Advanced Probability Questions
Cambridge International Curriculum (Ages 15–22)
Solutions Include Step-by-Step Logic and Common Pitfalls
Question 1: Conditional Probability & Bayes’ Theorem
Exam Context: Frequently tested in AS/A-Level (Paper 6) and IB HL exams.
Problem Statement
A rare disease affects 1 in 10,000 people. A diagnostic test has:
99% accuracy for infected patients (true positive).
95% accuracy for healthy patients (true negative).
If a person tests positive, what is the probability they actually have the disease?
Key Concepts Needed
Bayes’ Theorem: P(A∣B)=P(B∣A)⋅P(A)P(B)P(A∣B)=P(B)P(B∣A)⋅P(A).
Law of Total
Probability: P(B)=P(B∣A)⋅P(A)+P(B∣Ac)⋅P(Ac)P(B)=P(B∣A)⋅P(A)+P(B∣Ac)⋅P(Ac).
Step-by-Step Solution
Define events:
DD: Having the disease (P(D)=0.0001P(D)=0.0001).
T+T+: Testing positive.
Apply Bayes’ Theorem:
P(D∣T+)=P(T+∣D)⋅P(D)P(T+)P(D∣T+)=P(T+)P(T+∣D)⋅P(D)
Calculate P(T+)P(T+):
P(T+)=P(T+∣D)⋅P(D)+P(T+∣Dc)⋅P(Dc)=(0.99×0.0001)+(0.05×0.9999)=0.05009
4P(T+)=P(T+∣D)⋅P(D)+P(T+∣Dc)⋅P(Dc)=(0.99×0.0001)+(0.05×0.9999)=0.05009
4
, Final probability:
P(D∣T+)=0.99×0.00010.050094≈0.00198 P(D∣T+)=0.0500940.99×0.0001
≈0.00198
Why Students Struggle
Misinterpreting "99% accuracy" as P(D∣T+)=99%P(D∣T+)=99%.
Forgetting to account for the disease’s rarity (P(D)P(D) is very low).
Exam Tip
Use a probability tree to visualize true/false positives/negatives.
Question 2: Combinatorics & Overlapping Events
Exam Context: Common in Olympiad-style questions and STEP papers.
Problem Statement
How many ways can you arrange the letters of “PROBABILITY” such that:
All vowels (O, A, I, I) are not all together.
The two I’s are not adjacent.
Key Concepts Needed
Permutations with repeated letters: n!n1!⋅n2!⋯n1!⋅n2!⋯n!.
Complementary counting: Total arrangements – unwanted arrangements.
Step-by-Step Solution
Total arrangements:
Letters: P, R, O, B, A, B, I, L, I, T, Y (11 letters, with duplicates: 2 B’s, 2 I’s).
11!2!⋅2!=9,979,2002!⋅2!11!=9,979,200
Unwanted arrangements (vowels all together):
Cambridge International Curriculum (Ages 15–22)
Solutions Include Step-by-Step Logic and Common Pitfalls
Question 1: Conditional Probability & Bayes’ Theorem
Exam Context: Frequently tested in AS/A-Level (Paper 6) and IB HL exams.
Problem Statement
A rare disease affects 1 in 10,000 people. A diagnostic test has:
99% accuracy for infected patients (true positive).
95% accuracy for healthy patients (true negative).
If a person tests positive, what is the probability they actually have the disease?
Key Concepts Needed
Bayes’ Theorem: P(A∣B)=P(B∣A)⋅P(A)P(B)P(A∣B)=P(B)P(B∣A)⋅P(A).
Law of Total
Probability: P(B)=P(B∣A)⋅P(A)+P(B∣Ac)⋅P(Ac)P(B)=P(B∣A)⋅P(A)+P(B∣Ac)⋅P(Ac).
Step-by-Step Solution
Define events:
DD: Having the disease (P(D)=0.0001P(D)=0.0001).
T+T+: Testing positive.
Apply Bayes’ Theorem:
P(D∣T+)=P(T+∣D)⋅P(D)P(T+)P(D∣T+)=P(T+)P(T+∣D)⋅P(D)
Calculate P(T+)P(T+):
P(T+)=P(T+∣D)⋅P(D)+P(T+∣Dc)⋅P(Dc)=(0.99×0.0001)+(0.05×0.9999)=0.05009
4P(T+)=P(T+∣D)⋅P(D)+P(T+∣Dc)⋅P(Dc)=(0.99×0.0001)+(0.05×0.9999)=0.05009
4
, Final probability:
P(D∣T+)=0.99×0.00010.050094≈0.00198 P(D∣T+)=0.0500940.99×0.0001
≈0.00198
Why Students Struggle
Misinterpreting "99% accuracy" as P(D∣T+)=99%P(D∣T+)=99%.
Forgetting to account for the disease’s rarity (P(D)P(D) is very low).
Exam Tip
Use a probability tree to visualize true/false positives/negatives.
Question 2: Combinatorics & Overlapping Events
Exam Context: Common in Olympiad-style questions and STEP papers.
Problem Statement
How many ways can you arrange the letters of “PROBABILITY” such that:
All vowels (O, A, I, I) are not all together.
The two I’s are not adjacent.
Key Concepts Needed
Permutations with repeated letters: n!n1!⋅n2!⋯n1!⋅n2!⋯n!.
Complementary counting: Total arrangements – unwanted arrangements.
Step-by-Step Solution
Total arrangements:
Letters: P, R, O, B, A, B, I, L, I, T, Y (11 letters, with duplicates: 2 B’s, 2 I’s).
11!2!⋅2!=9,979,2002!⋅2!11!=9,979,200
Unwanted arrangements (vowels all together):
