Sampling without Replacement
Sampling without replacement means an item is not put back once it is chosen, so the numbers in the bag change after every draw and later probabilities depend on earlier ones. In NSW Year 10 Advanced maths you multiply the changing fractions along each path, as in drawing two faulty items from a box (NESA outcome MA5‑PRO‑C‑01).
You will find probabilities for two and three draws from jars, bags and decks of cards, including exactly one, all and at least one of a kind. Problems include raffle prizes, choosing a committee, vowels from letter tiles, a lotto-style game and testing phones for faults. Harder questions form an equation to find an unknown number of marbles or counters, and compare inspecting with and without replacement. It is part of Year 10 Advanced Probability.
Every question with a fully worked solution.
- Sampling without Replacement - Video - Tree Diagrams without Replacement Watch
Theory
In sampling without replacement, each item chosen is kept out, so the total and the count of that type both drop by 1 before the next draw. This Year 10 Advanced maths page covers two- and three-draw problems, exactly \(k\) and at least one, playing cards, raffles and committees, finding an unknown count, and how the answer compares with sampling with replacement.
Sampling without replacement means an item, once chosen, is not put back. Choosing a committee, dealing cards, drawing raffle prizes and taking items from a box to test are all without replacement.
The draws are dependent: each probability after the first is a conditional probability, worked out from what is left. If the first item was red, the second draw has one fewer red and one fewer item in total.
Drawing several items "at once" or "at random together" gives the same probabilities as drawing them one after another without replacement, so use the same method.
Each order of the same items has the same probability: \(RRB\), \(RBR\) and \(BRR\) all have the same numerators and denominators, just in a different order. So "exactly \(k\)" is still the number of orders times the probability of one order.
| With replacement | Without replacement | |
|---|---|---|
| second-draw total | same as the first | one less |
| draws are | independent | dependent |
| two of a kind, \(a\) of \(N\) | \(\left(\dfrac{a}{N}\right)^2\) | \(\dfrac{a}{N}\times\dfrac{a-1}{N-1}\) (smaller) |
| at least one of a kind | \(1-\left(\dfrac{N-a}{N}\right)^2\) | \(1-\dfrac{N-a}{N}\times\dfrac{N-a-1}{N-1}\) (larger) |
How to solve a without-replacement problem
- Write down the numbers of each type and the total.
- Follow one path: for each draw, the numerator is how many of that type are left, and the denominator is how many items are left.
- Multiply along the path. For "exactly \(k\)", multiply by the number of orders.
- Add different cases, such as "same colour" \(=RR+BB\), or "more red than blue" \(=RRR+\) two red.
- Use the complement for "at least one" and "not all". "At least one from each group" is \(1-P(\text{all from one group})\) summed over the groups.
Other question types
- Letters and labels: count repeated letters separately. A word with \(3\) A tiles has \(3\) chances of drawing an A.
- Unknown count: write the path with \(n\) (Example 4) and solve. A product of consecutive whole numbers, such as \(r(r-1)(r-2)=120\), can be solved by recognising \(4\times 5\times 6\).
- Comparing methods: for the same box, sampling without replacement makes "two of the same rare type" less likely and "at least one of a type" more likely than with replacement.
- Show-that algebra: with \(n\) of each of two colours, write both same-colour paths over \(2n\) and \(2n-1\), then simplify.
(i) \(P(\text{both faulty})\)
(ii) \(P(\text{exactly one faulty})\)
(iii) \(P(\text{at least one faulty})\)
(i) Multiply along the \(FF\) path (Diagrams tab).
| \(P(FF)\) | \(=\) | \(\dfrac{4}{15}\times\dfrac{3}{14}=\dfrac{12}{210}=\boldsymbol{\dfrac{2}{35}}\) |
(ii) Two orders: \(FG\) and \(GF\).
| \(P\) | \(=\) | \(\dfrac{4}{15}\times\dfrac{11}{14}+\dfrac{11}{15}\times\dfrac{4}{14}\) |
| \(\) | \(=\) | \(\dfrac{88}{210}=\boldsymbol{\dfrac{44}{105}}\) |
(iii) Use the complement, \(GG\).
