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(C) Let the total number of bulbs produced be $100$. Since the production ratio is $2:2:1$,the number of bulbs produced by $M_1, M_2$,and $M_3$ are $4

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$0.30$

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(C) Let the total number of bulbs produced be $100$. Since the production ratio is $2:2:1$,the number of bulbs produced by $M_1, M_2$,and $M_3$ are $40, 40$,and $20$ respectively.Given that $20\%$ of total bulbs are defective,total defective bulbs $= 20$.For $M_1$,$15\%$ of $40$ bulbs are defective,so $0.15 	imes 40 = 6$ bulbs are defective.Let $x$ be the number of defective bulbs produced by $M_3$. Then the number of defective bulbs produced by $M_2$ is $20 - 6 - x = 14 - x$.Given that the probability that a randomly chosen defective bulb was produced by $M_2$ is $\frac{2}{5}$,we have:$P(M_2 | 	ext{Defective}) = \frac{	ext{Defective bulbs from } M_2}{	ext{Total defective bulbs}} = \frac{14 - x}{20} = \frac{2}{5}$.Solving for $x$:$14 - x = \frac{2}{5} 	imes 20 = 8$$x = 14 - 8 = 6$.Thus,$M_3$ produces $6$ defective bulbs out of $20$.The probability that a bulb chosen from $M_3$ is defective is $\frac{6}{20} = 0.3$.

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