“…, if, at time t, the system is used for its intended purpose and is in the operable state S 2 , if, at time t, the system is used for its intended purpose and is in an inoperable state (unrevealed failure) S 3 , if, the system is not used for its intended purpose at the time t because of inspection S 4 , if, at timet, a false positive occurs and a repair of falsely rejected system is performed S 5 , if, at time t, a true positive occurs and a repair is performed S 6 , if, at time t, an unscheduled repair is carried out due to revealed failure (32) In the case of the exponential distribution of time to both unrevealed and revealed failure, F(t) = 1 − exp(−λt) and Φ(t) = 1 − exp(−λ 0 t), the mean times of the system staying in states S 1 , S 2 , S 3 , S 4 , S 5 , and S 6 in the interval (0, ∞), are as follows [27,35]:…”