Using Tree Diagrams
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Theory
A probability tree diagram shows every path through a multistage experiment. Multiply along a path, then add the paths you need. This Year 10 maths page covers branches that depend on the first stage, missing branch values, choosing a bag or moving an item between bags, three-stage trees, trees that stop early, and finding an unknown probability from a path total.
Each stage of a tree is one step of the experiment. Each branch is labelled with its probability, and the branches leaving any one point add to \(1\). A path runs from the start to an end point and is one outcome of the whole experiment.
When the stages are independent, every set of second-stage branches is the same. When they are dependent, the second-stage branches change with the first result: they are conditional probabilities such as \(P(\text{win}\mid\text{serve in})\).
A tree can stop early: if the experiment ends once something happens (passing a test, the first goal), that branch has no further stages. Paths of different lengths are fine.
The probabilities of all the end points add to \(1\). This is a quick check, and it gives the complement trick: \(P(\text{at least one})=1-P(\text{none})\).
| Question asks for | Paths to add |
|---|---|
| exactly one success in two stages | \(SF\) and \(FS\) |
| exactly two successes in three stages | \(SSF\), \(SFS\), \(FSS\) |
| at least one success | \(1-P(\text{all fail})\) |
| at least two in three | exactly two \(+\) all three |
| the second result only | every path whose second branch is that result |
How to use a tree diagram
- Draw one stage per step. Label each branch with its outcome and probability.
- Fill in missing branches with \(1-\) the other branch from the same point.
- Check dependence: if the second step depends on the first (without replacement, conditions, weather), write the second-stage probabilities separately for each first result.
- Multiply along each path you need, and add those paths.
- Check with the complement, or that all the end points add to \(1\).
Other question types
- Weather or day-to-day chains: the first day is given, so start the tree at the next day. Each day's branches depend on the day before.
- Moving an item between bags: stage one is the item moved; stage two is the draw from the second bag, whose contents depend on stage one.
- Choosing a bag first: stage one is the choice (a coin toss or die), stage two the draw from that bag.
- Unknown branch: write the path total with a pronumeral, set it equal to the given probability and solve. It may be quadratic (Example 4); reject values outside \(0\) to \(1\) or outside the stated range.
- Matches: "first to win two games" stops after \(WW\) or \(LL\); otherwise a third game is played. List the winning paths: \(WW\), \(WLW\), \(LWW\).
(i) Find the probability that she wins the point.
(ii) A tree has \(P(A)=0.4\), \(P(B\mid A)=0.5\) and \(P(B\mid A')=x\). If \(P(B)=0.56\), find \(x\).
(i) Let \(G\) be a first serve that is in and \(F\) a fault. Two paths end in a win \(W\) (Diagrams tab).
| \(P(W)\) | \(=\) | \(0.6\times 0.7+0.4\times 0.45\) |
| \(\) | \(=\) | \(0.42+0.18=\boldsymbol{0.6}\) |
(ii) The missing first branch is \(P(A')=0.6\). Add the two paths that end in \(B\) and set the sum to \(0.56\).
| \(0.4\times 0.5+0.6x\) | \(=\) | \(0.56\) |
| \(0.6x\) | \(=\) | \(0.36\) |
| \(x\) | \(=\) | \(\boldsymbol{0.6}\) |
(i) \(P(\text{exactly two green})\)
(ii) \(P(\text{at least one green})\)
Let \(G\) be green and \(N\) not green. The "not green" probabilities are \(0.5\), \(0.6\) and \(0.1\).
(i) Exactly two green: \(GGN\), \(GNG\) or \(NGG\). These paths have different products, so work out each one.
| \(P(GGN)\) | \(=\) | \(0.5\times 0.4\times 0.1=0.02\) |
| \(P(GNG)\) | \(=\) | \(0.5\times 0.6\times 0.9=0.27\) |
| \(P(NGG)\) | \(=\) | \(0.5\times 0.4\times 0.9=0.18\) |
| \(P\) | \(=\) | \(0.02+0.27+0.18=\boldsymbol{0.47}\) |
(ii) Only \(NNN\) has no green light.
| \(P\) | \(=\) | \(1-0.5\times 0.6\times 0.1\) |
| \(\) | \(=\) | \(1-0.03=\boldsymbol{0.97}\) |
(i) \(P(\text{starts on exactly the second pull})\)
(ii) \(P(\text{starts within three pulls})\)
(i) Fail, then start: \(0.6\times 0.5=\boldsymbol{0.3}\).
(ii) Add the three success paths (Diagrams tab).
| \(P\) | \(=\) | \(0.4+0.6\times 0.5+0.6\times 0.5\times 0.6\) |
| \(\) | \(=\) | \(0.4+0.3+0.18=\boldsymbol{0.88}\) |
Check with the complement: it fails all three times with probability \(0.6\times 0.5\times 0.4=0.12\), and \(1-0.12=0.88\).
Same colour is \(RR\) or \(BB\).
| \(p^2+(1-p)^2\) | \(=\) | \(0.68\) |
| \(2p^2-2p+1\) | \(=\) | \(0.68\) |
| \(2p^2-2p+0.32\) | \(=\) | \(0\) |
| \(p^2-p+0.16\) | \(=\) | \(0\) |
| \((p-0.2)(p-0.8)\) | \(=\) | \(0\) |
\(p=0.2\) or \(p=0.8\). Since \(p\lt 0.5\), \(\boldsymbol{p=0.2}\).
Check: \(0.2^2+0.8^2=0.04+0.64=0.68\).
Common pitfalls
Frequently asked questions
How do you draw a probability tree diagram?
Draw one set of branches for each stage, label each branch with its outcome and probability, and make sure the branches from each point add to \(1\). For dependent stages, the later branches change with the earlier result.
When do you multiply and when do you add on a tree diagram?
Multiply along a path to find the probability of that sequence of outcomes (and). Add the probabilities of different paths when the event can happen in more than one way (or).
How do you find a missing probability on a tree?
Branches from the same point add to \(1\), so subtract the known branch from \(1\). If a later total is given, write it as a sum of paths with the unknown and solve the equation.
What is a dependent tree diagram?
It is a tree where the probabilities at a later stage depend on what happened earlier, for example drawing without replacement or a second serve after a fault. Each later set of branches is a conditional probability.
How many paths does a three-stage tree have?
If each stage has \(2\) outcomes, there are \(2\times 2\times 2=8\) paths. Trees that stop early have fewer.