Probability of exactly $7$ unreplaced draws from $\{1,...,10\}$ to get $1,2,3$? A box contains tickets numbered $1$ to $10$. Tickets are drawn at random without replacement until the numbers $1, 2$ and $3$ are drawn. Find the probability that exactly $7$ draws are required.
I am a little confused with this question. My first thought was to define the numbers $1,2,$ and $3$ as successes and the rest of the numbers as failures. Then,
$$P(\text{in $6$ draws you get $2$ successes}) = \frac{\binom{3}{2} \binom{7}{4}}{\binom{10}{6}}.$$ But I don't know where to go next or if this is even needed.
Can anyone help me? Thanks.

Probability $1,2,3$ are in the first $7$ draws is the probability none in last $3$ are $1,2,3$: $\frac{7 \choose 3}{10 \choose 3}$
Probability $1,2,3$ are in the first $6$ draws is the probability none in last $4$ are $1,2,3$: $\frac{7 \choose 4}{10 \choose 4}$
Probability exactly $7$ draws are required is the difference $\frac{7 \choose 3}{10 \choose 3}-\frac{7 \choose 4}{10 \choose 4} = \frac18$
