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.
Best Answer
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 $$\frac{{3\choose 2}{7\choose 4}}{{10\choose 6}}\cdot \frac14=\frac18$$