1. **Problem statement:** We are given the probability generating function (PGF) of a discrete random variable $Y$ as $$G_Y(t) = \frac{a + b t^3}{l}$$ where $a$ and $b$ are constants and $l$ is a normalizing constant. We want to find the probability that the sum of 10 independent observations of $Y$ equals $-7$. 2. **Recall the properties of PGFs:** The PGF of the sum of independent identically distributed random variables is the product of their PGFs. For 10 observations, the PGF of the sum $S = \sum_{i=1}^{10} Y_i$ is $$G_S(t) = (G_Y(t))^{10} = \left(\frac{a + b t^3}{l}\right)^{10}.$$ 3. **Interpreting the PGF:** The PGF is defined as $$G_Y(t) = E(t^Y) = \sum_{k} P(Y=k) t^k.$$ The coefficient of $t^k$ in the expansion of $G_Y(t)$ gives $P(Y=k)$. 4. **Sum of 10 observations:** The PGF of $S$ is $$G_S(t) = \left(\frac{a + b t^3}{l}\right)^{10} = \frac{1}{l^{10}} (a + b t^3)^{10}.$$ 5. **Finding $P(S = -7)$:** Since $Y$ takes values in multiples of 3 (due to $t^3$ term), the sum $S$ takes values in multiples of 3 times 10, i.e., multiples of 3. The powers of $t$ in $G_S(t)$ are multiples of 3 from 0 to 30. 6. **Check if $-7$ is possible:** The sum $S$ can only take values $0, 3, 6, ..., 30$ (multiples of 3). Since $-7$ is not a multiple of 3, $$P(S = -7) = 0.$$ **Final answer:** $$\boxed{0}.$$