Why am I getting a different answer than marking scheme in part ii?
Question.
Each week a sports team plays one home match and one away match. In their home matches they score goals at a constant average rate of $2.1$ goals per match. In their away matches they score goals at a constant average rate of $0.8$ goals per match. You may assume that goals are scored at random times and independently of one another.
(a) A week is chosen at random.
(i) Find the probability that the team scores a total of $4$ goals in their two matches.
My Answer:
$H \sim \operatorname{Poisson}(\lambda = 2.1), \\ A \sim \operatorname{Poisson}(\lambda = 0.8).$
$H + A \sim \operatorname{Poisson}(\lambda = 2.9)$.
$\Pr[H + A = 4] = e^{-2.9} \frac{(2.9)^4}{4!} \approx \boxed{0.162154}.$
(ii) Find the probability that the team scores a total of $4$ goals, with more goals scored in the home match than in the away match.
My Answer:
$\Pr[H > A]\; \text{and} \; \Pr[H + A = 4] = \Pr[H - A > 0] \times 0.162154.$
$H - A \sim \operatorname{Poisson}(\lambda = 1.3)$.
$\Pr[H - A > 0] = 1 - \Pr[H - A = 0] = 1 - e^{-1.3}$.
$\Pr[H > A]; \text{and} \; \Pr[H + A = 4] = (1 - e^{-1.3})(0.162154) \approx \boxed{0.118}.$
Marking Scheme Answer: $$\begin{array}{c|c} H & A \\ \hline 0 & 4 \\ 1 & 3 \\ 2 & 2 \\ \boxed{3} & \boxed{1} \\ \boxed{4} & \boxed{0} \end{array}$$
$$\begin{align} \Pr[H = 3, A = 1] + \Pr[H = 4 , A = 0] &= e^{-2.1} \frac{(2.1)^3}{3!} e^{-0.8} \frac{(0.8)^1}{1!} + e^{-2.1} \frac{(2.1)^4}{4!} e^{-0.8} \\ &\approx \boxed{0.1125}. \end{align}$$
The first reason why your approach is incorrect is because $H-A$ is not Poisson distributed. The sum of independent Poisson random variables is Poisson, but the difference is not. For instance, if $H < A$, how could $H - A$ still be Poisson with nonnegative support?
The second reason is that the events $H > A$ and $H + A = 4$ are not necessarily independent; thus the joint probability is not the product of their individual probabilities. With a computer, we can compute $$\Pr[H > A] = \sum_{a=0}^\infty \Pr[H > A \mid A = a]\Pr[A = a] = \sum_{a=0}^\infty \sum_{h=a+1}^\infty \Pr[H = h]\Pr[A = a] \approx 0.67437854965.$$ But we can see that $(0.674378)(0.162154) = 0.109353 \ne 0.11253$; this confirms that they are not independent events.