Let $\{N(t) :t\geq0\}$ be a Poisson process with rate $\lambda$ and let $T_1,T_2,\dots$ be the arrival times.
a)What is the probability that there are three arrivals in the time interval $[2,6]$and two arrivals in the time interval$[4,7]$?
b)Compute $P(N(2)+N(3) = 1)$
c)Compute $P(T_1> t_1,T_2> t_2)$
a) I know how to do it: $$ P(X+Y=3,Y+Z=2)=P(X=1,Y=2,Z=0)+P(X=3,Y=0,Z=2)+ P(X=2,Y=1,Z=1)=P(X=1)·P(Y=2)·P(Z=0)+...\\= λe^{-λ}·(λ^2·e^{-λ})/2 ·e^{-2λ}+..... $$
but b) and c) I don't know how to do them
Thank you
a) There are three cases. If you split the intervals into $[2,4][4,6][6,7]$, the cases are $1, 2, 0$ arrivals; $2, 1, 1$ arrivals; and $3, 0, 2$ arrivals, respectively. Since Poisson process is memoryless, the probability is
$$\frac{(2\lambda)^1e^{-2\lambda}}{1!}\frac{(2\lambda)^2e^{-2\lambda}}{2!}\frac{\lambda^0e^{-\lambda}}{0!}+ \frac{(2\lambda)^2e^{-2\lambda}}{2!}\frac{(2\lambda)^1e^{-2\lambda}}{1!}\frac{\lambda^1e^{-\lambda}}{1!}+ \frac{(2\lambda)^3e^{-2\lambda}}{3!}\frac{(2\lambda)^0e^{-2\lambda}}{0!}\frac{\lambda^2e^{-\lambda}}{2!}$$
b) This is the same as saying there is $1$ arrival between times $2$ and $3$ and none between times $0$ and $2$. We have a $\text{Poisson}(2\lambda)$ distribution with probability $$\frac{(2\lambda)^0e^{-2\lambda}}{0!}$$
and $\text{Poisson}(\lambda)$ distribution with probability $$\frac{\lambda^1 e^{-\lambda}}{1!}$$ for there being one occurrence; multiply them.
c) The distribution of times between arrivals in a Poisson process follow an exponential distribution so $T_1\sim\text{Exponential}(\lambda)$ so 1-cdf gives $Pr(T_1>t_1)=e^{-\lambda t_1}$. Another fact is that $T_2\sim \text{Gamma}(2, \lambda)$ so $$Pr(T_2>t_2)=\int_{t_2}^\infty\frac{\lambda^2}{\Gamma(2)}x^{2-1}e^{-\lambda x}dx$$
You can compute the integral to get $$e^{-\lambda t_2}\left(\lambda t_2+1\right)$$
and then multiply the two quantities to get the answer.