Hard A-Level Conditional probability Questions

Challenging, exam-style A-Level Conditional probability questions with worked solutions. Stretch yourself on the hardest conditional-probability, bayes, two-way-table, without-replacement problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Events AA and BB satisfy P(A)=710P(A)=\frac{7}{10}, P(B)=35P(B)=\frac{3}{5} and P(A∩B)=25P(A\cap B)=\frac{2}{5}. Are AA and BB independent?
Show worked solution

Worked solution

  1. Recall the independence condition

    P(A∩B)=P(A) P(B)P(A\cap B)=P(A)\,P(B)

    Two events are independent exactly when this equality holds.

  2. Compare the two sides

    P(A∩B)=25, P(A)P(B)=2150P(A\cap B)=\frac{2}{5},\ P(A)P(B)=\frac{21}{50}

    They are unequal, so the events are not independent.

  3. List the probabilities involved

    P(A)=710, P(B)=35, P(A∩B)=25P(A)=\frac{7}{10},\ P(B)=\frac{3}{5},\ P(A\cap B)=\frac{2}{5}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=35×23=25P(A\cap B)=P(B)\,P(A\mid B)=\frac{3}{5}\times \frac{2}{3}=\frac{2}{5}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=25710=47P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{2}{5}}{\frac{7}{10}}=\frac{4}{7}

    Conditioning on A instead of B reverses the roles of the events.

  6. Apply the addition law

    P(A∪B)=P(A)+P(B)−P(A∩B)=910P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{9}{10}

    The addition law subtracts the overlap so it is not counted twice.

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−710=310P(A')=1-P(A)=1-\frac{7}{10}=\frac{3}{10}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−23=13P(A'\mid B)=1-P(A\mid B)=1-\frac{2}{3}=\frac{1}{3}

    Given B, the events A and A' partition the restricted sample space.

  9. Test for independence

    P(A) P(B)=710×35=2150P(A)\,P(B)=\frac{7}{10}\times \frac{3}{5}=\frac{21}{50}

    Independent events must satisfy P(A\cap B)=P(A)P(B).

  10. Find P(A\cap B')

    P(A∩B′)=P(A)−P(A∩B)=710−25=310P(A\cap B')=P(A)-P(A\cap B)=\frac{7}{10}-\frac{2}{5}=\frac{3}{10}

    Splitting A according to whether B occurs gives this region.

  11. Find P(A'\cap B')

    P(A′∩B′)=1−P(A∪B)=1−910=110P(A'\cap B')=1-P(A\cup B)=1-\frac{9}{10}=\frac{1}{10}

    Outside both events is the complement of their union.

  12. Check the value is a valid probability

    0≤23≤10\le \frac{2}{3}\le 1

    Every probability, conditional or not, lies between 0 and 1.

  13. Confirm the joint probability is symmetric

    P(A) P(B∣A)=25=P(B) P(A∣B)P(A)\,P(B\mid A)=\frac{2}{5}=P(B)\,P(A\mid B)

    Both forms of the multiplication law give the same joint probability.

  14. Recall the definition of conditional probability

    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

    Conditioning on B restricts attention to the outcomes lying in B.

  15. State the conclusion

    P(A∩B)≠P(A)P(B)P(A\cap B)\ne P(A)P(B)

    The independence test settles the question.

Answer
Not independent, since P(A∩B)≠P(A)P(B)\text{Not independent, since } P(A\cap B)\ne P(A)P(B)
Question 2
8 markschallenging
Events AA and BB satisfy P(A)=58P(A)=\frac{5}{8}, P(B)=25P(B)=\frac{2}{5} and P(A∩B)=14P(A\cap B)=\frac{1}{4}. Are AA and BB independent?
Show worked solution

Worked solution

  1. Recall the independence condition

    P(A∩B)=P(A) P(B)P(A\cap B)=P(A)\,P(B)

    Two events are independent exactly when this equality holds.

  2. Compare the two sides

    P(A∩B)=14, P(A)P(B)=14P(A\cap B)=\frac{1}{4},\ P(A)P(B)=\frac{1}{4}

    They are equal, so the events are independent.

