Due to stress of coping with business, Harry begins to experience migraine headaches of random severities. Headaches are instantaneous and has zero duration. The times when headaches occur follow a Poisson process of rate λ. Headache severities are independent of times of occurrences and are i.i.d. random variables with common exponential distribution given by: $$ \mathbb{P}[H \le x] = 1 − e^{-x}, \quad x > 0 $$ Harry decides to commit himself to the hospital if a headache of severity greater than $c > 0$ occurs in the time period $[0, t]$.
Compute the probability that Harry does not commit himself in $[0, t]$.
I think that probability of not committing is $\mathbb{P}[H(t)=0]$
$\mathbb{P}[H(t)=0]$ should be equal to $e^{-λt \mathbb{P}[H_s]}$ where $H_s$ is the occurrence probability of a severe headache.
But how can I calculate $\mathbb{P}[H_s]$? Should I find $\mathbb{P}[H_s>t] = e^{-tm}$ ??
Let $N(t)$ be the number of headaches experienced in $(0,t]$. Then $N(t)\sim\mathsf{Pois}(\lambda t)$, so $$\mathbb P(N(t)=n) = \frac{(\lambda t)^n}{n!}e^{-\lambda}, n=0,1,2,\ldots. $$ The probability that a given headache exceeds severity $c$ is $$\mathbb P(H_1>c) = e^{-c}.$$ Let $N_1(t)$ be the number of non-critical headaches in $(0,t]$ and $N_2(t)$ be the number of critical headaches in $(0,t]$. Then $N(t)=N_1(t)+N_2(t)$ and $N_1(t)$, $N_2(t)$ are independent Poisson processes with rates $\lambda(1-e^{-c})$ and $\lambda e^{-c}$. Therefore the probability that Harry does not commit himself in $(0,t]$ is $$\mathbb P(N_2(t)=0) = e^{-\lambda t e^{-c}} $$ and the probability that he does is $1- e^{-\lambda te^{-c}}$.