Suppose we have 10 units in service, each with an MTBF (Mean Time Between Failure) of 2,000 hours. Each operates for 2 months per year. When one fails it is returned to the maintenance shop for repair, and spends 1 week there. What is the probability that two units will be in the maintenance shop at the same time for repair? Assuming the MTBF is quoted in operational hours, the failure distribution is exponential, and the "2 months per year" operational time is spread evenly through the year (e.g., 4 hours per day), it follows that the mean calender time between failures of a specific unit is 12,000 hours. Also, a unit spends 168 hours (=1 week) in the shop each time it fails. Therefore, each unit has a mean duty cycle of 12168 hours, of which 12000 are spent in the field and 168 are spent in the shop. The periods in the shop are random and uncorrelated for each of the 10 units. The probability of any particular unit being in the shop at a randomly chosen instant is just p = 168/12168. The probability of two specific units being in the shop at a random instant is p^2. In general, the probability of exactly k out of 10 units being in the shop at a given instant is 10! Pr{k} = ---------- p^k (1-p)^(10-k) (10-k)! k! so the probability of finding exactly 2 units in the shop is 0.007675, whereas the probability of TWO OR MORE in the shop is 0.007970. Of course, the above analysis makes several tacit assumptions, and in reality the probability could be significantly higher or lower, for any of several reasons. For example, we didn't specify that all the operational periods were independent and uniformly distributed over the year, so it's possible that all the units are operational for the same 2-month period each year, which would increase the probability of having two in the shop at some point during the "busy season". On the other hand, the real probability could also be much smaller. Strictly speaking, within the stated conditions of the original problem, the probability of two or more units in the shop simultaneously could be anything from 0 to about 0.3874. For example, if each unit operates for EXACTLY 2000 hours between failures, we could stagger their repair cycles so that one fails every 200 hours, which is greater than the one week repair time of 168 hours. Thus, the probability of two in the shop would be 0. (One could also manipulate the probability by assuming the units operate in more or less mutually exclusive 2-month periods during the year.) On the other hand, if the units are operated in synchronized pairs (hopefully not two engines of a twin-engine airplane), and each fails every 2000 hours exactly, then each pair is in the shop for 168 hours out of every 2168 hours, giving a probability of 0.0774. If the five pairs of units are staggerred, then the fraction of [active] time with two units in the shop would be 0.3874. It might be interesting to figue out the maximum probability that would be strictly consistent with the stated conditions of the problem. In many casually-stated probability problems it turns out the answer can be anything on the interval from 0 to 1 (inclusive).

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