Hint
The two count processes are independent. Add the independent Poisson means.
Worked solution
(a)
\(\displaystyle \lambda_1+\lambda_2=\frac{37}{6}\)
Answer: \(\displaystyle \frac{37}{6}\).
Answer: \(\displaystyle \frac{37}{6}\)
(b)
\(\displaystyle \operatorname{Var}(X)+\operatorname{Var}(Y)=\frac{37}{6}\)
Answer: \(\displaystyle \frac{37}{6}\).
Answer: \(\displaystyle \frac{37}{6}\)
(c)
\(e^{-(\lambda_1+\lambda_2)}=0.0020982184180809\)
Answer: \(0.0020982184180809\).
Answer: \(\approx 0.0020982184180809\)
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: Add the independent Poisson means.
- 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: A difference of Poisson counts is not a nonnegative Poisson variable.
(b) · 2 marks
- 1 mark: Square both coefficients in the variance calculation.
- 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: Variances add for an independent difference.
(c) · 2 marks
- 1 mark: Use the zero-count probability of the combined process.
- 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: A difference of Poisson counts is not a nonnegative Poisson variable.
Total: 6 marks.