Three-Step Experiments and Compound Events
Theory
A three-step experiment has three parts, such as three draws from a bag or three shots at goal, and each outcome is a compound event. This Year 9 maths page shows how to list all the outcomes with a three-stage tree diagram, multiply along a path, count the paths for "exactly", use the complement for "at least one", and change the branches at every stage when there is no replacement. It is Year 9 content in the Australian Curriculum. In the Victorian Curriculum three-step experiments are Level 10, so Victorian students can use this page as extension. Curriculum: VC2M10P02 (Level 10, extension) / AC9M9P01.
A simple event is one result of one step, such as rolling a \(6\). A compound event joins two or more simple events with "and" or "or", such as "a \(6\) on every roll" or "at least one \(6\)".
A three-step (three-stage) experiment has three parts. Each outcome is a list of three results, such as \(\text{RRB}\) for red, red, blue. The order matters: \(\text{RRB}\) and \(\text{RBR}\) are different outcomes.
The number of outcomes is the product of the number of results at each step. Three steps with \(2\) results each give \(2 \times 2 \times 2 = 8\) outcomes; steps with \(3\), \(2\) and \(4\) results give \(3 \times 2 \times 4 = 24\).
A three-stage tree diagram has a set of branches for each step. Every path crosses three branches. Multiply along a path; add the paths that make up the event.
With replacement (or for independent trials such as dice and coins) every stage has the same branches. Without replacement the total goes down by one at each stage, and so does the count of the item just taken.
Some trees stop early. If the experiment ends as soon as something happens, such as passing a test, that path has no more branches.
Multiply along a path, then add the paths in the event:
Counting orders for three trials with success \(S\) and failure \(F\):
| Three trials | Paths | Number of orders |
|---|---|---|
| all three succeed | SSS | 1 |
| exactly two succeed | SSF, SFS, FSS | 3 |
| exactly one succeeds | SFF, FSF, FFS | 3 |
| none succeed | FFF | 1 |
How to solve a three-step problem
- Write the probabilities for each step. Check whether they stay the same (with replacement, independent trials) or change (without replacement, or given as different).
- Draw the tree, or list the paths you need, such as SSF, SFS and FSS.
- Multiply along each path.
- Add the paths in the event. If they all have the same probability, multiply one path by the number of orders.
- Use the complement for "at least one": \(1 - P(\text{none})\).
Other question types
- Listing outcomes: the tree has one end branch for each outcome. Read the outcomes along the paths.
- Working backwards: write the probability with an unknown, set it equal to the value given and solve. For \(1 - (1-p)^3\), find \((1-p)^3\) first, then take the cube root.
- More of one type than another: list which counts work (for example \(3\) red, or \(2\) red and \(1\) blue) and add their probabilities.
- One of each of three types: the three results can come in \(3 \times 2 \times 1 = 6\) orders (for colours R, B and G: RBG, RGB, BRG, BGR, GRB, GBR). Every order has the same probability, so find one order and multiply by \(6\).
(i) List all the possible codes.
(ii) Find the probability that all three symbols are the same.
(iii) Find the probability that the first and last symbols match.
(i) Read along each path: XXX, XXO, XOX, XOO, OXX, OXO, OOX and OOO. There are \(2 \times 2 \times 2 = 8\) equally likely codes.
(ii) All the same: XXX and OOO.
| \(P(\text{all the same})\) | \(=\) | \(\dfrac{2}{8}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{4}}\) |
(iii) First and last match: XXX, XOX, OXO and OOO.
| \(P\) | \(=\) | \(\dfrac{4}{8}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{2}}\) |
(i) scores all three
(ii) scores exactly two
(iii) scores the first two and misses the third.
Score \(S\): \(0.7\). Miss \(M\): \(1 - 0.7 = 0.3\). Every stage is the same.
(i) One path, SSS.
| \(P(\text{SSS})\) | \(=\) | \(0.7^3\) |
| \(\) | \(=\) | \(\boldsymbol{0.343}\) |
(ii) The miss can be first, second or third: MSS, SMS, SSM. Each path has two \(0.7\)s and one \(0.3\).
| \(P\) | \(=\) | \(3 \times 0.7^2 \times 0.3\) |
| \(\) | \(=\) | \(3 \times 0.147\) |
| \(\) | \(=\) | \(\boldsymbol{0.441}\) |
(iii) This is one path only, SSM, because the order is fixed.
| \(P(\text{SSM})\) | \(=\) | \(0.7 \times 0.7 \times 0.3\) |
| \(\) | \(=\) | \(\boldsymbol{0.147}\) |
(i) \(P(\text{RRB})\), red then red then blue
(ii) the probability of exactly two red
(iii) the probability of at least one blue.
(i) After each draw, take one off the total and one off the colour drawn.
| \(P(\text{RRB})\) | \(=\) | \(\dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8}\) |
| \(\) | \(=\) | \(\dfrac{120}{720} = \boldsymbol{\dfrac{1}{6}}\) |
(ii) Exactly two red is RRB, RBR or BRR. Each path has the same numbers on top (\(6, 5, 4\)) and underneath (\(10, 9, 8\)), just in a different order, so each is \(\dfrac{1}{6}\).
| \(P(\text{exactly two red})\) | \(=\) | \(3 \times \dfrac{1}{6}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{2}}\) |
(iii) The complement of "at least one blue" is RRR.
| \(P(\text{RRR})\) | \(=\) | \(\dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8}\) |
| \(\) | \(=\) | \(\dfrac{1}{6}\) |
| \(P(\text{at least one B})\) | \(=\) | \(1 - \dfrac{1}{6} = \boldsymbol{\dfrac{5}{6}}\) |
(ii) In another game the chance of a prize on each turn is \(p\), and the probability of at least one prize in \(3\) turns is \(0.784\). Find \(p\).
(i) The complement of "at least one prize" is "no prize" on all three turns.
| \(P(\text{none})\) | \(=\) | \(0.8^3 = 0.512\) |
| \(P(\text{at least one})\) | \(=\) | \(1 - 0.512\) |
| \(\) | \(=\) | \(\boldsymbol{0.488}\) |
(ii) Write the same steps with \(p\), then solve.
| \(1 - (1 - p)^3\) | \(=\) | \(0.784\) |
| \((1 - p)^3\) | \(=\) | \(0.216\) |
| \(1 - p\) | \(=\) | \(\sqrt[3]{0.216} = 0.6\) |
| \(p\) | \(=\) | \(\boldsymbol{0.4}\) |
Check: \(1 - 0.6^3 = 1 - 0.216 = 0.784\) ✓
Common pitfalls
Frequently asked questions
How many outcomes does a three-step experiment have?
Multiply the number of results at each step. Three steps with \(2\) results each give \(2 \times 2 \times 2 = 8\) outcomes.
What is a compound event?
A compound event combines two or more simple events, such as "red on every draw" or "at least one six". Its probability is found from all the outcomes that make it up.
How do you find the probability of exactly two out of three?
Find the probability of one order, such as success, success, failure, then add the other orders. In three trials there are \(3\) orders for exactly two successes.
How do you find the probability of at least one?
Work out the probability of none, then subtract it from \(1\). This needs only one path instead of seven.
How does a three-stage tree change without replacement?
At each new stage the total goes down by one, and the count of the item already taken goes down by one, so the fractions on the branches change from stage to stage.