Hard A-Level Probability Questions

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

tree diagramwithout replacementconditionaltwo-way tablecomplementindependent
A-Level34 questionsStep-by-step solutions
Question 1
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})?
Show worked solution

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
Question 2
8 markschallenging
Three independent events have P(A)=12P(A)=\tfrac{1}{2}, P(B)=13P(B)=\tfrac{1}{3} and P(C)=14P(C)=\tfrac{1}{4}. Which value gives P(A∩B∩C)P(A\cap B\cap C)?
Show worked solution

Worked solution

  1. Extend independence to three events

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

    Mutual independence multiplies.

  2. Write down P(A)P(A)

    P(A)=12P(A)=\frac{1}{2}

    First probability.

  3. Write down P(B)P(B)

    P(B)=13P(B)=\frac{1}{3}

    Second probability.

  4. Write down P(C)P(C)

    P(C)=14P(C)=\frac{1}{4}

    Third probability.

  5. Multiply the first two

    P(A∩B)=12×13=16P(A\cap B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}

    Combine A and B.

  6. Multiply by the third

    P(A∩B∩C)=16×14P(A\cap B\cap C)=\frac{1}{6}\times\frac{1}{4}

    Include C.

  7. Evaluate the product

    =124=\frac{1}{24}

    Simplify.

  8. Check by a direct product

    12×13×14\frac{1}{2}\times\frac{1}{3}\times\frac{1}{4}

    Multiply all at once.

  9. Multiply the numerators

    1×1×1=11\times 1\times 1=1

    Top of the fraction.

  10. Multiply the denominators

    2×3×4=242\times 3\times 4=24

    Bottom of the fraction.

  11. Confirm the value

    124\frac{1}{24}

    Same as before.

  12. Contrast with the sum

    12+13+14=1312\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}

    Greater than 1, impossible.

  13. Contrast with the union

    P(A∪B∪C) is largerP(A\cup B\cup C)\text{ is larger}

    A different quantity.

  14. Reject the incorrect options

    only 124 fits\text{only }\frac{1}{24}\text{ fits}

    Eliminate the distractors.

  15. State the answer

    P(A∩B∩C)=124P(A\cap B\cap C)=\frac{1}{24}

    Required probability.

Answer
124\frac{1}{24}
Question 3
8 markschallenging
A bag contains 3 red and 5 blue balls. Two are drawn without replacement. Which of the following is the correct value of the probability that both balls are red?
Show worked solution

Worked solution

  1. State the composition and total

    n(S)=3+5=8n(S)=3+5=8

    Eight balls in the bag.

  2. Note there is no replacement

    second draw is conditional\text{second draw is conditional}

    The first ball is not returned.

  3. Probability the first is red

    P(R1)=38P(R_1)=\frac{3}{8}

    Three red out of eight.

  4. Conditional probability the second is red

    P(R2∣R1)=27P(R_2\mid R_1)=\frac{2}{7}

    One red removed.

  5. Multiply for both red

    P(RR)=38×27=328P(RR)=\frac{3}{8}\times\frac{2}{7}=\frac{3}{28}

    Product along the branch.

  6. Probability the first is blue

    P(B1)=58P(B_1)=\frac{5}{8}

    Five blue out of eight.

  7. Conditional probability the second is blue

    P(B2∣B1)=47P(B_2\mid B_1)=\frac{4}{7}

    One blue removed.

  8. Multiply for both blue

    P(BB)=58×47=514P(BB)=\frac{5}{8}\times\frac{4}{7}=\frac{5}{14}

    Second branch product.

  9. Set up the mixed probability

    P(mixed)=1−328−514P(\text{mixed})=1-\frac{3}{28}-\frac{5}{14}

    Complement of the two matches.

  10. Use a common denominator

    =1−328−1028=1-\frac{3}{28}-\frac{10}{28}

    Rewrite over 28.

  11. Combine the fractions

    =1−1328=1-\frac{13}{28}

    Add the subtracted terms.

  12. Evaluate the mixed probability

    P(mixed)=1528P(\text{mixed})=\frac{15}{28}

    One of each colour.

  13. Check the three cases sum to 1

    328+1028+1528=1\frac{3}{28}+\frac{10}{28}+\frac{15}{28}=1

    Exhaustive outcomes.

  14. Identify the requested probability

    P(both red)P(\text{both red})

    The question asks for two reds.

