Sampling with Replacement
Sampling with replacement means each item is put back before the next is chosen, so the draws are independent. In NSW Year 10 Advanced maths this applies to repeated trials such as rolling a die, tossing a coin or taking penalty shots, and the complement often gives the quickest route to \(P(\text{at least one})\) (NESA outcome MA5‑PRO‑C‑01).
You will find probabilities of all, none, exactly \(k\) or at least one success over several trials, such as hits on a target or sixes on a die, counting the orders each time. You will find the chance that a code has no repeated symbol or that the first success comes on a given trial, and the least number of trials that makes a success likely. Harder questions find an unknown probability. It is part of Year 10 Advanced Probability.
Every question with a fully worked solution.
- Sampling with Replacement - Video - Tree Diagrams with Replacement Watch
Theory
In sampling with replacement, each item is put back before the next draw, so every draw has the same probabilities and the draws are independent. This Year 10 Advanced maths page covers repeated trials, at least one by the complement, exactly \(k\) successes by counting paths, the first success on a given trial, colours in any order and working backwards to an unknown.
Sampling with replacement means each chosen item is returned before the next choice. The bag is the same every time, so each draw is independent of the others.
The same model covers any repeated independent trials with a fixed chance of success \(p\): rolling a die, tossing a coin, shooting at a target, digits of a code chosen with repeats allowed.
Multiply along a path: \(n\) successes in a row have probability \(p^n\), and \(n\) failures have \((1-p)^n\). An event that can happen in several orders, such as "exactly two out of three", is a sum of equal paths: number of orders \(\times\) probability of one order.
"At least one" is the complement of "none": \(P(\text{at least one})=1-(1-p)^n\). As \(n\) grows this gets closer to \(1\), but never reaches it.
| Trials | Successes wanted | Number of orders (paths) |
|---|---|---|
| \(2\) | exactly \(1\) | \(2\): \(SF\), \(FS\) |
| \(3\) | exactly \(1\) or exactly \(2\) | \(3\) |
| \(4\) | exactly \(1\) or exactly \(3\) | \(4\) |
| \(4\) | exactly \(2\) | \(6\): \(SSFF\), \(SFSF\), \(SFFS\), \(FSSF\), \(FSFS\), \(FFSS\) |
| \(n\) | exactly \(1\) or exactly \(n-1\) | \(n\) |
| \(3\) | three different things | \(3\times 2\times 1=6\) |
How to solve a with-replacement problem
- Find the probability of each outcome for one draw or trial. It stays the same every time.
- Describe the event as paths: "all", "none", "exactly \(k\)", "first on trial \(k\)".
- Multiply along one path. If several paths have the same mix of outcomes, multiply by the number of orders.
- Use the complement for "at least one" and for "not all".
- Round only at the end, using exact values or the calculator memory.
Other question types
- Smallest number of trials: solve \((1-p)^n\lt\) the target by trying values of \(n\) on a calculator (Example 1).
- Unknown probability: write the path total with \(p\) and solve. Take roots carefully, and reject values outside \(0\) to \(1\) or values that make no sense, such as a fraction of a marble.
- Taking turns: if players take turns until someone succeeds, list the winning paths for one player (miss, miss, success, …) and add them.
- Comparing games: work out \(1-(1-p)^n\) for each game and compare. Multiplying \(n\times p\) does not give the probability.
(i) A basketballer makes each free throw with probability \(0.8\), independently. In \(5\) throws, find \(P(\text{all } 5)\) and \(P(\text{misses at least one})\), to \(4\) decimal places.
(ii) A game square gives a prize with probability \(0.15\) on each turn. Find the smallest number of turns for which \(P(\text{at least one prize})\gt 0.75\).
(i) Independent trials, so multiply.
| \(P(\text{all } 5)\) | \(=\) | \(0.8^5=0.32768\approx\boldsymbol{0.3277}\) |
| \(P(\text{at least one miss})\) | \(=\) | \(1-0.8^5\approx\boldsymbol{0.6723}\) |
(ii) \(P(\text{at least one})=1-0.85^n\). This is more than \(0.75\) when \(0.85^n\lt 0.25\). Try values:
| \(0.85^8\) | \(=\) | \(0.272\ldots \quad (\text{too big})\) |
| \(0.85^9\) | \(=\) | \(0.231\ldots \quad (\lt 0.25)\) |
The smallest number is \(\boldsymbol{n=9}\) turns.
(i) \(P(\text{exactly } 3 \text{ heads})\)
(ii) \(P(\text{the first head is on the fourth toss})\)
(i) The single tail can be in any of the \(4\) positions: \(THHH\), \(HTHH\), \(HHTH\), \(HHHT\). Each path has three \(0.6\)s and one \(0.4\).
| \(P\) | \(=\) | \(4\times 0.6^3\times 0.4\) |
| \(\) | \(=\) | \(4\times 0.0864=\boldsymbol{0.3456}\) |
(ii) The first head on toss four is one path only: \(TTTH\).
| \(P\) | \(=\) | \(0.4^3\times 0.6=\boldsymbol{0.0384}\) |
(i) \(P(\text{three different colours})\)
(ii) \(P(\text{no green})\)
(iii) \(P(\text{exactly one green})\)
Each draw has \(P(R)=0.2\), \(P(B)=0.3\), \(P(G)=0.5\).
(i) One order, such as \(RBG\), has probability \(0.2\times 0.3\times 0.5=0.03\). The three colours can come in \(3\times 2\times 1=6\) orders.
| \(P\) | \(=\) | \(6\times 0.03=\boldsymbol{0.18}\) |
(ii) Not green each time: \(0.5^3=\boldsymbol{0.125}\).
(iii) Green in one of \(3\) positions: \(3\times 0.5\times 0.5^2=\boldsymbol{0.375}\).
(i) A spinner lands on red with probability \(p\). It is spun twice, and \(P(\text{at least one red})=0.91\). Find \(p\).
(ii) A bag has \(x\) red and \(3\) blue marbles. Two are drawn with replacement, and \(P(\text{both blue})=\dfrac{1}{16}\). Find \(x\).
(i) The complement is "no red" on both spins.
| \(1-(1-p)^2\) | \(=\) | \(0.91\) |
| \((1-p)^2\) | \(=\) | \(0.09\) |
| \(1-p\) | \(=\) | \(0.3 \quad (1-p\ge 0)\) |
| \(p\) | \(=\) | \(\boldsymbol{0.7}\) |
(ii) Each draw has \(P(\text{blue})=\dfrac{3}{x+3}\).
| \(\left(\dfrac{3}{x+3}\right)^2\) | \(=\) | \(\dfrac{1}{16}\) |
| \(\dfrac{3}{x+3}\) | \(=\) | \(\dfrac{1}{4}\) |
| \(x+3\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(\boldsymbol{9}\) |
Common pitfalls
Frequently asked questions
What does sampling with replacement mean?
Each item chosen is put back before the next choice, so the population is the same every time. The draws are independent and each has the same probabilities.
How do you find the probability of at least one success in n trials?
Find the probability of no successes, \((1-p)^n\), then subtract it from \(1\).
How do you find the probability of exactly k successes?
Work out the probability of one order, such as success, success, failure, then multiply by the number of different orders. For exactly \(2\) successes in \(4\) trials there are \(6\) orders.
Why does the probability of at least one success never reach 1?
Because the probability of no successes, \((1-p)^n\), gets smaller and smaller but is never zero when \(p\lt 1\).
Is rolling a die several times sampling with replacement?
Yes, in effect. Every roll has the same six equally likely outcomes, just as if each outcome were put back after it was chosen.