Two natural numbers $r, s$ are drawn one at a time, without replacement from the set $S=\{1, 2, 3, ...., n\}$ . Find $P[r\leq p|s\leq p]$ , where $p\in S$
The solution is given as $\frac{p-1}{n-1}$ in my reference.
I just know $P[A|B]=\frac{P(A\cap B)}{P(B)}$ but, I don't get how to approach this problem. Any help/hint 'd be appreciated.
On
Given that $s \leq p$, you now choose $r$ from among the remaining $(n-1)$ numbers. Of these numbers, $(p-1)$ of them are less than or equal to $p$ (originally there are of course $p$ numbers that are less than or equal to $p$ but you just picked one of them to be $s$ and did not replace it). So the probability that you chose a number that is at most $p$ is equal to $(p-1)/(n-1)$.
Here are a few hints. You're off to a good start, you just need to calculate the numerator and denominator. As an example, $P(\{r\leq p\}\cap \{s\leq p\})$ is the probability that both the numbers are less than or equal to $p$, i.e. $r,p\in\{1,2,\cdots,p\}$. If you start with $r$, there are $p$ choices for $r$ and $p-1$ choices for $s$. Overall, there are $n$ choices for $r$ and $n-1$ choices for $s$. Can you finish the problem from here? Note that you can also reason with binomial coefficients, e.g. $\binom{p}{2}$ and $\binom{n}{2}$.