Correct answer is B.
The probability of failure of the first component = 0.2 =P(A)
The probability of failure of second component = 0.3 = P(B)
The probability of failure of third component = 0.5 = P(C)
As the events are independent,
The machine will operate only when all the components work, i.e.,
(1-0.2)(1-0.3)(1-0.5) = P(A’)P(B’)P(C’)
In rest of the cases, it won’t work,
So P(AUBUC) = 1 – P(A’∩B’∩C’) = 1 – (0.8).(0.7).(0.5)
\(\Rightarrow\) 1 – 0.28 = 0.72