| \(P\) | \(=\) | \(1-\dfrac{11}{15}\times\dfrac{10}{14}=1-\dfrac{110}{210}=\boldsymbol{\dfrac{10}{21}}\) |
(i) \(P(\text{all red})\)
(ii) \(P(\text{exactly two red})\)
(iii) \(P(\text{at least one blue})\)
(i) The totals go \(10\), \(9\), \(8\).
| \(P(RRR)\) | \(=\) | \(\dfrac{6}{10}\times\dfrac{5}{9}\times\dfrac{4}{8}=\dfrac{120}{720}=\boldsymbol{\dfrac{1}{6}}\) |
(ii) Exactly two red: \(RRB\), \(RBR\), \(BRR\). Each order has numerators \(6, 5, 4\) in some order over \(10\times 9\times 8\), so each equals \(\dfrac{120}{720}\).
| \(P\) | \(=\) | \(3\times\dfrac{120}{720}=\boldsymbol{\dfrac{1}{2}}\) |
(iii) At least one blue is everything except \(RRR\): \(1-\dfrac{1}{6}=\boldsymbol{\dfrac{5}{6}}\).
(i) \(P(\text{both diamonds})\)
(ii) \(P(\text{a king, then a queen})\)
(iii) \(P(\text{one king and one queen, in either order})\)
(i) \(13\) diamonds, then \(12\) of the \(51\) cards left.
| \(P\) | \(=\) | \(\dfrac{13}{52}\times\dfrac{12}{51}=\boldsymbol{\dfrac{1}{17}}\) |
(ii) Drawing a king does not remove a queen, so there are still \(4\) queens out of \(51\).
| \(P\) | \(=\) | \(\dfrac{4}{52}\times\dfrac{4}{51}=\boldsymbol{\dfrac{4}{663}}\) |
(iii) Add the queen-then-king order, which has the same probability: \(2\times\dfrac{4}{663}=\boldsymbol{\dfrac{8}{663}}\).
(i) Find \(n\).
(ii) Compare with the probability of two greens if the first counter is replaced.
(i) There are \(n+2\) counters, then \(n+1\).
| \(\dfrac{n}{n+2}\times\dfrac{n-1}{n+1}\) | \(=\) | \(\dfrac{2}{5}\) |
| \(5n(n-1)\) | \(=\) | \(2(n+2)(n+1)\) |
| \(5n^2-5n\) | \(=\) | \(2n^2+6n+4\) |
| \(3n^2-11n-4\) | \(=\) | \(0\) |
| \((3n+1)(n-4)\) | \(=\) | \(0\) |
\(n\) is a positive whole number, so \(\boldsymbol{n=4}\). Check: \(\dfrac{4}{6}\times\dfrac{3}{5}=\dfrac{2}{5}\).
(ii) With replacement: \(\left(\dfrac{4}{6}\right)^2=\dfrac{4}{9}\), which is more than \(\dfrac{2}{5}\). Removing a green counter makes a second green less likely.
Common pitfalls
Frequently asked questions
What does sampling without replacement mean?
Each item chosen is kept out, so it cannot be chosen again. The total number of items, and the number of the type chosen, both go down by \(1\) before the next draw.
Are draws without replacement independent?
No. Each draw changes what is left, so the probabilities for the next draw depend on what was drawn before. They are dependent events.
Is drawing two items at once the same as drawing one after the other?
Yes. Taking two items together gives the same probabilities as taking one and then another without putting the first back.
How do you find the probability of exactly two of a kind in three draws without replacement?
Work out one order, such as two of the kind then one other, multiplying the counts left over the totals left. The other orders have the same probability, so multiply by \(3\).
Why is the probability of two of the same kind smaller without replacement?
After one item of that kind is removed, there are fewer of that kind left compared with the total, so the second draw is less likely to match.