William Feller Vol I, P149, example (d) Power supply problem "A power supply problem:
Suppose that n = 10 workers are to use electric power intermittently, and we are interested in estimating the total load to be expected. For a crude approximation imagine that at any given time each worker has the same probability p of requiring a unit of power. If they work independently, the probability of exactly k workers requiring power at the same time should be b(k; n, p). If, on the average, a worker uses power for 12 minutes per hour, we would put p = 1/5. The probability of seven or more workers requiring current at the same time is then b(7; 10, 0.2) + ... + b(10; 10, 0.2) = 0.0008643584. "
If we are randomly putting 10 balls into 5 slots, the probability of 7 balls in the same cell is $\: {{C_1^{5} C_7^{10} 4^3} \over 5^{10}}$, which is the same as $\: C_1^{5} b(7; n, p)$. On the other hand, if I divide the hour into 5 equal slots, and ask the probability of k out of 10 people choosing one specific slot during the hour, that clearly is 10 binomial trials with p = 1/5.
Question: Is Feller implicitly indicating he's fixing a time slot, and viewing the power supply problem from that point of view? It surely doesn't make sense, that the probability of 7 workers coincide in any time slot is the same as that of coinciding in a given time slot.
 A: The ball/slot model you are using is not analogous to the power allocation problem, because in the former, the slots are discrete and non-overlapping, whereas in the power allocation problem, the time during which a worker needs power is not specified within a predefined partition of the hour.
This difference is why the ball/slot model has a higher probability than the power allocation problem:  compare a situation in which six balls occupy the same slot.  The conditional probability that the seventh ball also occupies that same slot is $1/5$.  But in the case where there is a moment in time $t^* \in [0,1]$ during which six workers need power, this does not mean that all six workers need power during the same interval--i.e., if worker $W_i$ needs power during $[t_i, u_i]$ for $i \in \{1, 2, \ldots, 7\}$, the fact that $$t^* \in \bigcap_{i=1}^6 [t_i, u_i] \tag{1}$$ does not in general imply that $[t_1, u_1] = [t_2, u_2] = \ldots = [t_6, u_6]$; in fact, this is a very unlikely outcome.  Thus the measure of the intersection of intervals $(1)$ in which the seventh worker would need to also demand power is in general smaller than $1/5$ of an hour.

A model that is analogous to the power allocation problem would be something like this:  a car is exactly $1/5$ of the length of a curb.  Each day for $10$ days, the car's driver picks uniformly at random a spot along the curb to park.  For the entire $10$ day period, what is the probability that there was a point along the curb for which the car had been parked next to it for $7$ out of the $10$ days?
The resulting probability in this case is the same as the power allocation problem, but there is one difference:  the car's length relative to the curb, which is analogous to the average time a worker needs power, is a fixed quantity; moreover, the car occupies a continuous subinterval of the curb's length on any given day--it doesn't, for instance, split into different parts.  The power allocation problem does not require that each worker needs the power in a single continuous interval of time.
Interestingly, a minor modification of the car parking problem turns it into your ball/slot model:  just paint parking lines on the road so that there are now five parking spaces, and the car must be parked within a given space each day.  (Assume there is no gap between spaces and each space fits the car exactly.)  Then the driver must park entirely within the space, and there are only five discrete choices each day.
