A (random) number $X$ of people in a given year that are bitten by poisonous snakes follows a Poisson distribution with parameter $\lambda$. Whenever someone is bitten there is a chance $p_1$ that they make a full recovery, a probability $p_2$ that they survive but become blind, and a probability $p_3$ that they die as a result of the bite – where of course $p_1 + p_2 + p_3 = 1$. We may assume that which of the three outcomes a particular bite victim has is not influenced by what happens to the other victims, or the total number of victims that year, etc. etc. Let $X_1$ denote the number of people that are bitten in the upcoming year and make a full recovery; let $X_2$ denote the number of people that are bitten and become blind; and let $X_3$ denote the number of bite victims that die.
(i) Show that $X_i ∼ \mathrm{Poi}(\lambda p_i)$. (Hint: It suffices that $\mathbb{P}(X_i = k)$ has the required form for each $k$. What is $\mathbb{P}(X_i = k\mid X = n)$?)
Thanks for any hints or help in advance!!!
The computation is done only for the index $1$. We use the complementary probability weight $q_1=1-p_1$. $$ \begin{aligned} \Bbb P(X_1=k) &= \Bbb P(X_1=k\text{ and }X\ge k) \\ &= \sum_{N\ge k}\Bbb P(X_1=k\text{ and }X=N) \\ &= \sum_{N\ge k}\Bbb P(X=N)\cdot \Bbb P(X_1=k\ |\ X=N) \\ &= e^{-\lambda} \sum_{N\ge k}\frac{\lambda^N}{N!} \cdot \binom Nk p_1^kq_1^{N-k} \\ &= e^{-\lambda} p_1^k \sum_{N\ge k}\frac{\lambda^N}{N!} \cdot \frac {N!}{k!(N-k)!} \cdot q_1^{N-k} \\ &= \frac{e^{-\lambda}}{k!} p_1^k \sum_{N\ge k}\frac{\lambda^N}{(N-k)!} \cdot q_1^{N-k} \\ &\qquad\text{ Substitution: $M=N-k$, then $N\ge k$ becomes $M\ge 0$,} \\ &= \frac{e^{-\lambda}}{k!} p_1^k \sum_{M\ge 0}\lambda ^k\frac{\lambda^M}{M!} \cdot q_1^{M} \\ &= \frac{e^{-\lambda}}{k!} (\lambda p_1)^k \underbrace{\sum_{M\ge 0}\frac{(q_1\lambda)^M}{M!}}_{=\exp(q_1\lambda)} \qquad\text{(edited)}\\ &= \frac{e^{-\lambda(1-q_1)}}{k!} (\lambda p_1)^k \\ &= \frac{e^{-\lambda p_1}}{k!} (\lambda p_1)^k \ . \end{aligned} $$