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**Focus** In that case just use the formula $\displaystyle \mathbb{P}(X_t=n)=e^{-\lambda t}\frac{(t \lambda)^n}{n!}$.

a) You want $\displaystyle \mathbb{P}(X_1 \geq 3)=1-\mathbb{P}(X_1 \leq 2)$

b) If the intervals do not overlap, then they are independent, so first find $\displaystyle \mathbb{P}(X_1 \leq 3)$, then use independence (and stationary increments) to find the result. By that I mean if s_1<t_1<s_2<t_2, then $\displaystyle \mathbb{P}(X_{t_1}-X_{s_1},X_{t_2}-X_{s_2})=\mathbb{P}(X_{t_1-s_1})\mathbb{P}(X_{t_2-s_1})$

c) $\displaystyle \mathbb{P}(X_2=2)$

d),e) You need to look at some stopping times for this, let $\displaystyle \tau_n=\inf\{t \geq 0: X_t=n\}$. Now what is $\displaystyle \mathbb{P}(\tau_n \geq t)$? Well, that means by time t, you have had less than n arrivals.

I have no idea to which level you have done things, a little indications of which bits you don't understand would be nice.