The problem as posed does not have an answer. But a closely related problem does.
We modify the problem to specify that we choose an $n$-digit number at random, for fixed $n$. First we find how many $n$-digit integers there are. The first digit can be anything but $0$, so there are $9$ choices for the first digit. For every choice of first digit, there are $10$ choices for second digit, and for every choice of first digit and second digit, there are $10$ choices for the third digit, and so on for a total of $9\cdot 10^{n-1}$ choices. A simpler way of getting at the same thing is to note that we are, for say $n=5$, looking at the numbers from $10000$ to $99999$ inclusive, and there are $99999-10000+1$ of these, which is $90000$. The same idea works for more digits.
All these numbers are equally likely. Now we ask how many of our numbers have exactly $x$ odd digits.
The analysis is a little different for $n=x$ than for $n>x$, so we first deal with the case $n=x$. All digits must then be odd. There are $5$ choices for the first digit, and for each choice there are $5$ choices for the second digit, and so on, for a total of $5^x$. The required probability is then $\frac{5^x}{9\cdot 10^{x-1}}.$ There is a fair bit of cancellation, and the probability simplifies to $5/(9\cdot 2^{x-1})$. Not really surprising! The probability that the first digit is odd is $5/9$, and once we are past the first digit, the probability that any subsequent digit is odd is $1/2$.
Next we look at the case $n>x$. Maybe (i) the first digit is odd or (ii) maybe it is even.
In case (i), we must choose where the remaining $x-1$ odd digits will be. This can be done in $\binom{n-1}{x-1}$ ways. Now that we have decided on the locations of the odd digit, we must choose which odd digits we will use. That can be done in $5^x$ ways. Finally, we must fill up the remaining $n-x$ places with even digits. This can be done in $5^{n-x}$ ways, for a total of $\binom{n-1}{x-1}5^x5^{n-x}=\binom{n-1}{x-1}5^{n}$ ways.
In case (ii), we must choose where the $x$ odd digits will go, from the $n-1$ places available. There are $\binom{n-1}{x}$ ways to do the choosing. Now there are $5^x$ ways of deciding which odd digits we use. For the even digits, we have $4$ choices at the first place, and $5$ choices at the remaining $n-x-1$ places. Thus the number of choices for case (ii) is $\binom{n-1}{x}5^x5^{n-x-1}\cdot 4=\binom{n-1}{x}5^{n-1}\cdot 4.$ Add up. The total number of choices is $\binom{n-1}{x-1}5^{n}+\binom{n-1}{x}5^{n-1}\cdot 4.$ Divide by $9\cdot 10^{n-1}$ for the probability. A good deal of further simplification can be done.
Comment: The problem as stated (the number has $x$ or more digits, nothing else specified) cannot be solved, for there is no probability distribution that assigns equal probabilities to all numbers with $x$ or more digits. We could use the idea above to solve the problem for randomly chosen numbers with between $m$ and $n$ digits, where $x \le m \le n$.