Hint
Each count describes the same population. Divide the intersection count by the B count.
Worked solution
(a)
\(P(A\mid B)=14/40=\frac{7}{20}\)
Answer: \(\frac{7}{20}\).
Answer: \(\frac{7}{20}\)
(b)
\(P(A\cup B)=(26+40-14)/92=\frac{13}{23}\)
Answer: \(\frac{13}{23}\).
Answer: \(\frac{13}{23}\)
(c)
\(P(B\mid A^c)=(40-14)/(92-26)=\frac{13}{33}\)
Answer: \(\frac{13}{33}\).
Answer: \(\frac{13}{33}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Divide the intersection count by the B count.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not interchange P(A|B) and P(B|A).
(b) · 2 marks
- 1 mark: Subtract the intersection once in the addition law.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Being disjoint is different from being independent.
(c) · 2 marks
- 1 mark: Restrict the population to people outside A.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not interchange P(A|B) and P(B|A).
Total: 6 marks.