Probability of lottery jackpot with or without bonus ball I understand that for a lottery where you choose 6 numbers in a draw from 49 balls the probability is 
$$\frac{1}{{n\choose k}}=\frac{1}{{49\choose 6}}.$$ 
I am not sure if the following calculation is correct:
We introduce a bonus ball with a bigger jackpot, whereby you select a number that is independent from the above draw, and is drawn independent from the above draw from a set of 49 balls. Then the probability of selecting 6 balls and the bonus correctly.
$$\frac{1}{{49\choose 6}} * \frac{1}{49}$$
Now, the probability of winning either with or without the bonus ball. That is to choose 6 numbers correctly OR choose 6 numbers AND the bonus ball correctly
$$ \frac{1}{{49\choose 6}}+\frac{1}{{49\choose 6}} * \frac{1}{49}$$
Is that correct?

Not correct.  Note that the event of choosing $6$ regular balls and bonus ball correctly is a subset of the event of choosing $6$ regular balls correctly.  Thus the probability of winning with or without the bonus ball is just $\frac{1}{49\choose 6}$.
If you want to break it out by cases: $6$ correct without bonus or $6$ correct with bonus, you can rewrite as
$$\frac{1}{{49\choose 6}}\cdot \frac{48}{49}+\frac{1}{{49\choose 6}}\cdot \frac{1}{49}$$ which conditions on whether you don't or do get the bonus ball (along with the $6$ regular balls).
