= Solution
The <law of total probability> gives
$$
\mathbb P(D)=\frac15\alpha+\frac25\frac\alpha2+\frac25\frac\alpha4=\frac\alpha2.
$$
For $\alpha>0$, <Bayes' theorem> therefore yields
$$
\boxed{\mathbb P(A\mid D)=\frac{\alpha/5}{\alpha/2}=\frac25.}
$$
At the allowed endpoint $\alpha=0$, no bulb is defective: $\mathbb P(D)=0$. This <conditional probability> is then \b[undefined], rather than $2/5$. The value $2/5$ is its limit as $\alpha\to0^+$, but does not define conditioning on the impossible event within the stated discrete model.
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