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Probability questions often hide their method in two words: "with replacement" or "without replacement". Miss them and the whole calculation changes. This post solves one problem both ways, shows a result that surprises most students, and explains when the two answers come out almost the same.
Where is this in the syllabus?
The NCERT Class 12 chapter Probability (Reprint 2026-27) covers the multiplication rule, independent events and Bayes' theorem, and its exercises use both kinds of drawing. The JEE (Main) 2026 syllabus lists "addition and multiplication theorems of probability", and the JEE (Advanced) 2026 brochure lists conditional probability and independence. Check your exam year's documents.
What changes when you replace?
The open textbook OpenStax Introductory Statistics 2e puts it simply. With replacement, each draw is considered independent: the first result does not change the probabilities for the second. Without replacement, each member can be chosen only once, the probabilities for the second pick depend on the first, and the events are dependent.
In symbols, NCERT's multiplication rule is P(E ∩ F) = P(E) P(F|E). With replacement, P(F|E) = P(F), so you multiply the same probability each time. Without replacement, you must work out P(F|E) from what is left.
NCERT's Example 8 shows the second case: an urn has 10 black and 5 white balls, two are drawn without replacement, and P(both black) = (10/15) × (9/14) = 3/7.
Worked example: one box, two ways
Sample numbers. A box has 12 bulbs, 4 defective (D) and 8 good (G). Three bulbs are drawn.
All three good.
With replacement: (8/12)³ = (2/3)³ = 8/27 ≈ 0.296.
Without replacement: (8/12) × (7/11) × (6/10) = 336/1320 = 14/55 ≈ 0.255.
Exactly one defective.
With replacement: the defective can be first, second or third, so 3 × (1/3) × (2/3)² = 12/27 = 4/9 ≈ 0.444.
Without replacement, count with combinations: C(4, 1) × C(8, 2) / C(12, 3) = 4 × 28 / 220 = 112/220 = 28/55 ≈ 0.509.
At least one defective. Use the complement in both cases.
With replacement: 1 − 8/27 = 19/27 ≈ 0.704.
Without replacement: 1 − 14/55 = 41/55 ≈ 0.745.
Notice that "without replacement" makes a mixed sample more likely here. Once you draw a good bulb, the share of defective bulbs left goes up.
Why is the second draw still 1/3?
What is the probability that the second bulb drawn, without replacement, is defective? Many students say "it depends on the first". Work it out by the two cases:
First defective, then defective: (4/12) × (3/11) = 12/132.
First good, then defective: (8/12) × (4/11) = 32/132.
Total: 44/132 = 1/3, exactly the same as the first draw, and the same as with replacement.
If you do not know what the first draw was, every position is equally likely to hold any particular bulb. Without replacement changes conditional probabilities, not the unconditional probability of a single draw. This is the theorem of total probability at work; our post on Bayes' theorem and conditional probability builds on it.
When do the two methods give nearly the same answer?
When the population is large compared with the sample, removing a few items hardly changes the proportions. Scale the box up to 1200 bulbs with 400 defective, keeping the same one-third share. Then P(three good, without replacement) = (800/1200) × (799/1199) × (798/1198) ≈ 0.2959, against 8/27 ≈ 0.2963 with replacement. The difference is in the fourth decimal place. That is why surveys of large populations are often treated as if they were done with replacement.
Which mistakes cost marks?
Multiplying the same fraction without replacement. The denominators must fall: 12, 11, 10.
Forgetting the arrangements. "Exactly one defective" with replacement needs the factor 3 for the three positions. With combinations, the counting already includes them.
Assuming draws without replacement are independent. OpenStax's advice is to assume dependence until you can show otherwise.
Thinking the second draw's probability changes when the first is unknown. It does not, as shown above.
Adding probabilities for "at least one". Use 1 − P(none).
For more on telling independent events apart, see our post on independent events in probability.
A practice routine
Solve NCERT Exercise 13.2 Q3 (the box of 15 oranges with 3 bad ones). Check that you get 44/91.
Solve NCERT Exercise 13.2 Q13, which draws with replacement, then redo it without replacement and compare.
For any "without" problem, compute the answer both by conditional products and by combinations.
Change one problem from small to large population and watch the two answers converge.
When you cannot tell which method a question needs, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see how it reads the wording before it calculates.
Today, find the probability that two cards drawn without replacement from a 52-card pack are both hearts, then do it with replacement.
