How to Solve Probability Problems Without Replacement
Solve probability problems without replacement using sequential conditioning and the combination cross-check, with a worked example.
Expected Interview Answer
Without replacement means each drawn item is kept out of the pool, so both the total count and the favorable count shrink after every draw, making successive draws dependent — you must condition each draw’s probability on what was already removed.
Unlike with-replacement draws, P(event on draw 2) depends on the outcome of draw 1, since the pool composition changed. The correct approach is sequential conditioning: P(first favorable) × P(second favorable | first favorable) × ... , where each subsequent probability’s denominator (and sometimes numerator) is reduced by what was already removed. Equivalently, simultaneous-selection problems can be solved with combinations — nCr over the reduced groups — which gives the identical answer to the sequential method, since order of removal does not affect the final probability of the whole unordered outcome. The single most tested trap is applying the with-replacement constant-probability shortcut (p^n) when the problem explicitly removes items, which silently overstates or understates the true probability.
- Sequential conditioning correctly captures the shrinking pool
- The combination method offers a fast cross-check against sequential multiplication
- Explicitly flagging “without replacement” prevents the most common interview trap
AI Mentor Explanation
Selecting a captain and then a vice-captain from a 15-man squad, without putting the first pick back, means the vice-captain choice is drawn from only 14 remaining players — the pool shrinks. P(both are all-rounders), given 4 all-rounders in the squad, is (4/15) × (3/14), not (4/15)^2, because the second pick’s favorable count and total both dropped by one after removing an all-rounder on the first pick. This shrinking, dependent-draw structure is exactly what 'without replacement' means.
Worked example (without replacement)
Sequential method
- (4/10) × (3/9) = 2/15
Combination cross-check
- 4C2 / 10C2 = 6/45 = 2/15
Result
- P(both red) = 2/15
Step-by-Step Explanation
Step 1
Confirm removal
Each drawn item is kept out, so both count and total shrink for later draws.
Step 2
Condition each subsequent draw
P(2nd | 1st) uses the reduced pool after removing the first item.
Step 3
Multiply the conditioned chain
P(all n) = P(1st) × P(2nd|1st) × ... × P(nth|previous).
Step 4
Cross-check with combinations
nCr over the reduced groups gives the same answer as sequential multiplication.
What Interviewer Expects
- Correct identification that removal shrinks both count and total for later draws
- Correct sequential conditioning P(2nd|1st) rather than treating draws as independent
- Ability to cross-check with the combination method
- Recognition of the trap of applying the with-replacement p^n shortcut here
Common Mistakes
- Using the same probability for every draw as if replacement occurred
- Forgetting the numerator (favorable count) also shrinks, not just the total
- Mixing sequential and combination methods inconsistently within one solution
- Not recomputing the pool after each removed item in a 3+ draw problem
Best Answer (HR Friendly)
“Without replacement, every draw actually changes the pool for the next one, so I never reuse the same probability twice — I condition each subsequent draw on what was already removed. I like cross-checking sequential multiplication against the combination method, nCr over the reduced groups, since both must give the same answer.”
Follow-up Questions
- How would the same problem change if it were with replacement instead?
- How do you compute the probability of exactly one favorable item in 2 draws without replacement?
- Why does the combination method give the same answer as sequential conditioning?
- How would a 3-draw without-replacement problem be structured?
MCQ Practice
1. A bag has 3 red and 7 blue balls. Two balls are drawn without replacement. P(both red) is?
P = (3/10) × (2/9) = 6/90 = 1/15.
2. A box has 5 defective and 15 good items. Two items are drawn without replacement. P(both good) is?
Without replacement, second draw pool is 19 with 14 good remaining: (15/20)×(14/19).
3. Which method gives the same result as sequential conditioning for a simultaneous draw of r items from n?
The combination method (favorable nCr / total nCr) matches sequential without-replacement conditioning exactly.
Flash Cards
What does “without replacement” mean? — The drawn item is not returned, so pool size and composition shrink for later draws.
Are without-replacement draws independent? — No — each draw is conditioned on the outcomes of prior draws.
Sequential formula for 2 draws without replacement? — P(1st) × P(2nd | 1st), using the reduced pool for the second term.
Alternative cross-check method? — Combinations: favorable nCr / total nCr over the whole group at once.