Concentrations of mixed solutions

If I mix a $\pu{2 M}$ solution with a different $\pu{1 M}$ solution, do the individual solutions reduce in molarity to $\pu{1.33 M}$ and $\ce{0.33 M}$ respectively?




You need to know how much of each solution you add and the final total volume. The molarity of each one would be 


$$\frac{C\_1 \cdot V\_1}{V\_1 + V\_2} \quad \text{and} \quad \frac{C\_2 \cdot V\_2}{V\_1 + V\_2}$$


where $C\_1 = \pu{2 M}$, $C\_2 = \pu{1 M}$, $V\_1$, $V\_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.


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