Multistage Events: The Product Rule
Master the product (multiplication) rule for multistage events in NSW Year 12 Mathematics Standard 2. For independent events the probability that \(A\) and \(B\) both happen is \(P(A \text{ and } B)=P(A)\times P(B)\): on a probability tree you simply multiply along the branches from the start to the outcome.
You will learn to apply the product rule to combined outcomes, tell apart draws with replacement (independent, branch probabilities repeat) from without replacement (the branch probabilities change), and find the probability of "at least one" using the complement \(1 - P(\text{none})\) — a core Standard 2 probability skill for weather, quality control, cards and games.
Theory
The product (multiplication) rule finds the probability of a combined outcome across two or more stages by multiplying. This Year 12 Standard 2 (NSW) guide shows how to multiply along the branches of a probability tree, use \(P(A \text{ and } B)=P(A)\times P(B)\) for independent events, handle draws with and without replacement, and find "at least one" with the complement \(1 - P(\text{none})\).
A multistage event happens in two or more stages — such as tossing a coin then rolling a die, or drawing two counters from a bag. The product (multiplication) rule finds the probability of a combined outcome by multiplying the probabilities of the separate stages.
When the stages are independent (one does not change the next), \(P(A \text{ and } B) = P(A)\times P(B)\). On a probability tree this is exactly "multiply along the branches" from the start of the tree to the outcome you want.
If an item is drawn with replacement the stages stay independent, so the branch probabilities repeat. If it is drawn without replacement there is one fewer item, so the second-stage probabilities change — you still multiply along the path. For an "at least one" question the fastest route is the complement, \(P(\text{at least one}) = 1 - P(\text{none})\). This is a core Year 12 Standard 2 (NSW) probability skill.
For two independent events the product rule is:
Along any path of a probability tree you multiply every branch you pass through:
For an "at least one" outcome, use the complement of the product:
How to use the product rule
- Identify the stages and whether they are independent (with replacement or genuinely separate) or without replacement.
- Write each branch probability. For without replacement, reduce the totals — one fewer item overall and one fewer of the colour already taken.
- Multiply along the path to the outcome you want: \(P(A \text{ and } B)=P(A)\times P(B)\).
- For "at least one", find \(P(\text{none})\) as a product and subtract from \(1\); simplify the final fraction or give the decimal.
The events are independent, so multiply the two probabilities.
| \(P(\text{green})\) | \(=\) | \(0.4\) |
| \(P(\text{on time})\) | \(=\) | \(0.7\) |
| \(P(\text{green and on time})\) | \(=\) | \(0.4\times 0.7\) |
| \(\) | \(=\) | \(0.28\) |
The probability is \(0.4\times 0.7 = \mathbf{0.28}\).
Replacing keeps the draws independent, so \(P(G)=\dfrac{4}{7}\) each time — multiply.
| \(P(GG)\) | \(=\) | \(P(G)\times P(G)\) |
| \(\) | \(=\) | \(\dfrac{4}{7}\times\dfrac{4}{7}\) |
| \(\) | \(=\) | \(\dfrac{16}{49}\) |
\(P(\text{both green}) = \mathbf{\dfrac{16}{49}}\).
One sock is kept, so the second draw is from \(8\) socks with one fewer black — the branch probability changes.
| \(P(\text{1st black})\) | \(=\) | \(\dfrac{5}{9}\) |
| \(P(\text{2nd black})\) | \(=\) | \(\dfrac{4}{8}\) |
| \(P(BB)\) | \(=\) | \(\dfrac{5}{9}\times\dfrac{4}{8}\) |
| \(\) | \(=\) | \(\dfrac{20}{72}=\dfrac{5}{18}\) |
\(P(\text{both black}) = \mathbf{\dfrac{5}{18}}\).
Use the complement: "at least one miss" is the opposite of "scores all three".
| \(P(\text{scores all three})\) | \(=\) | \(\dfrac{2}{3}\times\dfrac{2}{3}\times\dfrac{2}{3}\) |
| \(\) | \(=\) | \(\dfrac{8}{27}\) |
| \(P(\text{at least one miss})\) | \(=\) | \(1-\dfrac{8}{27}\) |
| \(\) | \(=\) | \(\dfrac{19}{27}\) |
\(P(\text{at least one miss}) = 1 - \dfrac{8}{27} = \mathbf{\dfrac{19}{27}}\).
Common pitfalls
Frequently asked questions
What is the product rule in probability?
For independent events the product rule says the probability that A and B both happen is the probability of A multiplied by the probability of B, written P(A and B) = P(A) x P(B). On a probability tree it means multiplying the probabilities along the branches from the start to the outcome you want.
When can you multiply probabilities together?
You multiply when you want two or more things to happen together and the stages are independent, meaning one does not change the next. Tossing a coin and rolling a die, or drawing with replacement, are independent, so their probabilities multiply.
What is the difference between with replacement and without replacement?
With replacement the item is put back, so the totals are unchanged and the stages are independent — the branch probabilities repeat. Without replacement the item is kept, so there is one fewer item and the second-stage probabilities change; you still multiply along the path.
How do you find the probability of at least one?
Use the complement. Instead of adding up all the ways of getting at least one, work out the probability of getting none as a single product, then subtract from 1: P(at least one) = 1 - P(none).
Do you add or multiply probabilities on a tree diagram?
You multiply along the branches of a single path to get the probability of that combined outcome. You only add when you combine several different complete paths, for example to find the probability of one of each colour.
Does the product rule need the events to be independent?
The rule P(A and B) = P(A) x P(B) is stated for independent events, which is exactly the with-replacement case. For a without-replacement tree you still multiply along the path, but the second branch probability is read off the tree after one item has been removed.