# Thread:

1. ## Probability distribution

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.

2. Originally Posted by somestudent2
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.
From the wording of the question, it is reasonable to assume that the probability can be modeled by a geometric distribution with $p = \frac {1}{5}$. That is assuming that a piece of heavy equipment is leased one day then the events, Y, are how days until the next lease: $Y = 0,1,2,3, \cdots$.
Thus: $P(Y = n) = \frac{{4^n }}{{5^{n + 1} }}$.

3. 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)?

thanks

4. =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?
That is counterintuitive.
Intuitively it should decrease. Don’t you think?
If not, why do you think otherwise?

5. 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.

6. Originally Posted by somestudent2
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.
The operative phrase here is “on the average”. That does not mean that there is a rental every five days. If fact, the company could have a run of 15 days with no rentals. But experience has told them that in 100 day period they can expect to have on average 20 rentals.

7. 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

Thanks again.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9584082365036011, "perplexity_flag": "middle"}