In addition to $7$ ordinary numbers (chosen out of 34 in this case), there are drawn 3 bonus numbers. In order to win the second prize, one needs $6$ numbers plus $1$ from the 3 bonus ones. I am wondering what is the probability of winning the second prize in this case?
I suppose there are $7$ ways to pick $6$ numbers from the initial slot, and therefore $27$ ways to pick the last number on the ticket($34-7$). We then have $3$ ways to choose $1$ number out of the bonus ones, and thus a total of $34$ numbers to choose for that one. I am not sure, however, how to calculate the probability of winning the prize here...
So maybe $34\cdot27$ times more likely to win the second prize than the first prize. But then I need to figure out the probability of winning the first prize...
Approach the problem combinatorically.
The probability of coming in $2$nd place is
$$\frac{\binom{7}{6} \times \binom{3}{1}}{\binom{34}{7}}. \tag1 $$
The explanation is:
The denominator, $~\displaystyle \binom{34}{7}~$ represents the total number of ways of filling out the ticket.
Without loss of generality, assume that the winning numbers (in some order) are $\{1,2,3,4,5,6,7\}.$
Also assume that the $(3)$ bonus numbers, in some order, are $\{8,9,10\}$.
In order to determine how many of the $~\displaystyle \binom{34}{7}~$ tickets will specifically come in $2$nd place, you have to determine how many such tickets will have exactly $(6)$ of the numbers from $\{1,2,\cdots,7\}$ and exactly $(1)$ of the numbers from $\{8,9,10\}$. This explains the numerator in (1) above.