A factory has a total of three manufacturing units $M_{1}$, $M_{2}$ and $M_{3}$, which produce bulbs independent of each other. The units $M_{1}$, $M_{2}$ and $M_{3}$ produce bulbs in the proportions of $2 : 2 : 1$, respectively. It is known that $20\%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_{1}$, $15\%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_{2}$ is $\frac{2}{5}$. If a bulb is chosen randomly from the bulbs produced by $M_{3}$, then the probability that it is defective is ______.

$P(M_1):P(M_2):P(M_3)=2:2:1$

$P(D)=20\%,\quad P(D\mid M_1)=15\%,\quad P(M_2\mid D)=\dfrac{2}{5}$

$\text{To find: } P(D\mid M_3)$

$P(M_1)=\dfrac{2}{5},\quad P(M_2)=\dfrac{2}{5},\quad P(M_3)=\dfrac{1}{5}$

$P(M_1)+P(M_2)+P(M_3)=1$

$\text{Each } M_i \text{ branches into } D \text{ (defective) and } D' \text{ (sound)}$

$P(D)=P(M_1)P(D\mid M_1)+P(M_2)P(D\mid M_2)+P(M_3)P(D\mid M_3)$

$P(M_2\mid D)=\dfrac{P(D\mid M_2)\,P(M_2)}{P(D)}$

$\dfrac{2}{5}=\dfrac{P(D\mid M_2)\cdot\dfrac{2}{5}}{\dfrac{1}{5}}$

$\Rightarrow P(D\mid M_2)=\dfrac{1}{5}$

$0.2=\dfrac{2}{5}(0.15)+\dfrac{2}{5}\left(\dfrac{1}{5}\right)+\dfrac{1}{5}\,x$

$0.2=0.06+0.08+0.2x=0.14+0.2x$

$0.2x=0.06\ \Rightarrow\ x=0.3$

$P(D\mid M_3)=0.3=\dfrac{3}{10}$

$\dfrac{P(D\mid M_2)P(M_2)}{P(D)}=\dfrac{(1/5)(2/5)}{0.2}=\dfrac{2}{5}$

$\boxed{P(D\mid M_3)=\dfrac{3}{10}}$