It should be possible to do it directly with Poisson distribution. But you can also do it directly, be specific.

P{ equipment (0, t) is normal} = P{ equipment (0, delta_t) is normal} * P{ equipment (delta_t, 2*delta_t) is normal | equipment (0, delta_t) is normal} *...* P{ equipment (t-delta_t)

That is, P{ equipment (0, t) is normal} = [1-lambda (0) * delta _ t] * ... [1-λ (t-delta _ t) * delta _ t].

At the same time, the above formula takes the logarithm of the natural base and then uses Ln( 1+x) = x+o(x) to get p = exp (the negative integral of λ on [0, t]) as the probability.