Poisson Process Question - arrival times Car crashes at a traffic stop arrive follows a Poisson Process with rate lambda = 2/hr. 
(a) Expected amount of time until the 2nd car crash arrives
(b) P(See at most 2 car crashes during rush hour) , rush hour = 7am-8am and 4pm-5pm
(c) P( first car crash comes after waiting 2 hours)
(d) If car crashes arrive at another traffic stop according to Poisson Process with rate 3/hr, find expected time we wait until a car crash occurs at either traffic stop
(e) Car crashes at our original traffic stop explode independently with prob. 0.3, find the probability we see our 1st explode car crash after 5 hours
Attempt:
for (a) , Let T2 =  amount of time until 2nd car crash arrives, T2~ exp (2), so E(T2) = 1/2
(d) E(time of car crash occurs at either traffic stop ) = 1/(2+3) = 1/5
Don't know how to start on the others, anything helps! Thanks so much!
 A: (a) No


*

*Linearity of expectation says: $\mathsf E(T_{2^{nd}}) = \mathsf E(T_{1^{st}})+\mathsf E(T_{2^{nd}}-T_{1^{st}})$

*Memorilessness of waiting time says: $ \mathsf E(T_{2^{nd}}-T_{1^{st}}) = \mathsf E(T_{1^{st}})$
(d) Yes


*

*If the events are arriving at one stop at $2/{\rm hr}$ and at the other at $3/{\rm hr}$ then events are arriving at both(either) stops at $5/{\rm hr}$.  This is itself a Poisson process. $T_{a\vee b} \sim \mathcal {Exp}(\lambda_{a\vee b})$


$$\lambda_{a\vee b} = \lambda_a+\lambda_b = 5\big/{\rm hr}
\\ \text{Find: } \mathsf E(T_{a\vee b})$$
A: (a) The expected time until the second crash is $1$ hour. For let $X_1$ be the time until the first crash, and $X_2$ the time between the first and the second. Then we want $E(X_1+X_2)$, which is $E(X_1)+E(X_2)$. Each of these is $\frac{1}{2}$.
(b) The number $X$ of crashes in a $2$ hour period has Poisson distribution with parameter $4$. The probability that $X\le 2$ is $\Pr(X=0)+\Pr(X=1)+\Pr(X=2)$. Now use the standard formula.
(c) The probability that the waiting time is $\gt 2$ is the probability that an exponentially distributed random variable with parameter $2$ is greater than $2$. This is $e^{-4}$. Alternately, the number $Y$ of crashes in $2$ hours has Poisson distribution with parameter $4$. The probability that $Y=0$ is $e^{-4}$.
(d) Your answer is correct. An explanation should be added.
(e) Hint: The number of explosions in an hour has Poisson distribution with parameter $(0.3)(2)$.
