A-Level Conditional probability Practice Questions

Free A-Level Conditional probability practice questions with full step-by-step worked solutions. Covers conditional-probability, definition, multiplication-law, addition-law. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Two events AA and BB satisfy P(A∩B)=15P(A\cap B)=\frac{1}{5} and P(B)=25P(B)=\frac{2}{5}. Find 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)}

    Conditional probability divides the joint probability by P(B).

  2. Substitute the given probabilities

    P(A∣B)=1525=12P(A\mid B)=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{2}

    Dividing the joint probability by P(B) gives the conditional probability.

  3. State the conditional probability

    P(A∣B)=12P(A\mid B)=\frac{1}{2}

    This is the required probability.

Answer
12\frac{1}{2}
Question 2
2 markseasy
For events AA and BB, P(A∩B)=16P(A\cap B)=\frac{1}{6} and P(A)=13P(A)=\frac{1}{3}. Find P(B∣A)P(B\mid A).
Show worked solution

Worked solution

  1. Write the conditional probability formula

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

    Conditioning on A divides the joint probability by P(A).

  2. Substitute the given probabilities

    P(B∣A)=1613=12P(B\mid A)=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}

    Dividing the joint probability by P(A) gives the conditional probability.

  3. State the conditional probability

    P(B∣A)=12P(B\mid A)=\frac{1}{2}

    This is the required probability.

Answer
12\frac{1}{2}
Question 3
3 marksintermediate
Events AA and BB satisfy P(A∩B)=25P(A\cap B)=\frac{2}{5} and P(B)=45P(B)=\frac{4}{5}. Find 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)}

    Conditional probability divides the joint probability by P(B).

  2. Substitute the given probabilities

    P(A∣B)=2545=12P(A\mid B)=\frac{\frac{2}{5}}{\frac{4}{5}}=\frac{1}{2}

    Dividing the joint probability by P(B) gives the conditional probability.

  3. List the probabilities involved

    P(A)=12, P(B)=45, P(A∩B)=25P(A)=\frac{1}{2},\ P(B)=\frac{4}{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)=45×12=25P(A\cap B)=P(B)\,P(A\mid B)=\frac{4}{5}\times \frac{1}{2}=\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)=2512=45P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{2}{5}}{\frac{1}{2}}=\frac{4}{5}

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

  6. State the conditional probability

    P(A∣B)=12P(A\mid B)=\frac{1}{2}

    This is the required probability.

Answer
12\frac{1}{2}
Question 4
5 markshard
Events AA and BB satisfy P(A∩B)=310P(A\cap B)=\frac{3}{10} and P(B)=35P(B)=\frac{3}{5}. 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)=1−31035=12P(A'\mid B)=1-\frac{\frac{3}{10}}{\frac{3}{5}}=\frac{1}{2}

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

  3. List the probabilities involved

    P(A)=12, P(B)=35, P(A∩B)=310P(A)=\frac{1}{2},\ P(B)=\frac{3}{5},\ P(A\cap B)=\frac{3}{10}

    Collecting the known probabilities keeps the working organised.

  4. Apply the multiplication law

    P(A∩B)=P(B) P(A∣B)=35×12=310P(A\cap B)=P(B)\,P(A\mid B)=\frac{3}{5}\times \frac{1}{2}=\frac{3}{10}

    The multiplication law links the joint and conditional probabilities.

  5. Find P(B\mid A)

    P(B∣A)=P(A∩B)P(A)=31012=35P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{3}{10}}{\frac{1}{2}}=\frac{3}{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)=45P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{4}{5}

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

  7. Find the complement P(A')

    P(A′)=1−P(A)=1−12=12P(A')=1-P(A)=1-\frac{1}{2}=\frac{1}{2}

    Complementary events have probabilities that sum to 1.

  8. Find P(A'\mid B)

    P(A′∣B)=1−P(A∣B)=1−12=12P(A'\mid B)=1-P(A\mid B)=1-\frac{1}{2}=\frac{1}{2}

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

  9. Test for independence

    P(A) P(B)=12×35=310P(A)\,P(B)=\frac{1}{2}\times \frac{3}{5}=\frac{3}{10}

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

  10. State the conditional probability

    P(A′∣B)=12P(A'\mid B)=\frac{1}{2}

    This is the required probability.

Answer
12\frac{1}{2}
Question 5
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)

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