$\mathbf{Task}$: The machine's failure stream is the simplest, with an average failure interval of 300 hours of continuous operation. every day the machine is turned on for 8 hours. How long on average will it work until the first failure without a belt.
The answer is $$\frac{1}{e^\frac{2}{75} - 1} \approx 37 days$$
$\mathbf{My}$ $\mathbf{attempt}$:
In the first sentense we get intensity $$\lambda_1 = \frac{1}{300}$$ Since the stream is simple(I don't know English Analogue but it's have Poisson Distribution) then (The probability that the number of failures is equal to K)$$P(X(t) = k) = \frac{(\lambda t)^k}{k!}e^\left(-\lambda t\right)$$ So we can get the probability of at least one failure $$\sum_{k=1}^\infty \frac{(\lambda t)^k}{k!}e^\left(-\lambda t\right) = (\frac{(\lambda t)^1}{1!} + \frac{(\lambda t)^2}{2!} + \dots)e^\left(-\lambda t\right) = (e^\left(\lambda t\right) - 1)e^\left(-\lambda t\right)$$.
If we use $\lambda_1$ and $t = 8$ $$(e^\left(\frac{8}{300}\right) - 1)e^\left(-\frac{8}{300}\right)$$we get the probabily of at least one failure in one day.I think that stream in which the machine runs 8 hours a day is simplest too and mean time between failure will be $\frac{1}{\lambda_2}$. But I don't know how to get $\lambda_2$. I think I need to use the derived probability because it's very similar to the answer :)
I use a translator, so some terms may not match their English analogues. Sorry about that
$\mathbf{UPD}$: The solution to the problem is below in the comments