Two-step Experiments
A two-step experiment, such as tossing a coin and then rolling a die, has outcomes made of two parts. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑PRO‑C‑01, using arrays, tree diagrams and the rule \(P(A \text{ and } B) = P(A) \times P(B)\) for independent events.
You will count outcomes by multiplying, list sample spaces and use arrays for two dice, finding, for example, that a sum of \(10\) has probability \(\frac{1}{12}\). You will tell independent events from dependent ones, draw counters with and without replacement, as in \(\frac{5}{9} \times \frac{4}{8} = \frac{5}{18}\), find at least one six using the complement, and add probabilities when an outcome can happen along two branches, such as rain on exactly one of two days. It is part of Year 9 Core Probability.
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Theory
A two-step experiment has two stages, such as rolling a die and spinning a spinner, or drawing two counters from a bag. This Year 9 maths page shows how to list the outcomes with an array or tree diagram, multiply the probabilities of independent events, handle drawing with and without replacement, and find “at least one” with the complement.
A two-step experiment is two chance events, one after the other. If the first stage can happen in \(m\) ways and the second in \(n\) ways, there are \(m \times n\) outcomes altogether.
An array (a grid) or a tree diagram lists every outcome. Order matters: a head then a tail is a different outcome from a tail then a head.
Independent events do not affect each other. Two coins, a coin and a die, or a draw with replacement (the item goes back) are independent.
Dependent events do affect each other. Drawing without replacement is the usual case. Once an item is taken out, there is one fewer item, so the second probability changes.
The word “replaced” decides which kind of question it is. Look for it before writing anything down.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Die across the top, spinner (1, 2, 3) down the side
For two independent events \(A\) and \(B\), multiply:
For “at least one”, use the complement:
| Drawing from 9 counters | 1st draw | 2nd draw | Events are |
|---|---|---|---|
| With replacement | out of \(9\) | out of \(9\) | independent |
| Without replacement | out of \(9\) | out of \(8\) | dependent |
How to solve a two-step probability problem
- Decide whether the stages are independent. Look for “with replacement” or “without replacement”.
- List the outcomes. Use an array for two dice or a die and a spinner, and a tree diagram for draws from a bag.
- Write the probability on each branch. Without replacement, the second fraction has one less on the bottom, and one less on top if the same kind was taken.
- Multiply along each path you need, then add the paths that count.
- Use the complement for “at least one”: work out “none”, then subtract from \(1\).
- Check that all the paths of the tree add to \(1\).
Every spin can pair with every roll, so multiply the number of ways for each stage.
| \(\text{spinner}\) | \(=\) | \(5 \text{ ways}\) |
| \(\text{die}\) | \(=\) | \(6 \text{ ways}\) |
| \(\text{outcomes}\) | \(=\) | \(5 \times 6\) |
| \(\) | \(=\) | \(\mathbf{30}\) |
An array with \(5\) rows and \(6\) columns would show all \(30\) of them.
The spinner cannot affect the die, so the events are independent. Multiply.
“Greater than \(4\)” means \(5\) or \(6\), so \(2\) of the \(6\) numbers.
| \(P(\text{green})\) | \(=\) | \(\dfrac{1}{5}\) |
| \(P(>4)\) | \(=\) | \(\dfrac{2}{6} = \dfrac{1}{3}\) |
| \(P(\text{green and} >4)\) | \(=\) | \(\dfrac{1}{5} \times \dfrac{1}{3}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{1}{15}}\) |
(i) both are white,
(ii) there is one of each colour.
Draw the tree. After the first draw only \(8\) counters are left, so every second-stage fraction is out of \(8\).
(i) One white has gone, so \(4\) whites are left out of \(8\). Multiply along the branch:
| \(P(WW)\) | \(=\) | \(\dfrac{5}{9} \times \dfrac{4}{8}\) |
| \(\) | \(=\) | \(\dfrac{20}{72}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{5}{18}}\) |
(ii) One of each happens two ways, white then black or black then white. Find both red paths and add:
| \(P(WB)\) | \(=\) | \(\dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{20}{72}\) |
| \(P(BW)\) | \(=\) | \(\dfrac{4}{9} \times \dfrac{5}{8} = \dfrac{20}{72}\) |
| \(P(\text{one of each})\) | \(=\) | \(\dfrac{40}{72}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{5}{9}}\) |
The opposite of “at least one red” is “no red at all”. Find that, then subtract from \(1\).
| \(P(\text{not red})\) | \(=\) | \(\dfrac{4}{5}\) |
| \(P(\text{no red})\) | \(=\) | \(\dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{16}{25}\) |
| \(P(\text{at least one red})\) | \(=\) | \(1 - \dfrac{16}{25}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{9}{25}}\) |
Counting directly would need three cases: red then not red, not red then red, and red twice. The complement needs one.
Common pitfalls
Frequently asked questions
What is a two-step experiment in probability?
It is an experiment with two stages, one after the other, such as tossing a coin and then rolling a die. The total number of outcomes is the number of ways for the first stage times the number of ways for the second.
What is the difference between independent and dependent events?
Independent events do not affect each other, like two rolls of a die. Dependent events do, like drawing two cards without putting the first one back: the second draw is from a smaller pack.
How do you find the probability of two independent events both happening?
Multiply their probabilities: \(P(A \text{ and } B) = P(A) \times P(B)\). For example, \(\dfrac{1}{5} \times \dfrac{1}{3} = \dfrac{1}{15}\).
How does drawing without replacement change the probability?
The item is not put back, so the second draw has one fewer item in total. If the first item was the colour you want, there is one fewer of that colour too. Both the top and the bottom of the second fraction can change.
How do you work out the probability of at least one?
Find the probability of none, then subtract it from \(1\). For a spinner with one red section out of \(5\) spun twice, \(P(\text{no red}) = \dfrac{16}{25}\), so \(P(\text{at least one red}) = \dfrac{9}{25}\).