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Exam questions · Maths · Probability

Conditional Probability

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

    A bag contains 4 yellow counters and 6 green counters. One counter is taken at random and not replaced. Another counter is then taken. Given that the first counter is yellow, work out the probability that the second counter is yellow. [2 marks]

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    Model answer

    After the first counter is taken there are 9 left, 3 of them yellow, so \(\dfrac{3}{9} = \dfrac{1}{3}\).

    Mark scheme

    • 3 yellow out of 9 remaining — M1
    • \(\dfrac{1}{3}\) — A1
  2. 2 Work out [5 marks]

    The table shows the drinks chosen by 100 people. (a) A person is chosen at random. Work out the probability that the person chose lemonade. [1 mark] (b) A person under 18 is chosen at random. Work out the probability that the person chose cola. [2 marks] (c) A person who chose water is chosen at random. Work out the probability that the person is aged 18 or over. [2 marks]

    A two-way table showing the drinks chosen by 50 people under 18 and 50 people aged 18 and over.
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    Model answer

    (a) \(\dfrac{20}{100} = \dfrac{1}{5}\). (b) \(\dfrac{25}{50} = \dfrac{1}{2}\). (c) 40 people chose water and 30 of them are 18 or over, so \(\dfrac{30}{40} = \dfrac{3}{4}\).

    Mark scheme

    • (a) \(\dfrac{1}{5}\) or \(\dfrac{20}{100}\) — B1
    • (b) \(\dfrac{25}{50}\) — M1
    • (b) \(\dfrac{1}{2}\) — A1
    • (c) \(\dfrac{30}{40}\) — M1
    • (c) \(\dfrac{3}{4}\) — A1
  3. 3 Work out [3 marks]

    \(P(A) = 0.6\), \(P(B) = 0.5\) and \(P(A \text{ and } B) = 0.3\). (a) Work out \(P(B \text{ given } A)\). [2 marks] (b) Are \(A\) and \(B\) independent? Give a reason. [1 mark]

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    Model answer

    (a) \(P(B \text{ given } A) = \dfrac{0.3}{0.6} = 0.5\). (b) Yes, because \(P(B \text{ given } A) = 0.5 = P(B)\).

    Mark scheme

    • (a) \(\dfrac{0.3}{0.6}\) — M1
    • (a) \(0.5\) — A1
    • (b) Yes, because \(P(B \text{ given } A) = P(B)\) — Q1
  4. 4 Work out [4 marks]

    In a group of 50 students, 30 study Spanish, 25 study French and 10 study both. (a) A student who studies French is chosen at random. Work out the probability that the student also studies Spanish. [2 marks] (b) A student who does not study Spanish is chosen at random. Work out the probability that the student studies French. [2 marks]

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    Model answer

    (a) \(\dfrac{10}{25} = \dfrac{2}{5}\). (b) \(50 - 30 = 20\) do not study Spanish. French only is \(25 - 10 = 15\), so the probability is \(\dfrac{15}{20} = \dfrac{3}{4}\).

    Mark scheme

    • (a) \(\dfrac{10}{25}\) — M1
    • (a) \(\dfrac{2}{5}\) — A1
    • (b) \(\dfrac{15}{20}\) — M1
    • (b) \(\dfrac{3}{4}\) — A1
  5. 5 Work out [4 marks]

    A bag contains 6 red counters and 4 white counters. Two counters are taken at random without replacement. Given that the second counter is white, work out the probability that the first counter is also white. [4 marks]

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    Model answer

    \(P(\text{red then white}) = \dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\) and \(P(\text{white then white}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}\). The probability that the second is white is \(\dfrac{36}{90}\). So the answer is \(\dfrac{12}{36} = \dfrac{1}{3}\).

    Mark scheme

    • \(\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\) — M1
    • \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}\) — M1
    • \(\dfrac{12}{36}\) or \(\dfrac{12}{90} \div \dfrac{36}{90}\) — M1
    • \(\dfrac{1}{3}\) — A1
  6. 6 Work out [4 marks]

    1 in 10 people have a medical condition. A test gives a positive result for 90% of the people who have the condition. The test also gives a positive result for 20% of the people who do not have it. A person is chosen at random and has a positive result. Work out the probability that the person has the condition. [4 marks]

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    Model answer

    Has the condition and positive: \(0.1 \times 0.9 = 0.09\). Does not have it and positive: \(0.9 \times 0.2 = 0.18\). All positives: \(0.27\). The probability is \(\dfrac{0.09}{0.27} = \dfrac{1}{3}\).

    Mark scheme

    • \(0.1 \times 0.9 = 0.09\) — M1
    • \(0.9 \times 0.2 = 0.18\) — M1
    • \(\dfrac{0.09}{0.27}\) — M1
    • \(\dfrac{1}{3}\) — 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 \(A\)
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    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\(\dfrac{14}{30}\)
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    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\(\dfrac{3}{5}\)
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    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\(\dfrac{24}{40}\)
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    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\(\dfrac{3}{4}\)
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    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
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    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 \(0.5 \times 0.4 = 0.2\)
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    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\(0.5\)
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    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\(0.83\)
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    B: \(0.3\)

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