I am having a little confusion with finding $E[X^{2}]$ that perhaps can be cleared up relatively easily. Here, $A, B, C$ are Poisson random variables with parameters $2.6, 3,$ and $3.4$, respectively. Let $X$ be the number of errors typed in a manuscript. Each of three typists $A,B,C$ are equally likely to type the manuscript with error rates given. Examining the worked solutions, there is this solution $$ E[X^{2}] = E[X^{2}|A]P(A) + E[X^2|B]P(B) + E[X^{2}|C]P(C) $$
$$= (2.6 + 2.6^{2})\cdot 1/3 + (3 + 3^{2})\cdot 1/3 + (3.4 + 3.4^{2})\cdot 1/3 \approx 12.11.$$ However, in each, my instinct was to write only the $2.6^{2}$ term and not $2.6 + 2.6^{2}$. What am I not connecting here? In my research, I see, for discrete random variable $X$, $$ E[X^{n}] = \sum_{x : p(x) > 0} x^{n}p(x) $$ so I assume the conditioning affects the result, but I am not following this result above $E[X^{2}]$.
For the Poisson $X$ with parameter $\lambda$, the mean and the variance are both $\lambda$.
Recall the familiar useful formula $$\text{Var}(X)=E(X^2)-(E(X))^2\tag{1}$$ for computing variance. Here it is is being used to compute $E(X^2)$, when the mean and variance are known.
For from (1) we have, for the Poisson, $\lambda=E(X^2)-\lambda^2$.