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Exam questions · Maths · Probability

Conditional Probability

  • 6 exam questions
  • 22 marks
  • 9 quick checks
  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]

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    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.
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    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]

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    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]

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    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]

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    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]

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    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 \(A\)
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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: \(0.3\)

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