Poisson process breakdown(waiting times)

Question

Certain electrical disturbances occur according to a Poisson process with rate 3 per hour. These disturbances cause damage to a computer.

a) Assume that a single disturbance will cause the computer to crash. What is the probability that the system will crash in the coming 10 minutes?

b) Assume that the computer will survive a single disturbance, but the second such disturbance will cause it to crash. What is, now, the probability that the computer will crash in the coming 10 minutes?

c) Assume that a crash will not happen unless there are two disturbances within 5 minutes of each other. Calculate the probability that the computer will crash in the coming 10 minutes

My attempt a)$$ P(N(1)=3) = \frac{e^{-3} 3^1}{1!} = 0.149$$ b) $$ P(N(2)=3|N(1)=3$$ not really sure how to move on from there

c) $$ P(N(1)=3|N(2)=6)) $$

How far off am I from the correct answers?

Answer 1

Let $T_n$ denote the successive event times of the Poisson process. You are supposed to be able to see that the answer to (a) is $$ P(T_1\lt10), $$ and to compute this quantity, and to see that the answer to (b) is $$ P(T_2\lt10), $$ and to compute this quantity.

To solve (c), note that, according to these new rules, a crash can happen in the first $10$ minutes only when $T_1\lt5$. Then, starting from time $T_1$, the Poisson process begins anew hence the conditional probability of interest, namely, $P(T_2\lt T_1+5\mid T_1)$ is simply $P(T\lt5)$ where $T$ is distributed like $T_1$. Thus, a crash happens with probability $$ P(T_1\lt5)^2. $$

Answer 2

The rate is 3 per hour so the rate for 10 min is 3/6. So prob computer will crash is 10 min is $1-P(n=0) = 1-exp(-1/2)$. i.e. at least 1 (maybe more) disturbances.

b) prob will crash $=1-P(m=0)-P(m=1)$. i.e. at least two.

c) for a Poisson process the time between events follows an exponential distribution. so want $P(T<5min)=1-exp(-3 *(5/60)$.