Two-step Experiments
A two-step experiment, such as rolling two dice, has outcomes that can be listed in a table or tree diagram, and for independent steps \(P(A \text{ and } B) = P(A) \times P(B)\). This NSW Year 9 Advanced maths subtopic finds probabilities with and without replacement (NESA outcome MA5‑PRO‑C‑01).
You will find probabilities for dice sums, coins and spinners, use a product table, and compare drawing counters with and without replacement. You will find 'at least one' probabilities, use a tree diagram for a late bus, decide whether a dice product game is fair, draw balls until a white one appears, and find how many red marbles are in a bag. It is part of Year 9 Advanced Probability.
Every question with a fully worked solution.
- Two-step Experiments - Video - Two Step Experiments Watch
Theory
A two-step experiment has two stages, such as rolling two dice or drawing two counters. List the outcomes in an array or a tree diagram, multiply along each path and add the paths you want. This Year 9 Advanced maths page covers independent and dependent events, replacement, and three-step problems.
A multistage (two-step) experiment is done in stages. Its outcomes can be listed in an array (a grid, for two stages) or a tree diagram (for two or more stages).
Events are independent if one does not change the probability of the other: two dice, a coin and a spinner, or draws with replacement. They are dependent if it does: draws without replacement, where the bag changes after each draw.
On a tree diagram, the probabilities on the branches from any one point add to \(1\). Multiplying along a path gives the probability of that whole path. Different paths are mutually exclusive, so their probabilities add.
Order matters in the outcomes: \((1, 6)\) and \((6, 1)\) on two dice are different outcomes.
Independent events:
Dependent events: the second probability is worked out after the first has happened.
How to solve a two-step problem
- Decide: independent or dependent? Is anything put back? Does the first result change the second?
- Show the outcomes: an array for two fair steps (dice, spinners, coins), or a tree when the probabilities differ or there are three steps.
- Multiply along each path that fits the event.
- Add the paths. "One of each" usually has two paths: red then blue, and blue then red.
Shortcuts and checks
- "At least one": use \(1 - P(\text{none})\).
- Sums and products: fill the array with the results, then count the cells that fit.
- Expected number: multiply the probability by the number of trials.
- Check: the probabilities of all the paths add to \(1\).
(i) \(P(\text{sum} = 5)\);
(ii) \(P(\text{sum is prime})\).
Draw an array: \(3 \times 4 = 12\) equally likely outcomes.
| + | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 |
| 2 | 3 | 4 | 5 | 6 |
| 3 | 4 | 5 | 6 | 7 |
| \(\text{(i) } P(\text{sum} = 5)\) | \(=\) | \(\dfrac{3}{12} = \mathbf{\dfrac{1}{4}}\) |
(ii) Prime sums are \(2, 3, 5, 7\). Count the cells: \(2\) appears once, \(3\) twice, \(5\) three times, \(7\) once.
| \(P(\text{prime})\) | \(=\) | \(\dfrac{1 + 2 + 3 + 1}{12} = \mathbf{\dfrac{7}{12}}\) |
(i) \(P(\text{both dark})\);
(ii) \(P(\text{one of each})\).
The first chocolate is not put back, so the second draw is from \(11\) chocolates.
Multiply along a path; add the paths that fit.
| \(\text{(i) } P(DD)\) | \(=\) | \(\dfrac{5}{12} \times \dfrac{4}{11} = \dfrac{20}{132} = \mathbf{\dfrac{5}{33}}\) |
| \(\text{(ii) } P(DM) + P(MD)\) | \(=\) | \(\dfrac{5}{12} \times \dfrac{7}{11} + \dfrac{7}{12} \times \dfrac{5}{11}\) |
| \(\) | \(=\) | \(\dfrac{35}{132} + \dfrac{35}{132} = \mathbf{\dfrac{35}{66}}\) |
(i) Find the probability that the team wins a match.
(ii) How many wins would you expect in a \(16\)-match season?
(i) Two paths end in a win: rain then win, dry then win.
| \(P(\text{win})\) | \(=\) | \(0.25 \times 0.4 + 0.75 \times 0.7\) |
| \(\) | \(=\) | \(0.1 + 0.525 = \mathbf{0.625}\) |
| \(\text{(ii) } 0.625 \times 16\) | \(=\) | \(\mathbf{10}\text{ wins}\) |
(i) \(P(\text{at least one light is not green})\);
(ii) \(P(\text{exactly two are green})\).
(i) "At least one not green" is the complement of "all three green".
| \(P(GGG)\) | \(=\) | \(0.6 \times 0.6 \times 0.6 = 0.216\) |
| \(P(\text{at least one not green})\) | \(=\) | \(1 - 0.216 = \mathbf{0.784}\) |
(ii) Exactly two green happens in \(3\) ways: \(GGN\), \(GNG\), \(NGG\). Each has probability \(0.6 \times 0.6 \times 0.4 = 0.144\).
| \(P(\text{exactly two})\) | \(=\) | \(3 \times 0.144 = \mathbf{0.432}\) |
Common pitfalls
Frequently asked questions
How do you find the probability of two events happening?
If the events are independent, multiply their probabilities. If the first event changes the second, multiply the first probability by the new probability of the second.
What is the difference between with and without replacement?
With replacement, the item goes back, so the second draw has the same probabilities as the first. Without replacement, the item stays out, so there is one fewer item and the probabilities change.
How do you draw a tree diagram for probability?
Draw a branch for each outcome of the first step and write its probability. From the end of each branch, draw branches for the second step with their probabilities. Multiply along each path to get its probability.
How do you work out the probability of at least one?
Find the probability of none, then subtract it from \(1\). For example, the probability of at least one head in four tosses of a coin is \(1 - \left(\dfrac{1}{2}\right)^4\) \(= \dfrac{15}{16}\).
How many outcomes are there when two dice are rolled?
There are \(6 \times 6 = 36\) equally likely outcomes, because each of the \(6\) results on the first die can pair with each of the \(6\) results on the second.