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There is a random quantity $L$, which is evenly distributed from 0 to $l_{\max}$.

How to find the dependence of the probability that the variable $x$ (positive) will be less than the value of the random variable $L$?

Intuitively, this will be an exponential dependence from zero to $l_{\max}$. But I cannot prove it to myself. Any thoughts in which direction to think?

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  • $\begingroup$ $P(x<l)=1-\frac x {l_{max}}$ for $0 <x <l_{\max}$. $\endgroup$ Jan 29 '21 at 8:44
  • $\begingroup$ thanks for the answer, but why? $\endgroup$ Jan 29 '21 at 8:51
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The random quantity L has the following distribution functio:

$$ F_{L}(x)=\begin{cases} 0, & x<0\\ \frac{x}{l_{max}}, & 0\leq x<l_{max}\\ 1, & x\geq t_{stay} \end{cases} $$

The distribution function has the meaning of the probability that the value of a random variable l will be less than an arbitrary number x.

$$ F_{L}(x)=P(l<x). $$

$l < x$ event, forms a complete group with $l\geq x$ event. Then the probability of this event is found by the formula:

$$ P(x\leq l)=1-P(l<x). $$

Then the required probability is found by the formula:

$$ p(x)=1-\begin{cases} 0, & x<0\\ \frac{x}{l_{max}}, & 0\leq x<l_{max}\\ 1, & x\geq l_{max} \end{cases} $$

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