A-Level Probability Practice Questions

Free A-Level Probability practice questions with full step-by-step worked solutions. Covers sample space, equally likely, complement, mutually exclusive. Practise exam-style problems and check your method.

sample spaceequally likelycomplementmutually exclusiveaddition ruleconcept
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A fair six-sided die is rolled once. Find the probability of rolling an even number.
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Worked solution

  1. Count the outcomes in the sample space

    n(S)=6n(S)=6

    A fair die has six equally likely outcomes.

  2. Count the favourable outcomes

    n(E)=3n(E)=3

    The even numbers are 2, 4 and 6.

  3. Divide favourable by total and simplify

    P(E)=36=12P(E)=\frac{3}{6}=\frac{1}{2}

    Probability is favourable over total.

Answer
12\frac{1}{2}
Question 2
2 markseasy
Which pair of events is independent?
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Worked solution

  1. Recall independence

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

    Independent events do not affect one another.

  2. Check for a physical link

    coin ⊥ die\text{coin } \perp \text{ die}

    A coin toss cannot influence a die roll.

  3. Select the unlinked pair

    coin and die\text{coin and die}

    These outcomes are independent.

Answer
coin and die\text{coin and die}
Question 3
3 marksintermediate
What does the notation P(A∩B)P(A\cap B) mean?
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Worked solution

  1. Read the intersection symbol

    ∩\cap

    The cap sign joins two events.

  2. Translate the symbol to words

    ∩=and\cap=\text{and}

    It means both events together.

  3. Describe the outcomes

    in both A and B\text{in both }A\text{ and }B

    Outcomes common to A and B.

  4. Contrast with the union symbol

    ∪=or\cup=\text{or}

    Either event, a different meaning.

  5. Contrast with the complement

    A′=not AA'=\text{not }A

    Yet another piece of notation.

  6. State the meaning

    both A and B occur\text{both }A\text{ and }B\text{ occur}

    Required interpretation.

Answer
P(both occur)\text{P(both occur)}
Question 4
5 markshard
Which expression is equal to P(A′∩B′)P(A'\cap B'), the probability that neither event occurs?
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Worked solution

  1. Interpret 'neither'

    not A and not B\text{not }A\text{ and not }B

    Both events fail to occur.

  2. Write it with complements

    A′∩B′A'\cap B'

    Intersection of the two complements.

  3. Apply De Morgan's law

    A′∩B′=(A∪B)′A'\cap B'=(A\cup B)'

    Neither is the complement of the union.

  4. Apply the complement rule

    P((A∪B)′)=1−P(A∪B)P((A\cup B)')=1-P(A\cup B)

    One minus the union probability.

  5. State the resulting expression

    1−P(A∪B)1-P(A\cup B)

    The required form.

  6. Rule out 1−P(A∩B)1-P(A\cap B)

    complement of intersection\text{complement of intersection}

    A different event.

  7. Rule out P(A′)+P(B′)P(A')+P(B')

    overcounts ’neither’\text{overcounts 'neither'}

    Adds overlapping outcomes twice.

  8. Rule out 1−P(A)−P(B)1-P(A)-P(B)

    only valid if ME\text{only valid if ME}

    Ignores the overlap in general.

  9. Sanity check with numbers

    P(A∪B)=0.7⇒0.3P(A\cup B)=0.7\Rightarrow 0.3

    A consistent value.

  10. State the answer

    1−P(A∪B)1-P(A\cup B)

    Correct expression.

Answer
1−P(A∪B)1-P(A\cup B)
Question 5
8 markschallenging
On any day the probability of rain is 0.20.2. If it rains, the bus is late with probability 0.50.5; if it does not rain, the bus is late with probability 0.10.1. Which value gives P(late)P(\text{late})?
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Worked solution

  1. Set up the tree

    rain or no rain, then late or not\text{rain or no rain, then late or not}

    Two stages of the experiment.

  2. Probability of rain

    P(R)=0.2P(R)=0.2

    Given.

  3. Probability of no rain

    P(R′)=0.8P(R')=0.8

    Complement of rain.

  4. Late given rain

    P(L∣R)=0.5P(L\mid R)=0.5

    Given conditional.

  5. Late given no rain

    P(L∣R′)=0.1P(L\mid R')=0.1

    Given conditional.

  6. Path: rain and late

    P(R∩L)=0.2×0.5P(R\cap L)=0.2\times 0.5

    Multiply along the branch.

  7. Evaluate that path

    P(R∩L)=0.1P(R\cap L)=0.1

    First contribution.

  8. Path: no rain and late

    P(R′∩L)=0.8×0.1P(R'\cap L)=0.8\times 0.1

    Second branch.

  9. Evaluate that path

    P(R′∩L)=0.08P(R'\cap L)=0.08

    Second contribution.

  10. State the total probability rule

    P(L)=P(R∩L)+P(R′∩L)P(L)=P(R\cap L)+P(R'\cap L)

    Add the paths that give 'late'.

  11. Substitute the path probabilities

    P(L)=0.1+0.08P(L)=0.1+0.08

    Insert the values.

  12. Evaluate

    P(L)=0.18P(L)=0.18

    Add the contributions.

  13. Find the complement

    P(L′)=1−0.18=0.82P(L')=1-0.18=0.82

    Probability the bus is on time.

  14. Reject adding the conditionals

    0.5+0.1 is wrong0.5+0.1\text{ is wrong}

    Conditionals are not simply added.

  15. State the answer

    P(late)=0.18P(\text{late})=0.18

    Required probability.

Answer
0.180.18

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