P(x=0)=(2)⁰.(2.718)²

❌ Incorrect.

Your calculation $P(x=0)=(2)^0 \cdot (2.718)^2$ is incorrect.

For a Poisson distribution, the probability mass function for $P(X=k)$ is given by:

$$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$

In the previous problem (4c), we found $\lambda = 2$.
To find $P(X=0)$, we substitute $k=0$ and $\lambda=2$ into the formula:

$$P(X=0) = \frac{e^{-2} (2)^0}{0!}$$

Since $(2)^0 = 1$ and $0! = 1$:

$$P(X=0) = e^{-2}$$

Using $e \approx 2.718$:

$$P(X=0) \approx (2.718)^{-2}$$ $$P(X=0) \approx \frac{1}{(2.718)^2}$$ $$P(X=0) \approx \frac{1}{7.387524}$$ $$P(X=0) \approx \boxed{0.135335}$$

Your calculation used $(2.718)^2$ instead of $(2.718)^{-2}$, which resulted in a value greater than 1, which cannot be a probability.