$H_1$: The bulb is produced by unit $M_1$.

$H_2$: The bulb is produced by unit $M_2$.

$H_3$: The bulb is produced by unit $M_3$.

$E$: The bulb is defective.

The unit production proportions and known probabilities are:

$P(H_1) = \frac{2}{5}$, $P(H_2) = \frac{2}{5}$, $P(H_3) = \frac{1}{5}$

$P(E | H_1) = \frac{15}{100} = \frac{3}{20}$

The total probability of a defective bulb is given by:

$ P(E) = P(E | H_1) \cdot P(H_1) + P(E | H_2) \cdot P(H_2) + P(E | H_3) \cdot P(H_3) $

Given:

$ P(E) = 0.20 = \frac{3}{20} \cdot \frac{2}{5} + P(E | H_2) \cdot \frac{2}{5} + P(E | H_3) \cdot \frac{1}{5} $

This simplifies to:

$ 2P(E | H_2) + P(E | H_3) = \frac{7}{10} \quad \text{...(i)} $

Additionally, the probability that a defective bulb is from $M_2$ is given as:

$ P(H_2 | E) = \frac{2}{5} = \frac{P(E | H_2) \cdot \frac{2}{5}}{\frac{1}{5}} $

Solving the equation above gives:

$ P(E | H_2) = \frac{1}{5} \quad \text{...(ii)} $

Using the equations (i) and (ii), we substitute $P(E | H_2)$ into (i):

$ 2 \cdot \frac{1}{5} + P(E | H_3) = \frac{7}{10} $

Solving for $P(E | H_3)$:

$ \frac{2}{5} + P(E | H_3) = \frac{7}{10} $

Converting $\frac{2}{5}$ to $\frac{4}{10}$ gives:

$ \frac{4}{10} + P(E | H_3) = \frac{7}{10} $

$ P(E | H_3) = \frac{7}{10} - \frac{4}{10} = \frac{3}{10} $

Therefore, the probability that a randomly chosen bulb from $M_3$ is defective is 0.30.