Answer: D)0.205
Explanation:
In binomial distribution, the probability of getting success in x trials is given by :-

, where n is the total number of trials , p is the probability of getting success in each trial .
Given : The probability that a component will fail is 0.2.
Let x be the number of components will stop working.
i.e. p=0.2
n=12
Since The machine will stop working if more than three components fail.
Then, the probability that the machine will stop working will be :-
![P(X>3)=1-P(X\leq3)\\\\=1-[P(0)+P(1)+P(2)+P(3)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/e1jexucfodblxu2rdfhc9b0wsvv8hs9bj5.png)
![=1-[^(12)C_0(0.2)^0(0.8)^(12)+^(12)C_1(0.2)^1(0.8)^(11)+^(12)C_2(0.2)^2(0.8)^(10)+^(12)C_3(0.2)^3(0.8)^(9)]](https://img.qammunity.org/2021/formulas/mathematics/high-school/mxfoq51exwrzvfwm8avddvt9v1rhaivftk.png)
![=1-[(1)(0.8)^(12)+(12)(0.2)(0.8)^(11)+(12!)/(2!10!)(0.2)^2(0.8)^(10)+(12!)/(3!9!)(0.2)^3(0.8)^(9)]\\\\=1-(0.068719476736+0.206158430208+0.283467841536+0.23622320128 )\\\\=1-0.79456894976\\\\=0.20543105024\approx0.205](https://img.qammunity.org/2021/formulas/mathematics/high-school/voi1hu7jc2m85hx5pkit5qu9cxdpm39nf0.png)
Hence, the probability that the machine will stop working is 0.205.
Thus , the correct answer is D)0.205