Consider the collective risk model $S = X_1 + \cdots + X_N$ where $N$ has distribution $\operatorname{Binom}(10, 0.15)$ and the $X_i$'s are independent and identically distributed with probability mass function $$\mathbb P(X = x) = \begin{cases}0.2 & \text{if}\, x = 0,\\ 0.4 & \text{if}\,x = 1,\\ 0.3 & \text{if}\, x = 2,\\ 0.1 & \text{if}\, x = 3,\\ 0 & \text{otherwise}\end{cases}$$ and are independent of $N$. Calculate $\mathbb P(S>3)$.
How would you do this? Would you need to consider all the possibilities for x combined with $N$ and work out each case and add them up?
Here is a way to proceed. Note that $$ P(S>3)=1-P(S=0)-P(S=1)-P(S=2)-P(S=3)\tag{0} $$ Now the probability generating function of $N$ is given by $G_{N}(t)=Et^N=(0.85+0.15t)^{10}$ while the probability generating function of $X$ is given by $ G_{X}(t)=0.2+0.4t+0.3t^2+0.1t^3 $ whence the probability generating function of of $S$ is given by $$ G_{S}(t)=Et^S=G_{N}(G_{X}(t))=(0.88+0.06t+0.045t^2+0.015t^3)^{10} $$ Then $P(S=0)=G_{S}(0)$ and in general $$ P(S=k)=\frac{G^{(k)}(0)}{k!}\tag{1} $$ for $k\geq 1$. You can use $(0)$ and $(1)$ together to solve the problem.