Calculate the molar fraction of a mixture component given only its pressure
Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.

As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.

I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$
We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three ideal gases, as Avogadro's law says.

This means that not only R and T are constant but that the volume will also not change, which in turn simplifies everything a lot.
$$n=p~\frac{V_m}{R~T}=p~k$$
Let's insert it into the mole fraction (as I told you to do) and see what happens.
$$\begin{align}
x_\ce{Ar}=&amp;~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\
=&amp;~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&amp;~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&amp;~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$

And this is the simple reason why you do not need more than your pressures.