Probability of choosing a of b numbers in exactly c guesses (This is my first question, and I have not formally learned much probability, so forgive me if this question is too easy.)
Background information: while watching a UK gameshow called The Code, I began wondering how many questions a contestant would need to answer, on average, to win. Relevant to this question, the only thing you need to know about the show is that 3 distinct numbers from 0-9 have been selected, and after answering a question correctly, the contestant chooses a number and gets to see whether it is one of the 3 they are looking for. They need to find all 3 to win.
I'm looking for a general formula for the probability of finding a numbers selected out of a character set of b in exactly c attempts (obviously $a,b,c \in \mathbb N$). In the example that inspired this question, $a = 3$ and $b = 10$ (and I would like to know the probability for all $c \in [3, 10]$, and/or some other way of calculating the median number of guesses necessary).
I have calculated some specific values of c as follows:


*

*For $c = 3$, the player needs to make three correct guesses in a row, with probability $\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{1}{8} = \frac{1}{120}$. I notice that this is the reciprocal of ${10 \choose 3}$, but the next example is not $1 \div {10 \choose 4} = \frac{1}{210}$, so I don't think it's that simple.

*For $c = 4$, the player will make three correct guesses and one incorrect one. There are seven incorrect numbers that could be chosen (before the incorrect guess), and there will be a similar 3, 2, 1 pattern to what we saw in the last example for the correct choices. The possible numbers to choose from at each stage will be 10, 9, 8, 7. And finally, there are three potential orders (the incorrect guess can be any of the first three positions, but not the fourth as this is covered by $c=3$). So this gives us the probability: $3 \cdot \frac{7\cdot3\cdot2\cdot1}{10\cdot9\cdot8\cdot7} = \frac{1}{40}$, which I believe is correct.


Are these calculations correct, and if so, how can they be generalised?

Note that this is can be modeled by a hypergeometric distribution for $a-1$ successes out of the first $c-1$ guesses, along with a $\frac{1}{b-c+1}$ chance of getting the last number on the next guess. Plugging in to the formula should give us $\frac{\binom{a}{a-1}\binom{b-a}{c-a}}{\binom{b}{c-1}} \times \frac{1}{b-c+1}$.
Alternatively, this can be seen as the probability of winning in at most $c$ turns but not winning in at most $c-1$ turns. There are $\binom{a}{a}\binom{b-a}{c-a}$ ways to win in $c$ turns, out of a total of $\binom{b}{c}$ possible turns, so this gives us
$$\frac{\binom{a}{a}\binom{b-a}{c-a}}{\binom{b}{c}} - \frac{\binom{a}{a}\binom{b-a}{c-a-1}}{\binom{b}{c-1}}$$
We check to make sure that this matches the above answer:
\begin{align*}
&= \frac{\frac{(b-a)!}{(c-a)!(b-c)!}}{\frac{b!}{c!(b-c)!}} - \frac{\frac{(b-a)!}{(c-a-1)!(b-c+1)!}}{\frac{b!}{(c-1)!(b-c+1)!}}\\
&=\frac{(b-a)!c!}{(c-a)!b!} - \frac{(b-a)!(c-1)!}{(c-a-1)!b!}\\
&=\frac{(b-a)!c!- (c-a)(b-a)!(c-1)!}{(c-a)!b!}\\
&=\frac{(b-a)!(c - (c-a))(c-1)!}{(c-a)!b!}\\
&=\frac{a(b-a)!(c-1)!}{(c-a)!b!}\times\frac{\frac{1}{(b-c)!}}{\frac{1}{(b-c)!}}\\
&=\frac{a\frac{(b-a)!}{(c-a)!(b-c)!}}{\frac{b!}{(c-1)!(b-c)!}}\\
&=\frac{a\binom{b-a}{c-a}}{\frac{b!}{(c-1)!(b-c+1)!}}\times\frac{1}{b-c+1}\\
&=\frac{a\binom{b-a}{c-a}}{\binom{b}{c-1}}\times\frac{1}{b-c+1}\\
\end{align*}
as desired.
