= Solution
Let $D$ be the number of days up to and including the first breakdown, and $M$ the event of a morning first breakdown. For $d=1,2,\ldots$, <conditional probability> gives
$$
\mathbb P(D=d\mid M)=\frac{\alpha q^{d-1}}{\alpha/(1-q)}=(1-q)q^{d-1}.
$$
Thus
$$
\boxed{D\mid M\sim\operatorname{Geom}(\alpha+\beta-\alpha\beta)\text{ on }\{1,2,\ldots\}}.
$$
This <geometric distribution> counts the day on which the first breakdown occurs. Its parameter is the probability of at least one breakdown during a day, rather than the morning breakdown probability $\alpha$ alone.
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