  3. List the probabilities involved

    P(A)=58, P(B)=25, P(A∩B)=14P(A)=\frac{5}{8},\ P(B)=\frac{2}{5},\ P(A\cap B)=\frac{1}{4}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=25×58=14P(A\cap B)=P(B)\,P(A\mid B)=\frac{2}{5}\times \frac{5}{8}=\frac{1}{4}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=1458=25P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{4}}{\frac{5}{8}}=\frac{2}{5}

    Conditioning on A instead of B reverses the roles of the events.

  6. Apply the addition law

    P(A∪B)=P(A)+P(B)−P(A∩B)=3140P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{31}{40}

    The addition law subtracts the overlap so it is not counted twice.

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−58=38P(A')=1-P(A)=1-\frac{5}{8}=\frac{3}{8}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−58=38P(A'\mid B)=1-P(A\mid B)=1-\frac{5}{8}=\frac{3}{8}

    Given B, the events A and A' partition the restricted sample space.

  9. Test for independence

    P(A) P(B)=58×25=14P(A)\,P(B)=\frac{5}{8}\times \frac{2}{5}=\frac{1}{4}

    Independent events must satisfy P(A\cap B)=P(A)P(B).

  10. Find P(A\cap B')

    P(A∩B′)=P(A)−P(A∩B)=58−14=38P(A\cap B')=P(A)-P(A\cap B)=\frac{5}{8}-\frac{1}{4}=\frac{3}{8}

    Splitting A according to whether B occurs gives this region.

  11. Find P(A'\cap B')

    P(A′∩B′)=1−P(A∪B)=1−3140=940P(A'\cap B')=1-P(A\cup B)=1-\frac{31}{40}=\frac{9}{40}

    Outside both events is the complement of their union.

  12. Check the value is a valid probability

    0≤58≤10\le \frac{5}{8}\le 1

    Every probability, conditional or not, lies between 0 and 1.

  13. Confirm the joint probability is symmetric

    P(A) P(B∣A)=14=P(B) P(A∣B)P(A)\,P(B\mid A)=\frac{1}{4}=P(B)\,P(A\mid B)

    Both forms of the multiplication law give the same joint probability.

  14. Recall the definition of conditional probability

    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

    Conditioning on B restricts attention to the outcomes lying in B.

  15. State the conclusion

    P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)

    The independence test settles the question.

Answer
Independent, since P(A∩B)=P(A)P(B)\text{Independent, since } P(A\cap B)=P(A)P(B)
Question 3
8 markschallenging
Events AA and BB satisfy P(A∩B)=514P(A\cap B)=\frac{5}{14} and P(B)=58P(B)=\frac{5}{8}. Which of the following equals P(A∣B)P(A\mid B)?
Show worked solution

Worked solution

  1. Write the conditional probability formula

    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

    Select the option produced by this definition.

  2. Substitute the given probabilities

    P(A∣B)=51458=47P(A\mid B)=\frac{\frac{5}{14}}{\frac{5}{8}}=\frac{4}{7}

    Evaluating the ratio identifies the correct option.

  3. List the probabilities involved

    P(A)=58, P(B)=58, P(A∩B)=514P(A)=\frac{5}{8},\ P(B)=\frac{5}{8},\ P(A\cap B)=\frac{5}{14}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=58×47=514P(A\cap B)=P(B)\,P(A\mid B)=\frac{5}{8}\times \frac{4}{7}=\frac{5}{14}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=51458=47P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{5}{14}}{\frac{5}{8}}=\frac{4}{7}

    Conditioning on A instead of B reverses the roles of the events.

  6. Apply the addition law

    P(A∪B)=P(A)+P(B)−P(A∩B)=2528P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{25}{28}

    The addition law subtracts the overlap so it is not counted twice.

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−58=38P(A')=1-P(A)=1-\frac{5}{8}=\frac{3}{8}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−47=37P(A'\mid B)=1-P(A\mid B)=1-\frac{4}{7}=\frac{3}{7}

    Given B, the events A and A' partition the restricted sample space.

