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NCERT Exemplar · Q17

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

Andhra Pradesh BieapShort· 3mImportance★★★★★
Appeared in past exams:WBJEE 2024· Set math-2024· 1mexact
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The key idea is to use conditional probability: P(r≤p∣s≤p)=P(r≤p∩s≤p)P(s≤p)P(r \le p \mid s \le p) = \frac{P(r \le p \cap s \le p)}{P(s \le p)}. Since draws are without replacement, the numerator counts ordered pairs where both numbers are at most pp, and the denominator counts ordered pairs where the second number is at most pp. The final result is p−1n−1\boxed{\frac{p-1}{n-1}}.

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

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/np/n, but here the dependence changes things.

1. Set up the conditional probability

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

P(r≤p∣s≤p)=P(r≤p and s≤p)P(s≤p)P(r \le p \mid s \le p) = \frac{P(r \le p \ \text{and} \ s \le p)}{P(s \le p)}

Both numerator and denominator are probabilities over the ordered pair (r,s)(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:

n×(n−1)n \times (n-1)

That is, nn choices for rr, then n−1n-1 remaining choices for ss.

3. Find P(s≤p)P(s \le p)

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

  • ss can be any of the pp numbers ≤p\le p.
  • rr can be any of the remaining n−1n-1 numbers (since r≠sr \neq s).

So the number of favorable outcomes is:

p×(n−1)p \times (n-1)

Thus:

P(s≤p)=p(n−1)n(n−1)=pnP(s \le p) = \frac{p(n-1)}{n(n-1)} = \frac{p}{n}

Note

Interestingly, P(s≤p)=p/nP(s \le p) = p/n is the same as if we drew with replacement. The marginal distribution of the second draw is uniform over {1,…,n}\{1,\dots,n\} — a symmetry property of sampling without replacement.

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

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

  • Choose rr from {1,…,p}\{1,\dots,p\}: pp choices.
  • Then choose ss from the same set, but s≠rs \neq r: p−1p-1 choices.

So the number of favorable ordered pairs is:

p×(p−1)p \times (p-1)

Therefore:

P(r≤p and s≤p)=p(p−1)n(n−1)P(r \le p \ \text{and} \ s \le p) = \frac{p(p-1)}{n(n-1)}

5. Compute the conditional probability

Now plug into the formula: …

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