Time between rainfalls is exponentially distributed with $\lambda=1 ~\mbox{month}$. Probability that the longest time > 3 months in a 10 year period?

Question

I am trying to solve part (c) of the problem above. I don't know how to solve it using the extreme value distribution, so I tried a Poisson process approach:

Let $N\sim\mbox{Pois}(120)$ be the number of heavy rainfalls over the next 10 years. Let $T_i\sim\mbox{Exp}(1)$ be the time (in months) between the $(i-1)$st and the $i$th heavy rainfall for $1\leq{}i\leq{}N$ (where it is understood that the "$0$th heavy rainfall" is just the start time, whether or not there is any rainfall at that time). All the $T_i$s are iid. Then the desired probability is:

$$\sum_{k=0}^\infty{}P(\mbox{max time with \(k\) heavy rainfalls in \(10\) years > \(3\) months})*P(N=k)$$

Now it should be clear from the definition of $N$ that:

$$P(N=k)=\frac{e^{-120}120^k}{k!}$$

The probability I am less sure about is $p(k)\equiv{}P(\mbox{max time with \(k\) heavy rainfalls in \(10\) years > \(3\) months})$. The "max time" term refers to the maximum of the following times:

1. The time from the starting time to the time of the first heavy rainfall which occurs within 10 years from the starting time (for $k\geq1$).
2. The $k-1$ times between the $k$ heavy rainfalls which occur within 10 years from the starting time (for $k\geq2$).
3. The time from the $k$th heavy rainfall which occurs within 10 years from the starting time to the time exactly 10 years from the starting time (for $k\geq0$). Note that the "$0$th heavy rainfall" is just the start time, whether or not there is any rainfall at that time.

Now it should be clear that $p(k)=1$ for all $0\leq{}k\leq{}39$ since if the period of $120$ months is subdivided into less than 40 intervals then at least one of those intervals must be longer than 3 months, and if there are exactly 40 intervals then the only way for none of those intervals to be longer than 3 months is if they are all exactly 3 months long (the probability of which is $0$). On the other hand, for all $k\geq40$ we have $p(k)=1-q(k)$ where:

$$q(k)\equiv{}P(\mbox{max time with \(k\) heavy rainfalls in \(10\) years < \(3\) months})$$ $$=P(T_1<3,\ldots,T_k<3|117<T_1+\cdots+T_k<120)$$

I am not sure how to proceed further, however, since the condition $117<T_1+\cdots+T_k<120$ means that the $T_i$s are no longer independent. Does anybody know if it is possible to get something useful out of the expression above? I simulated this problem using R and after running the simulation a million times I got $0.999089$ as the desired probability, but I am wondering if it is possible to get this answer numerically. I have posted the R code below:

n <- 1000000 # number of trials

trials <- replicate(n,0)

for (i in 1:n){
    times <- c()
    
    while (sum(times)<120){
        time <- rexp(1,1)
        if (time > (120-sum(times))) break
        else times <- c(times,time)
    }
    
    max <- max(max(times),(120-sum(times)))
    
    if (max>3) trials[i] <- 1
}

mean(trials)