Urn problem (draw two balls of different colours with replacement) e.g. 3 white balls and 7 black balls.
Prove that the probability to draw two different coloured balls is 0.42 
I get the intuition, but can't formally explain the probability.

Whenever you are doing a probability question you want to make sure you formulate it in the easiest way for you to work with. Often it is easier to work with the complement of a set than with the set itself. For instance, if you are asked what is the probability that at least 1 ball is black, it is easiest to do $1 - P(\text{none black})$. In this problem you do the same trick.
\begin{align}
P({\rm different}) &= 1 - P({\rm same}) \\
&= 1 - P(\text{both white}) - P(\text{both black}) \\
&= 1 - \frac{3}{10}\frac{3}{10} - \frac{7}{10}\frac{7}{10} \\
&= \frac{100 - 9 - 49}{100} = \frac{42}{100}
\end{align}
