Q.Two natural numbers $r$, $s$ are drawn one at a time, without replacement from the set $S = \{1, 2, 3, \ldots, n\}$. Find $P[\,r \le p \mid s \le p\,]$, where $p \in S$.

Concept: Conditional Probability — we need $P(r \le p \mid s \le p)$, which is the probability that the first draw is at most $p$, given that the second draw is at most $p$.

Step 1 – Interpret the event.

Since draws are without replacement, the pair $(r,s)$ is equally likely among all $n(n-1)$ ordered pairs of distinct numbers from $S$.

Step 2 – Find the numerator.

We want $P(r \le p \ \text{and} \ s \le p)$. Both numbers must be from $\{1,\dots,p\}$ and distinct. Number of ordered pairs: $p(p-1)$.

Step 3 – Find the denominator. …

We are drawing two natural numbers $r$ and $s$ one at a time, without replacement from $\{1,2,\dots,n\}$. The event we want is: given that the second draw $s$ is at most $p$, what is the probability that the first draw $r$ is also at most $p$?

This is a classic conditional probability problem. The phrase "without replacement" is crucial — it means the two draws are dependent. If we had replacement, the answer would simply be $p/n$, but here the dependence changes things.

1. Set up the conditional probability

We want $P(r \le p \mid s \le p)$. By definition:

Both numerator and denominator are probabilities over the ordered pair $(r,s)$ drawn without replacement.

2. Count the total number of outcomes

Since draws are without replacement and order matters, the total number of equally likely outcomes is:

That is, $n$ choices for $r$, then $n-1$ remaining choices for $s$.

3. Find $P(s \le p)$

The event $s \le p$ means the second draw is one of $\{1,2,\dots,p\}$. How many ordered pairs $(r,s)$ satisfy this?

• $s$ can be any of the $p$ numbers $\le p$.
• $r$ can be any of the remaining $n-1$ numbers (since $r \neq s$).

So the number of favorable outcomes is:

Thus:

4. Find $P(r \le p \ \text{and} \ s \le p)$

Here both draws are at most $p$. Since draws are without replacement, we need ordered pairs $(r,s)$ with $r \neq s$ and both $\le p$.

• Choose $r$ from $\{1,\dots,p\}$: $p$ choices.
• Then choose $s$ from the same set, but $s \neq r$: $p-1$ choices.

So the number of favorable ordered pairs is:

Therefore:

5. Compute the conditional probability

Now plug into the formula: …

Method: Conditional Probability by Counting Equally-Likely Ordered Pairs

Use the definition of conditional probability with careful counting when items are drawn without replacement.

Steps

Step 1: Fix the sample space of ordered draws.

Two items drawn one at a time without replacement give $n(n-1)$ equally likely ordered pairs $(r,s)$ with $r\neq s$.

Step 2: Count the numerator and denominator events.

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\#(A\cap B)}{\#B}.$ …

Common Mistakes

Mistake 1: Writing $P(r\le p \cap s\le p)=(p/n)^2$.

Why it's wrong: the two draws are without replacement (dependent), so they cannot both be squared as if independent; the correct joint count is $p(p-1)$ ordered pairs, giving $\tfrac{p(p-1)}{n(n-1)}$. Correct approach: count distinct ordered pairs, not a product of marginals.

Mistake 2: Cancelling the wrong way and losing the $-1$ shifts. …