Step-by-step variance evaluation:
For any Poisson variate, the variance equals the mean:

• Incorrect: 6 is obtained by calculating $3 \times 1 + 3 = 6$, which is the mean $E(3X+Y)$, not the variance.
• Incorrect: 9 is $3^2 \text{Var}(X)$.
• Correct: For independent random variables $X$ and $Y$:
  $$\text{Var}(3X + Y) = 3^2 \text{Var}(X) + 1^2 \text{Var}(Y) = 9(1) + 1(3) = 9 + 3 = 12$$
• Incorrect: 18 is obtained if the coefficients are squared incorrectly.

Evaluation of Poisson formula:
For a Poisson distribution with mean $\lambda = e$:

• Correct: Exactly represents $\frac{e^{-e} e^x}{x!} = \frac{\exp(x - e)}{x!}$.
• Incorrect: $e!$ is undefined in the denominator and numerator exponents are inverted.
• Incorrect: The denominator contains $e!$ instead of $x!$.
• Incorrect: $\exp(-x - e)$ is incorrect for $e^x \cdot e^{-e}$.

Step-by-step probability calculation:
Let $W$ denote a win ($P(W) = p = 0.80$) and $L$ denote a loss ($P(L) = q = 0.20$).
For the first win to occur on the third game, the first two games must be losses and the third game must be a win ($L_1 \cap L_2 \cap W_3$).

• Incorrect: 0.8 is the probability of winning any individual game.
• Incorrect: 0.2 is the probability of losing any individual game.
• Correct: By independence:
  $$P(\text{First win on game 3}) = P(L_1) \times P(L_2) \times P(W_3) = 0.20 \times 0.20 \times 0.80 = 0.04 \times 0.80 = 0.032$$
• Incorrect: $0.512 = 0.80^3$ is the probability of winning all first three games.