  9. Test for independence

    P(A) P(B)=58×58=2564P(A)\,P(B)=\frac{5}{8}\times \frac{5}{8}=\frac{25}{64}

    Independent events must satisfy P(A\cap B)=P(A)P(B).

  10. Compare P(A\cap B) with P(A)P(B)

    P(A∩B)=514, P(A)P(B)=2564P(A\cap B)=\frac{5}{14},\ P(A)P(B)=\frac{25}{64}

    They are unequal, so A and B are not independent.

  11. Find P(A\cap B')

    P(A∩B′)=P(A)−P(A∩B)=58−514=1556P(A\cap B')=P(A)-P(A\cap B)=\frac{5}{8}-\frac{5}{14}=\frac{15}{56}

    Splitting A according to whether B occurs gives this region.

  12. Find P(A'\cap B')

    P(A′∩B′)=1−P(A∪B)=1−2528=328P(A'\cap B')=1-P(A\cup B)=1-\frac{25}{28}=\frac{3}{28}

    Outside both events is the complement of their union.

  13. Check the value is a valid probability

    0≤47≤10\le \frac{4}{7}\le 1

    Every probability, conditional or not, lies between 0 and 1.

  14. Confirm the joint probability is symmetric

    P(A) P(B∣A)=514=P(B) P(A∣B)P(A)\,P(B\mid A)=\frac{5}{14}=P(B)\,P(A\mid B)

    Both forms of the multiplication law give the same joint probability.

  15. Select the matching option

    P(A∣B)=47P(A\mid B)=\frac{4}{7}

    This value matches the correct answer.

Answer
47\frac{4}{7}
Question 4
8 markschallenging
Events AA and BB satisfy P(A∩B)=715P(A\cap B)=\frac{7}{15} and P(B)=710P(B)=\frac{7}{10}. Which of the following equals P(A∣B)P(A\mid B)?
Show worked solution

Worked solution

  1. Write the conditional probability formula

    P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

    Select the option produced by this definition.

  2. Substitute the given probabilities

    P(A∣B)=715710=23P(A\mid B)=\frac{\frac{7}{15}}{\frac{7}{10}}=\frac{2}{3}

    Evaluating the ratio identifies the correct option.

  3. List the probabilities involved

    P(A)=710, P(B)=710, P(A∩B)=715P(A)=\frac{7}{10},\ P(B)=\frac{7}{10},\ P(A\cap B)=\frac{7}{15}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=710×23=715P(A\cap B)=P(B)\,P(A\mid B)=\frac{7}{10}\times \frac{2}{3}=\frac{7}{15}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=715710=23P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{7}{15}}{\frac{7}{10}}=\frac{2}{3}

    Conditioning on A instead of B reverses the roles of the events.

  6. Apply the addition law

    P(A∪B)=P(A)+P(B)−P(A∩B)=1415P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{14}{15}

    The addition law subtracts the overlap so it is not counted twice.

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−710=310P(A')=1-P(A)=1-\frac{7}{10}=\frac{3}{10}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−23=13P(A'\mid B)=1-P(A\mid B)=1-\frac{2}{3}=\frac{1}{3}

    Given B, the events A and A' partition the restricted sample space.

  9. Test for independence

    P(A) P(B)=710×710=49100P(A)\,P(B)=\frac{7}{10}\times \frac{7}{10}=\frac{49}{100}

    Independent events must satisfy P(A\cap B)=P(A)P(B).

  10. Compare P(A\cap B) with P(A)P(B)

    P(A∩B)=715, P(A)P(B)=49100P(A\cap B)=\frac{7}{15},\ P(A)P(B)=\frac{49}{100}

    They are unequal, so A and B are not independent.

  11. Find P(A\cap B')

    P(A∩B′)=P(A)−P(A∩B)=710−715=730P(A\cap B')=P(A)-P(A\cap B)=\frac{7}{10}-\frac{7}{15}=\frac{7}{30}

    Splitting A according to whether B occurs gives this region.

