$\text{A factory has a total of three manufacturing units,}{\mathrm{ M}}_1,M_2,\text{and}{M}_3,\text{which produce bulbs independent of each}$

$\text{other. The units}{\mathrm{ M}}_1,M_2,\text{and}{M}_3\text{produce bulbs in the proportions}\mathrm{ 2}:2:1,\text{respectively. It is known that}\mathrm{ 20}\%\text{of the}$

$\text{bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by}{\mathrm{ M}}_1,15\%\text{are}$

$\text{defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the}$

$\text{probability that it was produced by}{\mathrm{ M}}_2\text{ is }\frac{2}{5}.$
$\text{If a bulb is chosen randomly from the bulbs produced by}{\mathrm{ M}}_3,\text{then the probability that it is defective is}\text{\_\_\_\_\_\_}$

The correct answer is 0.30.

Let us define the events representing the selection of a bulb from each manufacturing unit:
Let $E_1$ be the event that the bulb is produced by unit $M_1$.
Let $E_2$ be the event that the bulb is produced by unit $M_2$.
Let $E_3$ be the event that the bulb is produced by unit $M_3$.

The three units produce bulbs in the proportions 2 : 2 : 1. Therefore, the probabilities of choosing a bulb from each unit are:
$P\left(E_1\right)=\frac{2}{2+2+1}=\frac{2}{5}=0.4$
$P\left(E_2\right)=\frac{2}{5}=0.4$
$P\left(E_3\right)=\frac{1}{5}=0.2$

Let $D$ be the event that a bulb is defective.
The overall probability of a bulb chosen from the factory being defective is 20%:
$P\left(D\right)=0.20$

It is given that 15% of the bulbs produced by unit $M_1$ are defective:
$P\left(D\right|E_1)=0.15$

We are also given that, if a randomly chosen bulb is found to be defective, the probability that it was produced by unit $M_2$ is $\frac{2}{5}$:
$P\left(E_2\right|D)=\frac{2}{5}=0.4$

Using Bayes' theorem, we know:
$P\left(E_2\right|D)=\frac{P\left(E_2\right)·P\left(D\right|E_2)}{P\left(D\right)}$

Substituting the known values into this equation:
$0.4=\frac{0.4·P\left(D\right|E_2)}{0.20}$

Dividing both sides by 0.4:
$1=\frac{P\left(D\right|E_2)}{0.20}\Rightarrow P\left(D\right|E_2)=0.20$

By the Law of Total Probability, the total probability of a defective bulb $P\left(D\right)$ is:
$P\left(D\right)=P\left(E_1\right)·P\left(D\right|E_1)+P(E_2)·P(D\left|E_2\right)+P\left(E_3\right)·P\left(D\right|E_3)$

Substituting all the known values:
$0.20=(0.4·0.15)+(0.4·0.20)+(0.2·P(D\left|E_3\right))$

Now we calculate the products:
$0.4·0.15=0.06$
$0.4·0.20=0.08$

Substituting these back into the equation:
$0.20=0.06+0.08+0.2·P\left(D\right|E_3)$
$0.20=0.14+0.2·P\left(D\right|E_3)$

Subtracting 0.14 from both sides:
$0.06=0.2·P\left(D\right|E_3)$

Solving for $P\left(D\right|E_3)$:
$P\left(D\right|E_3)=\frac{0.06}{0.2}=0.30$

Thus, the probability that a bulb produced by unit $M_3$ is defective is 0.30.