You ask whether the property you describe is true "in general for irrational numbers". That phrase can mean "all irrational numbers", or "most irrational numbers". Because you have accepted Geoffrey's answer of "Certainly not...", I guess you mean "all".

However, to be clear, I want to give treatment to "most" because the answer to that question is "Yes..."

Consider an irrational number like

$$x = 0.1280451740318436570487162...$$

that contains no 9s. We'll call such numbers *9-less*. From this single number, many 9-*full* irrationals can be created simply by inserting 9s in various places.

$x$ is non-terminating. Therefore, there is an infinitude of possible insertion points, which means an infinitude of 9-full irrationals can be generated in this way. (In fact, it can be shown that the set is uncountable.)

Each 9-less irrational generates a *distinct* infinite set of 9-full irrationals. Therefore, the ratio of 9-less irrationals to 9-full irrationals is zero.

Every irrational number is either 9-less or 9-full, but not both, so almost all irrational numbers are 9-full.

Addendum:

As you will find in the Wikipedia article, "almost all" has several different uses. In this case it means 9-less numbers have Lebesgue measure 0. In other words, they're "sprinkled" among the irrationals, no two "touching". It also means that if you pick an irrational number at random (from a finite interval), it has probability 0 of being 9-less, and probability 1 of being 9-full. This is *despite* the fact that both sets have the same size, or cardinality. Such is the weirdness of infinities.

Further addendum:

There's another proof of this involving the probability of selecting an infinite sequence of digits randomly and never getting a 9.