  12. Find P(A'\cap B')

    P(A′∩B′)=1−P(A∪B)=1−1415=115P(A'\cap B')=1-P(A\cup B)=1-\frac{14}{15}=\frac{1}{15}

    Outside both events is the complement of their union.

  13. Check the value is a valid probability

    0≤23≤10\le \frac{2}{3}\le 1

    Every probability, conditional or not, lies between 0 and 1.

  14. Confirm the joint probability is symmetric

    P(A) P(B∣A)=715=P(B) P(A∣B)P(A)\,P(B\mid A)=\frac{7}{15}=P(B)\,P(A\mid B)

    Both forms of the multiplication law give the same joint probability.

  15. Select the matching option

    P(A∣B)=23P(A\mid B)=\frac{2}{3}

    This value matches the correct answer.

Answer
23\frac{2}{3}
Question 5
8 markschallenging
Events AA and BB satisfy P(A∩B)=715P(A\cap B)=\frac{7}{15} and P(B)=710P(B)=\frac{7}{10}. Find P(A′∣B)P(A'\mid B).
Show worked solution

Worked solution

  1. Use the complement of a conditional probability

    P(A′∣B)=1−P(A∣B)P(A'\mid B)=1-P(A\mid B)

    Given B, either A or A' occurs, so their conditional probabilities sum to 1.

  2. Substitute the conditional probability

    P(A′∣B)=1−715710=13P(A'\mid B)=1-\frac{\frac{7}{15}}{\frac{7}{10}}=\frac{1}{3}

    Subtracting P(A\mid B) from 1 gives the required conditional probability.

  3. List the probabilities involved

    P(A)=710, P(B)=710, P(A∩B)=715P(A)=\frac{7}{10},\ P(B)=\frac{7}{10},\ P(A\cap B)=\frac{7}{15}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=710×23=715P(A\cap B)=P(B)\,P(A\mid B)=\frac{7}{10}\times \frac{2}{3}=\frac{7}{15}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=715710=23P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{7}{15}}{\frac{7}{10}}=\frac{2}{3}

    Conditioning on A instead of B reverses the roles of the events.

  6. Apply the addition law

    P(A∪B)=P(A)+P(B)−P(A∩B)=1415P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{14}{15}

    The addition law subtracts the overlap so it is not counted twice.

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−710=310P(A')=1-P(A)=1-\frac{7}{10}=\frac{3}{10}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−23=13P(A'\mid B)=1-P(A\mid B)=1-\frac{2}{3}=\frac{1}{3}

    Given B, the events A and A' partition the restricted sample space.

  9. Test for independence

    P(A) P(B)=710×710=49100P(A)\,P(B)=\frac{7}{10}\times \frac{7}{10}=\frac{49}{100}

    Independent events must satisfy P(A\cap B)=P(A)P(B).

  10. Compare P(A\cap B) with P(A)P(B)

    P(A∩B)=715, P(A)P(B)=49100P(A\cap B)=\frac{7}{15},\ P(A)P(B)=\frac{49}{100}

    They are unequal, so A and B are not independent.

  11. Find P(A\cap B')

    P(A∩B′)=P(A)−P(A∩B)=710−715=730P(A\cap B')=P(A)-P(A\cap B)=\frac{7}{10}-\frac{7}{15}=\frac{7}{30}

    Splitting A according to whether B occurs gives this region.

  12. Find P(A'\cap B')

    P(A′∩B′)=1−P(A∪B)=1−1415=115P(A'\cap B')=1-P(A\cup B)=1-\frac{14}{15}=\frac{1}{15}

    Outside both events is the complement of their union.

  13. Check the value is a valid probability

    0≤23≤10\le \frac{2}{3}\le 1

    Every probability, conditional or not, lies between 0 and 1.

  14. Confirm the joint probability is symmetric

    P(A) P(B∣A)=715=P(B) P(A∣B)P(A)\,P(B\mid A)=\frac{7}{15}=P(B)\,P(A\mid B)

    Both forms of the multiplication law give the same joint probability.

  15. State the conditional probability

    P(A′∣B)=13P(A'\mid B)=\frac{1}{3}

    This is the required probability.

Answer
13\frac{1}{3}

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