The uniform probability distribution of a specific value I am having this problem: The rainfall in April in Famagusta follows a uniform distribution between $0.5$ inches and $3.0$ inches. I am confused in finding the probability that the amount of rainfall will be exactly 1 inch of rain for the month. 
Based on the formula of a uniform probability distribution we have: $P(X=x)=\frac1{2.5}$ for every $x$ on the closed interval $[0.5; 3]$ and consequently $P(X=1)=\frac1{2.5}=0.4$.
On the other hand, based on the rectangular shape of the uniform probability distribution we have: $P(X=1)=$(height)(base)$=(\frac1{2.5}) (1-1)=0.$
But, $0$ is distinct from $0.4$. So what is the probability of exactly $1$ inch of rain? Which of the methods is wrong? 
Please give me a hand.
 A: Your problem is here:

Based on the formula of a uniform probability distribution we have:
  $P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.
A: You can show this regardless because of the definition of a continuous random variable. 
A continuous random variable has a density function $f(x)$ with $f(x) \geq 0$  with 
$$\int_{-\infty}^{\infty} f(x) dx = 1  $$
and 
$$ P(a < X < b)  =\int_{a}^{b} f(x) dx  $$
then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$ 
$$ P(X=c) = \int_{c}^{c} f(c) dx = 0$$
but you can show this specifically for the uniform distribution like this. The above
$X \sim U(a,b)$
specifically $ X \sim U(.5,3)$ this gives us a density function of 
$f(x) =\begin{align}\begin{cases} \frac{1}{3-.5} &  \textrm{ for  } \frac{1}{2} \leq x \leq 3\\ \\ 0  &  \textrm{ for  } x <\frac{1}{2} \textrm{ or } x > 3 \end{cases} \end{align}$
Then we have
$$ P(X=1)  = \int_{1}^{1} \frac{1}{2.5} dx = 0$$
Note the density function for uniform variables is simply
$f(x) =\begin{align}\begin{cases} \frac{1}{b-a} &  \textrm{ for  } a \leq x \leq b \\ \\ 0  &  \textrm{ for  } x <a \textrm{ or } x > b \end{cases} \end{align}$
