Hint
Each count describes the same population. Divide the intersection count by the B count.
Worked solution
(a)
\(\displaystyle P(A\mid B)= \frac{15}{42}= \frac{5}{14}\)
Answer: \(\displaystyle \frac{5}{14}\).
Answer: \(\displaystyle \frac{5}{14}\)
(b)
\(\displaystyle P(A\cup B)= \frac{27+ 42- 15}{96}= \frac{9}{16}\)
Answer: \(\displaystyle \frac{9}{16}\).
Answer: \(\displaystyle \frac{9}{16}\)
(c)
\(\displaystyle P(B\mid A^{c})= \frac{42- 15}{96- 27}= \frac{9}{23}\)
Answer: \(\displaystyle \frac{9}{23}\).
Answer: \(\displaystyle \frac{9}{23}\)
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.