  15. State the answer

    P(RR)=328P(RR)=\frac{3}{28}

    Required probability.

Answer
328\frac{3}{28}
Question 4
8 markschallenging
A standard pack of 52 playing cards contains 13 hearts. Two are drawn at random without replacement. Find the probability that both are hearts.
Show worked solution

Worked solution

  1. State the composition and total

    n(S)=13+39=52n(S)=13+39=52

    There are 52 items.

  2. Note there is no replacement

    second draw is conditional\text{second draw is conditional}

    The first item is not returned.

  3. Probability first is type X

    1352\frac{13}{52}

    13 of 52 are the first type.

  4. Conditional probability second is X

    1251\frac{12}{51}

    One of that type removed.

  5. Probability both are type X

    P(XX)=117P(XX)=\frac{1}{17}

    Product along the first branch.

  6. Probability first is type Y

    3952\frac{39}{52}

    39 of 52 are the second type.

  7. Conditional probability second is Y

    3851\frac{38}{51}

    One of that type removed.

  8. Probability both are type Y

    P(YY)=1934P(YY)=\frac{19}{34}

    Product along the second branch.

  9. Add for same colour

    P(same)=P(XX)+P(YY)=2134P(\text{same})=P(XX)+P(YY)=\frac{21}{34}

    The two same-colour branches are mutually exclusive.

  10. Find different colour by complement

    P(different)=1−2134=1334P(\text{different})=1-\frac{21}{34}=\frac{13}{34}

    The remaining outcomes are one of each.

  11. Probability of X then Y

    1352×3951=1368\frac{13}{52}\times\frac{39}{51}=\frac{13}{68}

    One mixed order.

  12. Probability of Y then X

    3952×1351=1368\frac{39}{52}\times\frac{13}{51}=\frac{13}{68}

    The other mixed order.

  13. Check different-colour probability

    1368+1368=1334\frac{13}{68}+\frac{13}{68}=\frac{13}{34}

    The two mixed paths add to P(different).

  14. Confirm all cases sum to 1

    P(XX)+P(YY)+P(different)=1P(XX)+P(YY)+P(\text{different})=1

    Exhaustive check.

  15. State the answer

    P=117P=\frac{1}{17}

    Required probability.

Answer
117\frac{1}{17}
Question 5
8 markschallenging
A bag contains 2 red and 4 blue counters. Two are drawn at random without replacement. Find the probability that both counters are the same colour.
Show worked solution

Worked solution

  1. State the composition and total

    n(S)=2+4=6n(S)=2+4=6

    There are 6 items.

  2. Note there is no replacement

    second draw is conditional\text{second draw is conditional}

    The first item is not returned.

  3. Probability first is type X

    26\frac{2}{6}

    2 of 6 are the first type.

  4. Conditional probability second is X

    15\frac{1}{5}

    One of that type removed.

  5. Probability both are type X

    P(XX)=115P(XX)=\frac{1}{15}

    Product along the first branch.

  6. Probability first is type Y

    46\frac{4}{6}

    4 of 6 are the second type.

  7. Conditional probability second is Y

    35\frac{3}{5}

    One of that type removed.

  8. Probability both are type Y

    P(YY)=25P(YY)=\frac{2}{5}

    Product along the second branch.

  9. Add for same colour

    P(same)=P(XX)+P(YY)=715P(\text{same})=P(XX)+P(YY)=\frac{7}{15}

    The two same-colour branches are mutually exclusive.

  10. Find different colour by complement

    P(different)=1−715=815P(\text{different})=1-\frac{7}{15}=\frac{8}{15}

    The remaining outcomes are one of each.

  11. Probability of X then Y

    26×45=415\frac{2}{6}\times\frac{4}{5}=\frac{4}{15}

    One mixed order.

  12. Probability of Y then X

    46×25=415\frac{4}{6}\times\frac{2}{5}=\frac{4}{15}

    The other mixed order.

  13. Check different-colour probability

    415+415=815\frac{4}{15}+\frac{4}{15}=\frac{8}{15}

    The two mixed paths add to P(different).

  14. Confirm all cases sum to 1

    P(XX)+P(YY)+P(different)=1P(XX)+P(YY)+P(\text{different})=1

    Exhaustive check.

  15. State the answer

    P=715P=\frac{7}{15}

    Required probability.

Answer
715\frac{7}{15}

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