To solve the problem of finding the probability \( P(r \leq p \,|\, s \leq p) \) where \( r \) and \( s \) are drawn from the set \( S = \{1, 2, 3, \ldots, n\} \) without replacement, we can follow these steps:
### Step 1: Understand the scenario
We are drawing two numbers \( r \) and \( s \) from the set \( S \) without replacement. We want to find the conditional probability that \( r \) is less than or equal to \( p \) given that \( s \) is less than or equal to \( p \).
### Step 2: Define the events
Let:
- Event A: \( r \leq p \)
- Event B: \( s \leq p \)
We need to find \( P(A | B) \), which is given by the formula:
\[
P(A | B) = \frac{P(A \cap B)}{P(B)}
\]
### Step 3: Calculate \( P(B) \)
To find \( P(B) \), we need to determine the total number of ways to choose \( s \) such that \( s \leq p \). The possible choices for \( s \) are from the numbers \( 1, 2, \ldots, p \), which gives us \( p \) choices.
Since we are drawing without replacement, after choosing \( s \), we have \( n - 1 \) choices left for \( r \). Therefore, the total number of favorable outcomes for event B is \( p \) (choices for \( s \)) multiplied by \( n - 1 \) (choices for \( r \)):
\[
P(B) = \frac{p(n-1)}{n(n-1)} = \frac{p}{n}
\]
### Step 4: Calculate \( P(A \cap B) \)
Next, we need to find \( P(A \cap B) \), which is the probability that both \( r \leq p \) and \( s \leq p \).
If \( s \) is chosen from the first \( p \) numbers, then \( r \) can also be chosen from these \( p \) numbers. Thus, the number of favorable outcomes for both events is \( p \) (choices for \( s \)) multiplied by \( p - 1 \) (choices for \( r \) since \( r \) cannot be equal to \( s \)):
\[
P(A \cap B) = \frac{p(p-1)}{n(n-1)}
\]
### Step 5: Substitute into the conditional probability formula
Now we can substitute \( P(A \cap B) \) and \( P(B) \) into the conditional probability formula:
\[
P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{p(p-1)}{n(n-1)}}{\frac{p}{n}} = \frac{(p-1)}{(n-1)}
\]
### Final Result
Thus, the final probability is:
\[
P(r \leq p \,|\, s \leq p) = \frac{p-1}{n-1}
\]
