OpenRevise

Exam questions · Maths · Probability

Tree Diagrams

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  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\(0.9\)
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    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\(0.49\)
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    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\(\dfrac{3}{4}\)
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    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\(\dfrac{3}{10}\)
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    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\(\dfrac{3}{10}\)
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    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\(\dfrac{3}{10}\)
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    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\(1.6\)
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    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\(\dfrac{1}{3}\)
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    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\(12\)
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    C: \(10\)

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