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 on rental on any other day, find the probability distribution of Y, the number of days between a pair of rentals.
Does this mean that probability of each sample point is 1/5? I am totally lost in this one, and how to solve it.
Hi, thanks for your quick response. However I do not completely understand the solution since we haven't covered geometric distribution yet. Is there any way to solve this problem just using the basics of probability distribution? Also by your formula we get p(0)=1/5; p(1)=4/25; p(2)=16/125 etc but If n is the number of days between two rentals, shouldn't the probability increase as n increases (just common sense)?
That is counterintuitive.=somestudent2;107401we get p(0)=1/5; p(1)=4/25; p(2)=16/125 etc but If n is the number of days between two rentals, shouldn't the probability increase as n increases?
Intuitively it should decrease. Don’t you think?
If not, why do you think otherwise?
Maybe then I just don't understand the problem. I thought of it as follows:
Let 1 represent a rent and 0 not a rent, the probability of 1 is 1/5, the prob of 0 is 4/5
110000 (means the rents are in two consecutive days) here n=0 and the probability must be the lowest?
100001 (means 2 rents 4 days apart ) here is n=4 and the probability must be 1?
since we are given that there is 1 rent in 5 days?
I am totally lost now.
Thanks for your help Plato, I finally get it, in fact I was misinterpreting given information in the problem.
I see how you arrived with this formula:
Since events are independent we we can multiply them to get the intersection.
Setting 1/5 for the first day * 4/5 * 4/5 *4/5....(for each day with no rental n times)
indeed it comes out to be 4^n/5^n+1