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Exam questions · Maths · Probability

Tree Diagrams

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  1. 1 Work out [2 marks]

    A fair dice is rolled and a fair coin is flipped. Work out the probability of getting a 6 and a head. [2 marks]

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    Model answer

    \(\dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}\).

    Mark scheme

    • \(\dfrac{1}{6} \times \dfrac{1}{2}\) — M1
    • \(\dfrac{1}{12}\) — A1
  2. 2 Work out [4 marks]

    A bag contains 3 yellow counters and 2 green counters. A counter is taken at random, and its colour is recorded. The counter is put back in the bag. A second counter is then taken. The incomplete tree diagram shows the first counter and the second counter. (a) Complete the tree diagram. [2 marks] (b) Work out the probability that the two counters are the same colour. [2 marks]

    A probability tree diagram for two counters taken from a bag, with the probability of yellow 3/5 on the first branch and the other probabilities missing.
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    Model answer

    (a) Green on the first counter is \(\dfrac{2}{5}\). All the second-counter branches are \(\dfrac{3}{5}\) for yellow and \(\dfrac{2}{5}\) for green, because the counter is replaced. (b) \(\dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{9}{25} + \dfrac{4}{25} = \dfrac{13}{25}\).

    Mark scheme

    • (a) \(\dfrac{2}{5}\) on the first green branch — B1
    • (a) \(\dfrac{3}{5}\) and \(\dfrac{2}{5}\) on all second-counter branches — B1
    • (b) \(\dfrac{3}{5} \times \dfrac{3}{5}\) and \(\dfrac{2}{5} \times \dfrac{2}{5}\) added — M1
    • (b) \(\dfrac{13}{25}\) — A1
  3. 3 Work out [3 marks]

    A box contains 8 chocolates. 5 are milk and 3 are plain. Bea takes two chocolates at random without replacement. Work out the probability that both are plain. [3 marks]

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    Model answer

    \(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}\).

    Mark scheme

    • \(\dfrac{3}{8}\) and \(\dfrac{2}{7}\) seen — M1
    • \(\dfrac{3}{8} \times \dfrac{2}{7}\) — M1
    • \(\dfrac{3}{28}\) or \(\dfrac{6}{56}\) — A1
  4. 4 Work out [4 marks]

    The probability that it rains on Saturday is 0.3. If it rains on Saturday, the probability that it rains on Sunday is 0.6. If it does not rain on Saturday, the probability that it rains on Sunday is 0.2. Work out the probability that it rains on at least one of the two days. [4 marks]

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    Model answer

    No rain on either day: \(0.7 \times 0.8 = 0.56\). So the probability of rain on at least one day is \(1 - 0.56 = 0.44\).

    Mark scheme

    • \(0.7\) and \(0.8\) seen — M1
    • \(0.7 \times 0.8 = 0.56\) — A1
    • \(1 - 0.56\) — M1
    • \(0.44\) — A1
  5. 5 Work out [3 marks]

    The probability that Raj is late for school on any day is 0.1. The days are independent. Work out the probability that he is late on exactly one of two days. [3 marks]

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    Model answer

    Late then on time: \(0.1 \times 0.9 = 0.09\). On time then late: \(0.9 \times 0.1 = 0.09\). The total is \(0.09 + 0.09 = 0.18\).

    Mark scheme

    • \(0.1 \times 0.9\) — M1
    • \(0.1 \times 0.9 + 0.9 \times 0.1\) — M1
    • \(0.18\) — A1
  6. 6 Work out [3 marks]

    A bag contains 4 red counters and 2 blue counters. Three counters are taken at random without replacement. Work out the probability that exactly one of them is blue. [3 marks]

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    Model answer

    One product is \(\dfrac{2}{6} \times \dfrac{4}{5} \times \dfrac{3}{4} = \dfrac{24}{120} = \dfrac{1}{5}\). There are three orders for the blue counter, and each has the same probability, so the total is \(3 \times \dfrac{1}{5} = \dfrac{3}{5}\).

    Mark scheme

    • One correct product such as \(\dfrac{2}{6} \times \dfrac{4}{5} \times \dfrac{3}{4}\) — M1
    • All three arrangements added or multiplied by 3 — M1
    • \(\dfrac{3}{5}\) — 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 \(n = 10\).