Conditional probability Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Conditional probability questions. See exactly how to solve problems on conditional-probability, definition, multiplication-law, addition-law.

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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).

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 two events AA and BB, P(A∩B)=16P(A\cap B)=\frac{1}{6} and P(B)=12P(B)=\frac{1}{2}. Find P(A∣B)P(A\mid B).

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)=1612=13P(A\mid B)=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}

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

  3. State the conditional probability

    P(A∣B)=13P(A\mid B)=\frac{1}{3}

    This is the required probability.

Answer
13\frac{1}{3}
Question 3
2 markseasy
Events AA and BB satisfy P(A∩B)=14P(A\cap B)=\frac{1}{4} and P(B)=12P(B)=\frac{1}{2}. Find P(A∣B)P(A\mid B).

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)=1412=12P(A\mid B)=\frac{\frac{1}{4}}{\frac{1}{2}}=\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 4
2 markseasy
Two events AA and BB satisfy P(A∩B)=15P(A\cap B)=\frac{1}{5} and P(A)=12P(A)=\frac{1}{2}. Find P(B∣A)P(B\mid A).

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)=1512=25P(B\mid A)=\frac{\frac{1}{5}}{\frac{1}{2}}=\frac{2}{5}

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

  3. State the conditional probability

    P(B∣A)=25P(B\mid A)=\frac{2}{5}

    This is the required probability.

Answer
25\frac{2}{5}
Question 5
2 markseasy
For events AA and BB, P(A)=12P(A)=\frac{1}{2} and P(B∣A)=25P(B\mid A)=\frac{2}{5}. Find P(A∩B)P(A\cap B).

Worked solution

  1. Write the multiplication law

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

    The joint probability is the marginal times the conditional.

  2. Substitute the given probabilities

    P(A∩B)=12×25=15P(A\cap B)=\frac{1}{2}\times \frac{2}{5}=\frac{1}{5}

    Multiplying P(A) by P(B\mid A) gives the probability of both events.

  3. State the joint probability

    P(A∩B)=15P(A\cap B)=\frac{1}{5}

    This is the probability that both events occur.

Answer
15\frac{1}{5}

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