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A rental agency, which leases heavy equipment by the day, has found that one expensive piece of equipment is leased, on the average, only one day in five. If rental on one day is independent of rental on any other day, find the probability distribution of Y, the number of days between a pair of rentals.

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Let R denote the event that a rental occurs on a given day and N represents the event that there is no rental. Thus, the sequence of interest is RR, RNR, RNNR, RNNNR, .... Let us consider the position immediately following R: it is filled by an R with probability 0.2 and by an N with probability 0.8. Thus, P(Y=0)=0.2P(Y = 0) = \boxed{0.2}, P(Y=1)=0.8×0.2=0.16P(Y = 1) = 0.8 \times 0.2 = \boxed{0.16}, P(Y=2)=0.128P(Y =2) = \boxed{0.128}. Using the rules of mathematical induction, we get:

P(Y=y)=0.2×0.8y;y=0,1,2,.......P(Y = y) = \boxed{0.2 \times {0.8}^y}; y = 0,1,2,.......

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