Mark (√) against the correct answer in each of the following:

A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third component are 0.2, 0.3 and 0.5, respectively. What is the probability that the machine will fail?


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)


⇒ 1 – 0.28 = 0.72

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