Hint
Objects are selected uniformly from those still available. Multiply the red-red conditional branch.
Worked solution
(a)
\(\displaystyle \frac{12}{15}\frac{11}{14}=\frac{22}{35}\)
Answer: \(\displaystyle \frac{22}{35}\).
Answer: \(\displaystyle \frac{22}{35}\)
(b)
\(\displaystyle 2\frac{12}{15}\frac{3}{14}=\frac{12}{35}\)
Answer: \(\displaystyle \frac{12}{35}\).
Answer: \(\displaystyle \frac{12}{35}\)
(c)
\(\displaystyle P(R_{1}\mid R_{2})= \frac{P(R_{1}\cap R_{2})}{P(R_{2})}= \frac{11}{14}\)
Answer: \(\displaystyle \frac{11}{14}\).
Answer: \(\displaystyle \frac{11}{14}\)
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: Multiply the red-red conditional branch.
- 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 keep the original colour counts after removing an object.
(b) · 2 marks
- 1 mark: Add red-blue and blue-red branches.
- 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: Order matters in a tree even when the final event does not.
(c) · 2 marks
- 1 mark: Divide the red-red probability by the marginal second-red probability.
- 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 keep the original colour counts after removing an object.
Total: 6 marks.