}\). (b) \(60 - 22 = 38\) do not swim. Of these, \(28 - 8 = 20\) cycle, so \(\dfrac{20}{38} = \dfrac{10}{19}\).

    Mark scheme

    • (a) \(\dfrac{8}{28}\) — M1
    • (a) \(\dfrac{2}{7}\) — A1
    • (b) \(\dfrac{20}{38}\) — M1
    • (b) \(\dfrac{10}{19}\) — A1
  5. 5 Calculate [4 marks]

    The probability that it rains on a day is 0.3. If it rains, the probability that Kim is late for school is 0.6. If it does not rain, the probability that Kim is late is 0.1. Given that Kim is late, calculate the probability that it rained. [4 marks]

    Show answerHide answer

    Model answer

    \(P(\text{rain and late}) = 0.3 \times 0.6 = 0.18\) and \(P(\text{no rain and late}) = 0.7 \times 0.1 = 0.07\). So \(P(\text{late}) = 0.25\), and the probability is \(\dfrac{0.18}{0.25} = \dfrac{18}{25}\).

    Mark scheme

    • \(0.3 \times 0.6 = 0.18\) — M1
    • \(0.7 \times 0.1 = 0.07\) — M1
    • \(\dfrac{0.18}{0.25}\) — M1
    • \(\dfrac{18}{25}\) or 0.72 — A1
  6. 6 Calculate [4 marks]

    A bag contains 3 red apples and 2 green apples. Two apples are taken at random without replacement. Given that at least one apple is green, calculate the probability that both are green. [4 marks]

    Show answerHide answer

    Model answer

    \(P(\text{both green}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{10}\). \(P(\text{no green}) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{3}{10}\), so \(P(\text{at least one green}) = \dfrac{7}{10}\). The probability is \(\dfrac{1}{10} \div \dfrac{7}{10} = \dfrac{1}{7}\).

    Mark scheme

    • \(\dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{10}\) — M1
    • \(1 - \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{7}{10}\) — M1
    • \(\dfrac{1}{10} \div \dfrac{7}{10}\) — M1
    • \(\dfrac{1}{7}\) — A1

Quick check

  1. 1

    What does \(P(A \mid B)\) mean?

    1. AThe probability of \(A\) and \(B\)
    2. BThe probability of \(A\) given that \(B\) has happened
    3. CThe probability of \(A\) or \(B\)
    4. DThe probability of \(B\) given \(A\)
    Show answerHide answer

    B: The probability of \(A\) given that \(B\) has happened

    The vertical line means “given that”.

  2. 2

    14 students play tennis, and 6 of them also play football. What is the probability that a tennis player also plays football?

    1. A\(\dfrac{3}{7}\)
    2. B\(\dfrac{6}{30}\)
    3. C\(\dfrac{1}{3}\)
    4. D\(\dfrac{14}{30}\)
    Show answerHide answer

    A: \(\dfrac{3}{7}\)

    The group is the 14 tennis players: \(\dfrac{6}{14} = \dfrac{3}{7}\).

  3. 3

    Of 50 people who drive to work, 20 are female. A driver is chosen at random. What is the probability that the person is male?

    1. A\(\dfrac{2}{5}\)
    2. B\(\dfrac{30}{100}\)
    3. C\(\dfrac{1}{2}\)
    4. D\(\dfrac{3}{5}\)
    Show answerHide answer

    D: \(\dfrac{3}{5}\)

    \(\dfrac{30}{50} = \dfrac{3}{5}\). The group is the 50 drivers.

  4. 4

    Of 100 people, 36 male drivers and 24 female drivers drive to work. A driver is chosen at random. What is the probability that the driver is female?

    1. A\(\dfrac{24}{100}\)
    2. B\(\dfrac{3}{5}\)
    3. C\(\dfrac{2}{5}\)
    4. D\(\dfrac{24}{40}\)
    Show answerHide answer

    C: \(\dfrac{2}{5}\)

    The group is \(36 + 24 = 60\) drivers, and \(\dfrac{24}{60} = \dfrac{2}{5}\).

  5. 5

    A bag has 3 red and 2 blue counters. One red counter is taken and not replaced. What is the probability that the next counter is red?

    1. A\(\dfrac{3}{5}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{2}{5}\)
    4. D\(\dfrac{3}{4}\)
    Show answerHide answer

    B: \(\dfrac{1}{2}\)

    2 red counters are left among 4, so \(\dfrac{2}{4} = \dfrac{1}{2}\).

  6. 6

    Is \(P(A \mid B)\) always equal to \(P(B \mid A)\)?

    1. ANo, because they use different groups
    2. BYes, always
    3. COnly for independent events with different probabilities
    4. DOnly when \(A\) and \(B\) cannot both happen
    Show answerHide answer

    A: No, because they use different groups

    The group on the bottom is different in each.

  7. 7

    \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \text{ and } B) = 0.2\). Are \(A\) and \(B\) independent?

    1. ANo, because \(0.5 + 0.4 \neq 0.2\)
    2. BYes, because \(0.5 > 0.4\)
    3. CNo, because \(0.2 < 0.5\)
    4. DYes, because \(0.5 \times 0.4 = 0.2\)
    Show answerHide answer

    D: Yes, because \(0.5 \times 0.4 = 0.2\)

    Independent events have \(P(A \text{ and } B) = P(A) \times P(B)\).

  8. 8

    \(P(A \text{ and } B) = 0.1\) and \(P(B) = 0.4\). What is \(P(A \mid B)\)?

    1. A\(0.04\)
    2. B\(4\)
    3. C\(0.25\)
    4. D\(0.5\)
    Show answerHide answer

    C: \(0.25\)

    \(\dfrac{0.1}{0.4} = 0.25\).

  9. 9

    \(P(A \mid B) = 0.5\) and \(P(B) = 0.6\). What is \(P(A \text{ and } B)\)?

    1. A\(1.1\)
    2. B\(0.3\)
    3. C\(0.1\)
    4. D\(0.83\)
    Show answerHide answer

    B: \(0.3\)

    \(P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.6 = 0.3\).