Solution
= Solution
Let $A,B,C$ denote the mutually exclusive supplier events and $D$ the defective-bulb event. Their <probabilities> and conditional defect rates are
$$
\begin{array}{c|ccc}
&A&B&C\\\hline
\mathbb P(\text{supplier})&1/5&2/5&2/5\\
\mathbb P(D\mid\text{supplier})&\alpha&\alpha/2&\alpha/4.
\end{array}
$$
These supplier events partition the possible bulbs. The following calculations use the <law of total probability> and <Bayes' theorem> with that partition.