Solution
= Solution
The joint <probability> of supplier $C$ and a non-defective bulb is $\mathbb P(C\cap D^c)=(2/5)(1-\alpha/4)$. Normalize by the positive non-defective <probability>:
$$
\boxed{\mathbb P(C\mid D^c)
=\frac{(2/5)(1-\alpha/4)}{1-\alpha/2}
=\frac{4-\alpha}{5(2-\alpha)}.}
$$
This is defined even at $\alpha=0$, where it reduces to the original supplier fraction $2/5$. At $\alpha=1$ it is $3/5$.