Sampling Without Replacement
Explore sampling without replacement for Victorian Year 11 Mathematical Methods (VCAA) — finding probabilities when each item selected is not put back, so the pool shrinks each time.
You will learn to use tree diagrams and combinations to find probabilities of compound events, handle at-least-one questions with the complement, and reason with a changing sample space stage by stage.
Theory
In Year 11 Mathematical Methods (Unit 2), sampling without replacement means each item drawn is not put back, so the pool shrinks and the probabilities change from one draw to the next. This page shows how to multiply conditional probabilities along a tree diagram, use the complement for ‘at least one’ questions, and switch to combinations when order does not matter.
Sampling without replacement means that once an item is selected it is set aside and cannot be chosen again. Each draw therefore removes one item, so both the number of favourable outcomes and the total number of items decrease as you go.
Because the pool changes, the draws are not independent. The probability of the second event depends on what happened first — a conditional probability written \(P(B\mid A)\), read ‘the probability of \(B\) given \(A\)’.
A tree diagram lays out the draws stage by stage. Each branch carries the probability for that step (using the reduced counts), and you multiply along a branch to get the probability of that whole sequence.
When order does not matter — for example choosing a committee — the same probability can be found with combinations \(\binom{n}{r}\), the number of ways to choose \(r\) items from \(n\).
Multiplication rule along a branch (multiply the conditional probabilities for each stage):
Complement rule for ‘at least one’:
Combinations (order does not matter): choosing \(k\) of one type and \(n-k\) of another from a group gives
How to work a without-replacement problem
- Read the counts: note how many of each type and the total, and how many items are drawn.
- Draw the tree: label the first-draw branches, then the second-draw branches using the reduced totals (total and type counts each drop by one).
- Multiply along the branch you want, or add the branches for outcomes that can happen more than one way (such as ‘exactly one’).
- For ‘at least one’, find \(P(\text{none})\) and subtract from \(1\).
- Check with combinations when order does not matter, using \(\dbinom{n}{r}\).
First draw — \(6\) red out of \(10\):
| \(P(R_1)\) | \(=\) | \(\dfrac{6}{10}\) |
| \(=\) | \(\dfrac{3}{5}\) |
That red marble is now gone, leaving \(5\) red among \(9\) marbles.
Second draw — conditional on the first being red:
| \(P(R_2\mid R_1)\) | \(=\) | \(\dfrac{5}{9}\) |
Multiply along the red–red branch:
| \(P(RR)\) | \(=\) | \(\dfrac{6}{10}\times\dfrac{5}{9}\) |
| \(=\) | \(\dfrac{30}{90}\) | |
| \(=\) | \(\dfrac{1}{3}\) |
Both red: \(P(RR)=\dfrac{1}{3}\).
‘At least one’ is easiest through its complement: find \(P(\text{no red})\), then subtract from \(1\). There are \(6\) non-red balls.
No red on either draw (\(6\) of \(8\), then \(5\) of \(7\)):
| \(P(\text{no red})\) | \(=\) | \(\dfrac{6}{8}\times\dfrac{5}{7}\) |
| \(=\) | \(\dfrac{30}{56}\) | |
| \(=\) | \(\dfrac{15}{28}\) |
Subtract from \(1\):
| \(P(\text{at least one red})\) | \(=\) | \(1-\dfrac{15}{28}\) |
| \(=\) | \(\dfrac{13}{28}\) |
At least one red: \(P=\dfrac{13}{28}\).
Exactly one king happens two ways: king then other (\(KO\)), or other then king (\(OK\)). Add the two branch probabilities.
King then other (\(KO\)):
| \(P(KO)\) | \(=\) | \(\dfrac{3}{8}\times\dfrac{5}{7}\) |
| \(=\) | \(\dfrac{15}{56}\) |
Other then king (\(OK\)):
| \(P(OK)\) | \(=\) | \(\dfrac{5}{8}\times\dfrac{3}{7}\) |
| \(=\) | \(\dfrac{15}{56}\) |
Add the two orders:
| \(P(\text{one }K)\) | \(=\) | \(\dfrac{15}{56}+\dfrac{15}{56}\) |
| \(=\) | \(\dfrac{30}{56}\) | |
| \(=\) | \(\dfrac{15}{28}\) |
Combinations check — choose \(1\) king from \(3\) and \(1\) other from \(5\), over all pairs from \(8\):
| \(\dfrac{\binom{3}{1}\binom{5}{1}}{\binom{8}{2}}\) | \(=\) | \(\dfrac{3\times5}{28}\) |
| \(=\) | \(\dfrac{15}{28}\) |
Exactly one king: \(P=\dfrac{15}{28}\).
Method 1 — multiply the all-women branch (pool shrinks \(10,9,8\)):
| \(P(WWW)\) | \(=\) | \(\dfrac{6}{10}\times\dfrac{5}{9}\times\dfrac{4}{8}\) |
| \(=\) | \(\dfrac{120}{720}\) | |
| \(=\) | \(\dfrac{1}{6}\) |
Method 2 — combinations (order does not matter): choose \(3\) women from \(6\), over all groups of \(3\) from \(10\).
| \(P(WWW)\) | \(=\) | \(\dfrac{\binom{6}{3}}{\binom{10}{3}}\) |
| \(=\) | \(\dfrac{20}{120}\) | |
| \(=\) | \(\dfrac{1}{6}\) |
All three women: \(P=\dfrac{1}{6}\).
Common pitfalls
Frequently asked questions
What does sampling without replacement mean?
Each item drawn is set aside and cannot be picked again, so the total — and the count of that type — drops by one for the next draw.
How is it different from sampling with replacement?
With replacement the item goes back, the pool stays the same, and the draws are independent. Without replacement the pool shrinks and the draws are dependent, so you use conditional probabilities.
How do I find the probability that both draws are the same type?
Multiply along that branch of the tree using the reduced counts, e.g. \(\dfrac{6}{10}\times\dfrac{5}{9}\).
How do I handle ‘at least one’?
Find the probability of none of that type and subtract from \(1\): \(P(\text{at least one})=1-P(\text{none})\).
When should I use combinations instead of a tree?
When order does not matter or the sample is larger: \(P=\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\).
Why does the denominator go down by one each draw?
Because one item has been removed and not replaced, so the total number of items to choose from is one fewer.