I recently learned that the prime omega function $\Omega(n)=\Omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=\alpha_1+\alpha_2...+\alpha_k$ is very well studied. In particular, we know that $\Omega(n)$ is equally often even and odd. This statement is, in fact, equivalent to the prime number theorem.

My question is, do we know anything about the distribution of parities of $\omega(n)=\omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=k$?

It is natural to assume that $\omega(n)$ is equally often even and odd, but perhaps it is much harder to show. From what I understand the reason that the distribution of $\Omega(n)$ is so much easier to analyze is that the Liouville lambda function $\lambda(n)=(-1)^{\Omega(n)}$ is very well understood and it's summary function $L(x)=\sum_{n<x}\lambda(n)$ can be related to the Mobius/Mertens function by

$$L(x)=\sum_{d^2<x}M\left(\frac{x}{d^2}\right)$$

The Mertens function is obviously very well studied, but no such inversion formulas are possible for $\omega(n)$ so we cannot use methods like this. I am curious about not only whether or not the result I ask for is known but whether or not the result is easier/harder to prove than the equivalent result for $\Omega(n)$.