How to Solve Probability Problems With Replacement
Solve probability problems with replacement using constant per-draw probability, independence, and the binomial formula, with examples.
Expected Interview Answer
With replacement means each item drawn is put back before the next draw, so the total pool and every probability stay identical across draws, making successive draws independent and letting you simply multiply the same per-draw probability repeatedly.
Because the pool composition never changes, P(event on draw 1) = P(event on draw 2) = ... = P(event on draw n), and the events are independent, so P(all n draws favorable) = p^n where p is the constant single-draw probability. This is the same structure as repeated coin flips or repeated die rolls — each trial is a fresh, identical experiment. With replacement problems are the natural home of the binomial distribution: the probability of exactly k successes in n independent trials is nCk × p^k × (1−p)^(n−k). Contrast this constantly with without-replacement problems, where the pool shrinks and probabilities change draw to draw — mixing the two up is the single most common interview mistake.
- Constant per-draw probability makes multiplication straightforward
- Independence unlocks the binomial formula for “exactly k successes” problems
- A clean contrast point against without-replacement conditional draws
AI Mentor Explanation
A bowling machine reloaded with the same 10-ball mix (say 3 marked 'yorker') after every delivery keeps the yorker chance at 3/10 on every single ball, since the mix never changes — that’s replacement. Getting a yorker on 2 consecutive machine-fed balls is (3/10) × (3/10), simple multiplication because each ball is an identical, independent trial. This constant-probability, independent-trial structure is exactly what 'with replacement' means in any probability problem.
Worked example (with replacement)
First draw
- P(red) = 4/10
Second draw (pool restored)
- P(red) = 4/10 again
Combined
- 4/10 × 4/10 = 4/25
Step-by-Step Explanation
Step 1
Confirm replacement
The item is returned to the pool before the next draw, so pool composition is unchanged.
Step 2
Compute the single-draw probability
p = favorable/total, and this value stays fixed for every draw.
Step 3
Multiply for independent successive draws
P(all n favorable) = p^n, since each draw is identical and independent.
Step 4
Use the binomial formula for “exactly k”
P(exactly k successes in n draws) = nCk × p^k × (1−p)^(n−k).
What Interviewer Expects
- Correct identification that replacement keeps the pool and probability constant
- Understanding that with-replacement draws are independent
- Correct use of p^n for repeated identical successes
- Correct application of the binomial formula for “exactly k successes”
Common Mistakes
- Reducing the denominator on later draws as if it were without replacement
- Forgetting that replacement makes draws independent, justifying simple multiplication
- Using p^n when the question actually asks for “exactly k out of n” (needs the binomial formula)
- Confusing with-replacement (constant pool) and without-replacement (shrinking pool) setups
Best Answer (HR Friendly)
“With replacement, the pool resets after every draw, so the probability is exactly the same every single time and the draws are independent — I just multiply the same probability n times for n favorable draws in a row. If the question asks for exactly k successes rather than all of them, I switch to the binomial formula, nCk times p to the k times (1-p) to the n-k.”
Follow-up Questions
- How does the calculation change if the draw is without replacement instead?
- How do you compute the probability of at least one success across n with-replacement draws?
- What is the binomial formula and when do you apply it here?
- How would you compute the expected number of successes in n with-replacement trials?
MCQ Practice
1. A bag has 5 red and 5 blue balls. Two balls are drawn with replacement. P(both blue) is?
With replacement, P(blue) = 5/10 = 1/2 each time; P(both blue) = 1/2 × 1/2 = 1/4.
2. A coin is flipped 4 times (with replacement is implicit for coins). P(exactly 2 heads) is?
Binomial: 4C2 × (1/2)^2 × (1/2)^2 = 6 × 1/16 = 6/16 = 3/8.
3. A bag has 2 defective and 8 good items. An item is drawn, noted, and replaced 3 times. P(all 3 defective) is?
With replacement, the probability stays 2/10 on every draw, so P = (2/10)^3.
Flash Cards
What does “with replacement” mean? — The drawn item is returned to the pool before the next draw, keeping composition unchanged.
Are with-replacement draws independent? — Yes — the same fixed probability applies to every draw.
Formula for n identical favorable draws with replacement? — p^n, where p is the constant single-draw probability.
Formula for exactly k successes in n with-replacement trials? — Binomial: nCk × p^k × (1−p)^(n−k).