Continuous random variables sound strange to me I have read some introductory probability theory textbooks and found that for a continuous random variable, $P(x=a) =0\;\forall a$ , that means, whatever what possible outcome I choose, the possibility of it happens is zero. I found it strange, because it stated that no outcome is possible. (But I have no problem in understanding that it does not implies $P(whole sample space) = 0$)
Maybe the above interpretation is flawed, if yes, please correct me, thank you.
 A: One way to think about it is to imagine a sequence of realizations of an experiment where no one outcome is repeated more than once. For example this is what you would expect if you drew a number uniformly at random from the interval $[0,1]$ infinitely many times: at any given stage of the experiment, the "intuitive probability" that you hit a number you've already hit is extremely small (the mathematical probability is zero, but that's what we're trying to justify). Since frequentists define probability as the number of times the event occurs divided by the number of times we run the experiment, as we run the experiment more times...that fraction goes to zero, if the event only occurs once.
You should try to think about it like this or in a similar way, instead of getting stuck on the idea that $P(A)=0$ implies $A$ is impossible.
A: The probability of an event occurring is typically viewed (in the frequentist approach) as the ratio of number of outcomes favoring the occurrence of the event to the total number of outcomes.
For example, let us consider the example of tossing a coin $N$ times, where $N\geq 1$. Suppose that we are interested in finding the probability of a head turning up, and towards this end, let us say we toss the coin $N$ times and get the following sequence of outcomes:
\begin{eqnarray}
H,~\underbrace{T,T,T,T,T,\ldots,T}_{N-1~\text{times}}.
\end{eqnarray}
Then, the frequentist approach requires us to divide the number of heads that have occurred in $N$ tosses by $N$, which is $\frac{1}{N}$, and look at this ratio as $N\rightarrow \infty$. This is clearly $0$, but it does not mean that a head never occurred (the first outcome was a head).
A: For a continuous probability distribution, probability is an area, 
$P(a \le X \le b) = \int_a^b p(x)dx$ It follows that 
$P(X=a) = \int_a^a p(x)dx = 0$ because the area under a point is $0$. 
That's not to say that $P(X=a)$  has to be $0$. For example If you have a random variable defined on the interval $[0,1]$ you could define  
$P[X=0] = P[X = 1] = \dfrac 13$ and, on $(0,1)$, $P[a\le X \le b] = \dfrac{b-a}3$
Note that $\displaystyle P[0 \le x \le 1] =
P[X=0] + \lim_{\delta \to 0^+} P[\delta \le X \le 1-\delta] + P[X=1] = 1$$
But, if you wanted, say $P[X=x] = \alpha$ for all x in $[0,1]$ then you would have to figure out how to solve
$$\displaystyle P[0\le x \le 1] =  \sum_{x\in[0,1]}\alpha = 1 $$ for $\alpha$.
This sort of thinking has some unintuitive, but necessary, consequences. for example. Pick a point, x,  on the real number line at random. Then
the probability that $x$ is a rational number is $0$ while the probability that $x$ is an irrational number is $1$.
A: I won't describe strictly. Lets consider this topic from more intuitive, practical point of view. Suppose you are targeting a circle with radius R with a bullet. As you may guess the probability of hitting in some figure is: $ \frac{A}{\pi R^2} $, where A is the area of figure you are targeting. On the other hand, what is the area of some particular point (as you know it is 0), so the probability is 0 also, but if you consider the vicinity of the point, it will have the probability to be hit, because its area is non-zero. Definetly the area of the circle is not zero, though it has of infitely many points with zero area, so is with the probalities.
