1. Set the production probabilities

The three units produce bulbs in the ratio $2:2:1$. So, if a bulb is chosen at random from the factory output, $P(M_1)=\frac{2}{5},\quad P(M_2)=\frac{2}{5},\quad P(M_3)=\frac{1}{5}.$

Let $D$ denote the event that a bulb is defective.

Given: $P(D)=0.20, \qquad P(D\mid M_1)=0.15.$

Also given: $P(M_2\mid D)=\frac{2}{5}.$

2. Find $P(D\mid M_2)$ using Bayes' theorem

By Bayes' theorem, $P(M_2\mid D)=\frac{P(M_2)P(D\mid M_2)}{P(D)}.$ Substitute the known values: $\frac{2}{5}=\frac{\left(\frac{2}{5}\right)P(D\mid M_2)}{0.20}.$ Since $0.20=\frac{1}{5}$, $\frac{2}{5}=\frac{\left(\frac{2}{5}\right)P(D\mid M_2)}{\frac{1}{5}}=2P(D\mid M_2).$ Hence, $P(D\mid M_2)=\frac{1}{5}=0.20.$

3. Use total probability to find $P(D\mid M_3)$

Let $P(D\mid M_3)=x.$

Using the law of total probability, $P(D)=P(M_1)P(D\mid M_1)+P(M_2)P(D\mid M_2)+P(M_3)P(D\mid M_3).$

Substitute the values: $0.20=\left(\frac{2}{5}\right)(0.15)+\left(\frac{2}{5}\right)(0.20)+\left(\frac{1}{5}\right)x.$

Compute: $\left(\frac{2}{5}\right)(0.15)=0.06, \qquad \left(\frac{2}{5}\right)(0.20)=0.08.$ So, $0.20=0.06+0.08+\frac{x}{5}.$ $0.20=0.14+\frac{x}{5}.$ $0.06=\frac{x}{5}.$ $x=0.30.$

4. Final answer

Therefore, the probability that a bulb chosen from $M_3$ is defective is $\boxed{0.30}.$

5. Comparison with stored correct answer

Stored correct answer: $0.27 \text{ TO } 0.33$

Our answer $0.30$ lies in this range, so it agrees with the stored answer.