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.
 A: Actually you're off to a good start. Your calculation the probability of $2$ successes in the first $6$ draws (namely $\frac12$) is correct. Then for the seventh draw you have a $\frac14$ (conditional) probability of getting the remaining number of $1,2$, or $3$.
So your overall probability becomes $$\frac{{3\choose 2}{7\choose 4}}{{10\choose 6}}\cdot \frac14=\frac18$$
A: 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 $\frac{7 \choose 3}{10 \choose 3}-\frac{7 \choose 4}{10 \choose 4} = \frac18$
A: In first six draws it will be two successes : $\frac{\binom{3}{2}\times \binom{7}{4}}{\binom{10}{6}}$
and for the seventh draw : $\frac{1}{4}$
$$\text{P}=\frac{\binom{3}{2}\times \binom{7}{4}}{\binom{10}{6}}\times \frac{1}{4}=\frac{1}{8}$$
A: Imagine $10$ slots. One of the three "good" numbers is to be in slot $7$, the other two are to be in the six preceding slots from the $9$ slots left.
Put them in, turn by turn.
$Pr = \dfrac3{10}\dfrac69\dfrac58 = \dfrac18$
A: Another approach: There are $10!$ possible sequences of draws. As @GregMartin suggests, let's count the number of good sequences.

*

*There are $3$ possibilities for what the seventh number can be (1, 2 or 3).

*The first six numbers contain the other two numbers from $\{1,2,3\}$ as well as four numbers from $\{4, 5, \ldots, 10\}$. There are $\binom74$ possibilites for choosing those four numbers.

*The first six numbers, now that they are fixed, can be arranged in $6!$ ways.

*The last three numbers are now fixed as well and can be arranged in $3!$ ways.

That gives us $3\cdot\binom74\cdot6!\cdot3!$ desirable sequences of draws, so the probability is $$\frac{3\cdot\binom74\cdot6!\cdot3!}{10!}=\frac18.$$
In general, there are $3\cdot\binom7{n-3}\cdot(n-1)!\cdot(10-n)!$ sequences that require exactly $n$ draws, so the probability for a general $3\leq n\leq10$ is $$\frac{3\cdot\binom7{n-3}\cdot(n-1)!\cdot(10-n)!}{10!}=\frac{(n-2)(n-1)}{240}.$$
